Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2 variables a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the split lists. Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value)) result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
Replace NAs in split lists
15 messages · Eric Berger, Ek Esawi, Jeff Newmiller +2 more
I just came up with a solution right after i posted the question, but i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK
On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote:
Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2 variables a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the split lists. Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value)) result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
Hi library(zoo) has function na.locf, which probably can do what you want. so something like (untested) sdf1.fill<-lapply(sdf1, na.locf) do.call(rbind, sdf1.fill) Cheers. Petr
-----Original Message----- From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ek Esawi Sent: Monday, January 8, 2018 4:36 AM To: r-help at r-project.org Subject: Re: [R] Replace NAs in split lists I just came up with a solution right after i posted the question, but i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<- lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote:
Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2 variables a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the split lists. Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value)) result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later.
Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com> wrote: >I just came up with a solution right after i posted the question, but >i figured there must be a better and shorter one.than my solution >sdf1[[1]][1,4]<-lapplyresults[[1]] >sdf1[[2]][1,4]<-lapplyresults[[2]] > >EK > >On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote: >> Hi all-- >> >> I stumbled on this problem online. I did not like the solution given >> there which was a long UDF. I thought why cannot split and l/s apply >> work here. My aim is to split the data frame, use l/sapply, make >> changes on the split lists and combine the split lists to new data >> frame with the desired changes/output. >> >> The data frame shown below has a column named ID which has 2 >variables >> a and b; i want to replace the NAs on the Value column by 2, which is >> the only numeric entry, for ID=a and by 5 for ID=b. >> >> I worked out the solution but could not replace the results in the >split lists. >> >> Original dataframe , df1 >> ID ID_2 Firist Value >> 1 a aa TRUE 2 >> 2 a ab FALSE NA >> 3 a ac FALSE NA >> 4 b aa TRUE 5 >> 5 b ab FALSE NA >> Sdf1 >> $a >> ID ID_2 Firist Value >> 1 a aa TRUE 2 >> 2 a ab FALSE NA >> 3 a ac FALSE NA >> $b >> ID ID_2 Firist Value >> 4 b aa TRUE 5 >> 5 b ab FALSE NA >> Desired results >> ID ID_2 Firist Value >> 1 a aa TRUE 2 >> 2 a ab FALSE 2 >> 3 a ac FALSE 2 >> >> $b >> ID ID_2 Firist Value >> 4 b aa TRUE 5 >> 5 b ab FALSE 5 >> >> My code >> >> sdf <- split(df1,df$ID) >> lapply(sdf, function(z) >ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value)) >> result: >> $ a: num [1:3] 2 2 2 >> $ b: num [1:2] 5 5 >> >> How could I put these two lists back in the split data frame, sdf1? >> Then I could use do.call to reassemble a data frame from the split >> lists, >> >> Thanks, >> EK > >______________________________________________ >R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code.
Upon closer examination I see that you are not using the split version of
df1 as I usually would, so here is a reproducible example:
#----
df1 <- read.table( text=
"ID ID_2 Firist Value
1 a aa TRUE 2
2 a ab FALSE NA
3 a ac FALSE NA
4 b aa TRUE 5
5 b ab FALSE NA
", header=TRUE, as.is=TRUE )
sdf <- split( df1, df1$ID )
# note the extra [ 1 ] in case you have more than one non-NA value
# per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][ 1 ]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1 ]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com> wrote:
I just came up with a solution right after i posted the question, but i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote:
Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
---------------------------------------------------------------------------
Jeff Newmiller The ..... ..... Go Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
Hi With the example, na.locf seems to be the easiest way.
library(zoo)
na.locf(df1)
ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 4 b aa TRUE 5 5 b ab FALSE 5 Cheers Petr
-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jeff
Newmiller
Sent: Monday, January 8, 2018 9:13 AM
To: r-help at r-project.org; Ek Esawi <esawiek at gmail.com>
Subject: Re: [R] Replace NAs in split lists
Upon closer examination I see that you are not using the split version of
df1 as I usually would, so here is a reproducible example:
#----
df1 <- read.table( text=
"ID ID_2 Firist Value
1 a aa TRUE 2
2 a ab FALSE NA
3 a ac FALSE NA
4 b aa TRUE 5
5 b ab FALSE NA
", header=TRUE, as.is=TRUE )
sdf <- split( df1, df1$ID )
# note the extra [ 1 ] in case you have more than one non-NA value # per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][ 1 ]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1 ]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com> wrote:
I just came up with a solution right after i posted the question, but i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote:
Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
---------------------------------------------------------------------------
Jeff Newmiller The ..... ..... Go Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.
Yes, you are right if the IDs are always sequentially-adjacent and the first non-NA value appears in the first record for each ID.
Sent from my phone. Please excuse my brevity.
On January 8, 2018 2:29:40 AM PST, PIKAL Petr <petr.pikal at precheza.cz> wrote:
>Hi
>
>With the example, na.locf seems to be the easiest way.
>> library(zoo)
>
>> na.locf(df1)
> ID ID_2 Firist Value
>1 a aa TRUE 2
>2 a ab FALSE 2
>3 a ac FALSE 2
>4 b aa TRUE 5
>5 b ab FALSE 5
>
>Cheers
>Petr
>
>> -----Original Message-----
>> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jeff
>> Newmiller
>> Sent: Monday, January 8, 2018 9:13 AM
>> To: r-help at r-project.org; Ek Esawi <esawiek at gmail.com>
>> Subject: Re: [R] Replace NAs in split lists
>>
>> Upon closer examination I see that you are not using the split
>version of
>> df1 as I usually would, so here is a reproducible example:
>>
>> #----
>> df1 <- read.table( text=
>> "ID ID_2 Firist Value
>> 1 a aa TRUE 2
>> 2 a ab FALSE NA
>> 3 a ac FALSE NA
>> 4 b aa TRUE 5
>> 5 b ab FALSE NA
>> ", header=TRUE, as.is=TRUE )
>>
>> sdf <- split( df1, df1$ID )
>> # note the extra [ 1 ] in case you have more than one non-NA value #
>per ID
>> sdf2 <- lapply( sdf
>> , function( z ) {
>> z$Value <- ifelse( is.na( z$Value )
>> , z$Value[ !is.na( z$Value ) ][ 1
>]
>> , z$Value
>> )
>> z
>> }
>> )
>> df2 <- do.call( rbind, sdf2 )
>> df2
>> #> ID ID_2 Firist Value
>> #> a.1 a aa TRUE 2
>> #> a.2 a ab FALSE 2
>> #> a.3 a ac FALSE 2
>> #> b.4 b aa TRUE 5
>> #> b.5 b ab FALSE 5
>>
>> # or using tidyverse methods
>>
>> library(dplyr)
>> #>
>> #> Attaching package: 'dplyr'
>> #> The following objects are masked from 'package:stats':
>> #>
>> #> filter, lag
>> #> The following objects are masked from 'package:base':
>> #>
>> #> intersect, setdiff, setequal, union
>> df3 <- ( df1
>> %>% group_by( ID )
>> %>% do({
>> mutate( .
>> , Value = ifelse( is.na( Value )
>> , Value[ !is.na( Value ) ][ 1 ]
>> , Value
>> )
>> )
>> })
>> %>% ungroup
>> )
>> df3
>> #> # A tibble: 5 x 4
>> #> ID ID_2 Firist Value
>> #> <chr> <chr> <lgl> <int>
>> #> 1 a aa T 2
>> #> 2 a ab F 2
>> #> 3 a ac F 2
>> #> 4 b aa T 5
>> #> 5 b ab F 5
>> #----
>>
>> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>>
>> > Why do you want to modify df1?
>> >
>> > Why not just reassemble the parts as a new data frame and use that
>> > going forward in your calculations? That is generally the preferred
>> > approach in R so you can re-do your calculations easily if you find
>a
>> > mistake later.
>> > --
>> > Sent from my phone. Please excuse my brevity.
>> >
>> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
>wrote:
>> >> I just came up with a solution right after i posted the question,
>but
>> >> i figured there must be a better and shorter one.than my solution
>> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
>> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
>> >>
>> >> EK
>> >>
>> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
>wrote:
>> >>> Hi all--
>> >>>
>> >>> I stumbled on this problem online. I did not like the solution
>given
>> >>> there which was a long UDF. I thought why cannot split and l/s
>apply
>> >>> work here. My aim is to split the data frame, use l/sapply, make
>> >>> changes on the split lists and combine the split lists to new
>data
>> >>> frame with the desired changes/output.
>> >>>
>> >>> The data frame shown below has a column named ID which has 2
>> >> variables
>> >>> a and b; i want to replace the NAs on the Value column by 2,
>which
>> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
>> >>>
>> >>> I worked out the solution but could not replace the results in
>the
>> >> split lists.
>> >>>
>> >>> Original dataframe , df1
>> >>> ID ID_2 Firist Value
>> >>> 1 a aa TRUE 2
>> >>> 2 a ab FALSE NA
>> >>> 3 a ac FALSE NA
>> >>> 4 b aa TRUE 5
>> >>> 5 b ab FALSE NA
>> >>> Sdf1
>> >>> $a
>> >>> ID ID_2 Firist Value
>> >>> 1 a aa TRUE 2
>> >>> 2 a ab FALSE NA
>> >>> 3 a ac FALSE NA
>> >>> $b
>> >>> ID ID_2 Firist Value
>> >>> 4 b aa TRUE 5
>> >>> 5 b ab FALSE NA
>> >>> Desired results
>> >>> ID ID_2 Firist Value
>> >>> 1 a aa TRUE 2
>> >>> 2 a ab FALSE 2
>> >>> 3 a ac FALSE 2
>> >>>
>> >>> $b
>> >>> ID ID_2 Firist Value
>> >>> 4 b aa TRUE 5
>> >>> 5 b ab FALSE 5
>> >>>
>> >>> My code
>> >>>
>> >>> sdf <- split(df1,df$ID)
>> >>> lapply(sdf, function(z)
>> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>> >>> result:
>> >>> $ a: num [1:3] 2 2 2
>> >>> $ b: num [1:2] 5 5
>> >>>
>> >>> How could I put these two lists back in the split data frame,
>sdf1?
>> >>> Then I could use do.call to reassemble a data frame from the
>split
>> >>> lists,
>> >>>
>> >>> Thanks,
>> >>> EK
>> >>
>> >> ______________________________________________
>> >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >> http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>>
>>
>---------------------------------------------------------------------------
>> Jeff Newmiller The ..... ..... Go
>Live...
>> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
>Go...
>> Live: OO#.. Dead: OO#..
>Playing
>> Research Engineer (Solar/Batteries O.O#. #.O#. with
>> /Software/Embedded Controllers) .OO#. .OO#.
>rocks...1k
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>________________________________
>Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou
>ur?eny pouze jeho adres?t?m.
>Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav?
>neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho
>kopie vyma?te ze sv?ho syst?mu.
>Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento
>email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat.
>Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou
>modifikacemi ?i zpo?d?n?m p?enosu e-mailu.
>
>V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?:
>- vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en?
>smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu.
>- a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn?
>p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky
>ze strany p??jemce s dodatkem ?i odchylkou.
>- trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve
>v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech.
>- odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za
>spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn?
>zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly
>adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje,
>p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen?
>zn?m?.
>
>This e-mail and any documents attached to it may be confidential and
>are intended only for its intended recipients.
>If you received this e-mail by mistake, please immediately inform its
>sender. Delete the contents of this e-mail with all attachments and its
>copies from your system.
>If you are not the intended recipient of this e-mail, you are not
>authorized to use, disseminate, copy or disclose this e-mail in any
>manner.
>The sender of this e-mail shall not be liable for any possible damage
>caused by modifications of the e-mail or by delay with transfer of the
>email.
>
>In case that this e-mail forms part of business dealings:
>- the sender reserves the right to end negotiations about entering into
>a contract in any time, for any reason, and without stating any
>reasoning.
>- if the e-mail contains an offer, the recipient is entitled to
>immediately accept such offer; The sender of this e-mail (offer)
>excludes any acceptance of the offer on the part of the recipient
>containing any amendment or variation.
>- the sender insists on that the respective contract is concluded only
>upon an express mutual agreement on all its aspects.
>- the sender of this e-mail informs that he/she is not authorized to
>enter into any contracts on behalf of the company except for cases in
>which he/she is expressly authorized to do so in writing, and such
>authorization or power of attorney is submitted to the recipient or the
>person represented by the recipient, or the existence of such
>authorization is known to the recipient of the person represented by
>the recipient.
You can enforce these assumptions by sorting on multiple columns, which leads to na.locf(df1[ order(df1$ID,df1$Value), ]) On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
Yes, you are right if the IDs are always sequentially-adjacent and the first non-NA value appears in the first record for each ID. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 2:29:40 AM PST, PIKAL Petr <petr.pikal at precheza.cz> wrote:
Hi With the example, na.locf seems to be the easiest way.
library(zoo)
na.locf(df1)
ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 4 b aa TRUE 5 5 b ab FALSE 5 Cheers Petr
-----Original Message----- From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Jeff Newmiller Sent: Monday, January 8, 2018 9:13 AM To: r-help at r-project.org; Ek Esawi <esawiek at gmail.com> Subject: Re: [R] Replace NAs in split lists Upon closer examination I see that you are not using the split
version of
df1 as I usually would, so here is a reproducible example: #---- df1 <- read.table( text= "ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA ", header=TRUE, as.is=TRUE ) sdf <- split( df1, df1$ID ) # note the extra [ 1 ] in case you have more than one non-NA value #
per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][ 1
]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1 ]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find
a
mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
wrote:
I just came up with a solution right after i posted the question,
but
i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
wrote:
Hi all-- I stumbled on this problem online. I did not like the solution
given
there which was a long UDF. I thought why cannot split and l/s
apply
work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new
data
frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2,
which
is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in
the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame,
sdf1?
Then I could use do.call to reassemble a data frame from the
split
lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
-----------------------------------------------------------
----------------
Jeff Newmiller The ..... ..... Go
Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code.
"Enforce" is overstating it... results will differ if there are no non-NA values for a given ID, and there is a potential further discrepancy if there are multiple non-NA values. But these issues were not identified by the OP, so may not be relevant in their case.
Sent from my phone. Please excuse my brevity.
On January 8, 2018 6:41:33 AM PST, Eric Berger <ericjberger at gmail.com> wrote:
>You can enforce these assumptions by sorting on multiple columns, which
>leads to
>
>na.locf(df1[ order(df1$ID,df1$Value), ])
>
>
>
>On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller
><jdnewmil at dcn.davis.ca.us>
>wrote:
>
>> Yes, you are right if the IDs are always sequentially-adjacent and
>the
>> first non-NA value appears in the first record for each ID.
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On January 8, 2018 2:29:40 AM PST, PIKAL Petr
><petr.pikal at precheza.cz>
>> wrote:
>> >Hi
>> >
>> >With the example, na.locf seems to be the easiest way.
>> >> library(zoo)
>> >
>> >> na.locf(df1)
>> > ID ID_2 Firist Value
>> >1 a aa TRUE 2
>> >2 a ab FALSE 2
>> >3 a ac FALSE 2
>> >4 b aa TRUE 5
>> >5 b ab FALSE 5
>> >
>> >Cheers
>> >Petr
>> >
>> >> -----Original Message-----
>> >> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of
>Jeff
>> >> Newmiller
>> >> Sent: Monday, January 8, 2018 9:13 AM
>> >> To: r-help at r-project.org; Ek Esawi <esawiek at gmail.com>
>> >> Subject: Re: [R] Replace NAs in split lists
>> >>
>> >> Upon closer examination I see that you are not using the split
>> >version of
>> >> df1 as I usually would, so here is a reproducible example:
>> >>
>> >> #----
>> >> df1 <- read.table( text=
>> >> "ID ID_2 Firist Value
>> >> 1 a aa TRUE 2
>> >> 2 a ab FALSE NA
>> >> 3 a ac FALSE NA
>> >> 4 b aa TRUE 5
>> >> 5 b ab FALSE NA
>> >> ", header=TRUE, as.is=TRUE )
>> >>
>> >> sdf <- split( df1, df1$ID )
>> >> # note the extra [ 1 ] in case you have more than one non-NA value
>#
>> >per ID
>> >> sdf2 <- lapply( sdf
>> >> , function( z ) {
>> >> z$Value <- ifelse( is.na( z$Value )
>> >> , z$Value[ !is.na( z$Value ) ][
>1
>> >]
>> >> , z$Value
>> >> )
>> >> z
>> >> }
>> >> )
>> >> df2 <- do.call( rbind, sdf2 )
>> >> df2
>> >> #> ID ID_2 Firist Value
>> >> #> a.1 a aa TRUE 2
>> >> #> a.2 a ab FALSE 2
>> >> #> a.3 a ac FALSE 2
>> >> #> b.4 b aa TRUE 5
>> >> #> b.5 b ab FALSE 5
>> >>
>> >> # or using tidyverse methods
>> >>
>> >> library(dplyr)
>> >> #>
>> >> #> Attaching package: 'dplyr'
>> >> #> The following objects are masked from 'package:stats':
>> >> #>
>> >> #> filter, lag
>> >> #> The following objects are masked from 'package:base':
>> >> #>
>> >> #> intersect, setdiff, setequal, union
>> >> df3 <- ( df1
>> >> %>% group_by( ID )
>> >> %>% do({
>> >> mutate( .
>> >> , Value = ifelse( is.na( Value )
>> >> , Value[ !is.na( Value ) ][ 1
>]
>> >> , Value
>> >> )
>> >> )
>> >> })
>> >> %>% ungroup
>> >> )
>> >> df3
>> >> #> # A tibble: 5 x 4
>> >> #> ID ID_2 Firist Value
>> >> #> <chr> <chr> <lgl> <int>
>> >> #> 1 a aa T 2
>> >> #> 2 a ab F 2
>> >> #> 3 a ac F 2
>> >> #> 4 b aa T 5
>> >> #> 5 b ab F 5
>> >> #----
>> >>
>> >> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>> >>
>> >> > Why do you want to modify df1?
>> >> >
>> >> > Why not just reassemble the parts as a new data frame and use
>that
>> >> > going forward in your calculations? That is generally the
>preferred
>> >> > approach in R so you can re-do your calculations easily if you
>find
>> >a
>> >> > mistake later.
>> >> > --
>> >> > Sent from my phone. Please excuse my brevity.
>> >> >
>> >> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
>> >wrote:
>> >> >> I just came up with a solution right after i posted the
>question,
>> >but
>> >> >> i figured there must be a better and shorter one.than my
>solution
>> >> >> sdf1[[1]][1,4]<-lapplyresults[[1]]
>> >> >> sdf1[[2]][1,4]<-lapplyresults[[2]]
>> >> >>
>> >> >> EK
>> >> >>
>> >> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
>> >wrote:
>> >> >>> Hi all--
>> >> >>>
>> >> >>> I stumbled on this problem online. I did not like the solution
>> >given
>> >> >>> there which was a long UDF. I thought why cannot split and l/s
>> >apply
>> >> >>> work here. My aim is to split the data frame, use l/sapply,
>make
>> >> >>> changes on the split lists and combine the split lists to new
>> >data
>> >> >>> frame with the desired changes/output.
>> >> >>>
>> >> >>> The data frame shown below has a column named ID which has 2
>> >> >> variables
>> >> >>> a and b; i want to replace the NAs on the Value column by 2,
>> >which
>> >> >>> is the only numeric entry, for ID=a and by 5 for ID=b.
>> >> >>>
>> >> >>> I worked out the solution but could not replace the results in
>> >the
>> >> >> split lists.
>> >> >>>
>> >> >>> Original dataframe , df1
>> >> >>> ID ID_2 Firist Value
>> >> >>> 1 a aa TRUE 2
>> >> >>> 2 a ab FALSE NA
>> >> >>> 3 a ac FALSE NA
>> >> >>> 4 b aa TRUE 5
>> >> >>> 5 b ab FALSE NA
>> >> >>> Sdf1
>> >> >>> $a
>> >> >>> ID ID_2 Firist Value
>> >> >>> 1 a aa TRUE 2
>> >> >>> 2 a ab FALSE NA
>> >> >>> 3 a ac FALSE NA
>> >> >>> $b
>> >> >>> ID ID_2 Firist Value
>> >> >>> 4 b aa TRUE 5
>> >> >>> 5 b ab FALSE NA
>> >> >>> Desired results
>> >> >>> ID ID_2 Firist Value
>> >> >>> 1 a aa TRUE 2
>> >> >>> 2 a ab FALSE 2
>> >> >>> 3 a ac FALSE 2
>> >> >>>
>> >> >>> $b
>> >> >>> ID ID_2 Firist Value
>> >> >>> 4 b aa TRUE 5
>> >> >>> 5 b ab FALSE 5
>> >> >>>
>> >> >>> My code
>> >> >>>
>> >> >>> sdf <- split(df1,df$ID)
>> >> >>> lapply(sdf, function(z)
>> >> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>> >> >>> result:
>> >> >>> $ a: num [1:3] 2 2 2
>> >> >>> $ b: num [1:2] 5 5
>> >> >>>
>> >> >>> How could I put these two lists back in the split data frame,
>> >sdf1?
>> >> >>> Then I could use do.call to reassemble a data frame from the
>> >split
>> >> >>> lists,
>> >> >>>
>> >> >>> Thanks,
>> >> >>> EK
>> >> >>
>> >> >> ______________________________________________
>> >> >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
>see
>> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> >> PLEASE do read the posting guide
>> >> >> http://www.R-project.org/posting-guide.html
>> >> >> and provide commented, minimal, self-contained, reproducible
>code.
>> >> >
>> >> > ______________________________________________
>> >> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
>see
>> >> > https://stat.ethz.ch/mailman/listinfo/r-help
>> >> > PLEASE do read the posting guide
>> >> > http://www.R-project.org/posting-guide.html
>> >> > and provide commented, minimal, self-contained, reproducible
>code.
>> >> >
>> >>
>> >>
>> >-----------------------------------------------------------
>> ----------------
>> >> Jeff Newmiller The ..... .....
>Go
>> >Live...
>> >> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#.
>Live
>> >Go...
>> >> Live: OO#.. Dead: OO#..
>> >Playing
>> >> Research Engineer (Solar/Batteries O.O#. #.O#.
>with
>> >> /Software/Embedded Controllers) .OO#. .OO#.
>> >rocks...1k
>> >>
>> >> ______________________________________________
>> >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> >> https://stat.ethz.ch/mailman/listinfo/r-help
>> >> PLEASE do read the posting guide
>> >http://www.R-project.org/posting-guide.html
>> >> and provide commented, minimal, self-contained, reproducible code.
>> >
>> >________________________________
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>> PLEASE do read the posting guide http://www.R-project.org/
>> posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
Thank you Jeff. Your code works, as usual , perfectly. I am just
wondering why if i put the whole code in one line, i get an error
message.
sdf2 <- lapply( sdf, function(z){z$Value
<-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
error. unexpected symbol in sdf2
Thanks again
EK
On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
Upon closer examination I see that you are not using the split version of
df1 as I usually would, so here is a reproducible example:
#----
df1 <- read.table( text=
"ID ID_2 Firist Value
1 a aa TRUE 2
2 a ab FALSE NA
3 a ac FALSE NA
4 b aa TRUE 5
5 b ab FALSE NA
", header=TRUE, as.is=TRUE )
sdf <- split( df1, df1$ID )
# note the extra [ 1 ] in case you have more than one non-NA value # per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][ 1 ]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1 ]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that going forward in your calculations? That is generally the preferred approach in R so you can re-do your calculations easily if you find a mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com> wrote:
I just came up with a solution right after i posted the question, but i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote:
Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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---------------------------------------------------------------------------
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DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
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---------------------------------------------------------------------------
I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session?
Sent from my phone. Please excuse my brevity.
On January 8, 2018 8:03:45 AM PST, Ek Esawi <esawiek at gmail.com> wrote:
>Thank you Jeff. Your code works, as usual , perfectly. I am just
>wondering why if i put the whole code in one line, i get an error
>message.
>sdf2 <- lapply( sdf, function(z){z$Value
><-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
>error. unexpected symbol in sdf2
>
>Thanks again
>
>EK
>
>
>On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller
><jdnewmil at dcn.davis.ca.us> wrote:
>> Upon closer examination I see that you are not using the split
>version of
>> df1 as I usually would, so here is a reproducible example:
>>
>> #----
>> df1 <- read.table( text=
>> "ID ID_2 Firist Value
>> 1 a aa TRUE 2
>> 2 a ab FALSE NA
>> 3 a ac FALSE NA
>> 4 b aa TRUE 5
>> 5 b ab FALSE NA
>> ", header=TRUE, as.is=TRUE )
>>
>> sdf <- split( df1, df1$ID )
>> # note the extra [ 1 ] in case you have more than one non-NA value #
>per ID
>> sdf2 <- lapply( sdf
>> , function( z ) {
>> z$Value <- ifelse( is.na( z$Value )
>> , z$Value[ !is.na( z$Value ) ][ 1 ]
>> , z$Value
>> )
>> z
>> }
>> )
>> df2 <- do.call( rbind, sdf2 )
>> df2
>> #> ID ID_2 Firist Value
>> #> a.1 a aa TRUE 2
>> #> a.2 a ab FALSE 2
>> #> a.3 a ac FALSE 2
>> #> b.4 b aa TRUE 5
>> #> b.5 b ab FALSE 5
>>
>> # or using tidyverse methods
>>
>> library(dplyr)
>> #>
>> #> Attaching package: 'dplyr'
>> #> The following objects are masked from 'package:stats':
>> #>
>> #> filter, lag
>> #> The following objects are masked from 'package:base':
>> #>
>> #> intersect, setdiff, setequal, union
>> df3 <- ( df1
>> %>% group_by( ID )
>> %>% do({
>> mutate( .
>> , Value = ifelse( is.na( Value )
>> , Value[ !is.na( Value ) ][ 1 ]
>> , Value
>> )
>> )
>> })
>> %>% ungroup
>> )
>> df3
>> #> # A tibble: 5 x 4
>> #> ID ID_2 Firist Value
>> #> <chr> <chr> <lgl> <int>
>> #> 1 a aa T 2
>> #> 2 a ab F 2
>> #> 3 a ac F 2
>> #> 4 b aa T 5
>> #> 5 b ab F 5
>> #----
>>
>>
>> On Sun, 7 Jan 2018, Jeff Newmiller wrote:
>>
>>> Why do you want to modify df1?
>>>
>>> Why not just reassemble the parts as a new data frame and use that
>going
>>> forward in your calculations? That is generally the preferred
>approach in R
>>> so you can re-do your calculations easily if you find a mistake
>later.
>>> --
>>> Sent from my phone. Please excuse my brevity.
>>>
>>> On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
>wrote:
>>>>
>>>> I just came up with a solution right after i posted the question,
>but
>>>> i figured there must be a better and shorter one.than my solution
>>>> sdf1[[1]][1,4]<-lapplyresults[[1]]
>>>> sdf1[[2]][1,4]<-lapplyresults[[2]]
>>>>
>>>> EK
>>>>
>>>> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
>wrote:
>>>>>
>>>>> Hi all--
>>>>>
>>>>> I stumbled on this problem online. I did not like the solution
>given
>>>>> there which was a long UDF. I thought why cannot split and l/s
>apply
>>>>> work here. My aim is to split the data frame, use l/sapply, make
>>>>> changes on the split lists and combine the split lists to new data
>>>>> frame with the desired changes/output.
>>>>>
>>>>> The data frame shown below has a column named ID which has 2
>>>>
>>>> variables
>>>>>
>>>>> a and b; i want to replace the NAs on the Value column by 2, which
>is
>>>>> the only numeric entry, for ID=a and by 5 for ID=b.
>>>>>
>>>>> I worked out the solution but could not replace the results in the
>>>>
>>>> split lists.
>>>>>
>>>>>
>>>>> Original dataframe , df1
>>>>> ID ID_2 Firist Value
>>>>> 1 a aa TRUE 2
>>>>> 2 a ab FALSE NA
>>>>> 3 a ac FALSE NA
>>>>> 4 b aa TRUE 5
>>>>> 5 b ab FALSE NA
>>>>> Sdf1
>>>>> $a
>>>>> ID ID_2 Firist Value
>>>>> 1 a aa TRUE 2
>>>>> 2 a ab FALSE NA
>>>>> 3 a ac FALSE NA
>>>>> $b
>>>>> ID ID_2 Firist Value
>>>>> 4 b aa TRUE 5
>>>>> 5 b ab FALSE NA
>>>>> Desired results
>>>>> ID ID_2 Firist Value
>>>>> 1 a aa TRUE 2
>>>>> 2 a ab FALSE 2
>>>>> 3 a ac FALSE 2
>>>>>
>>>>> $b
>>>>> ID ID_2 Firist Value
>>>>> 4 b aa TRUE 5
>>>>> 5 b ab FALSE 5
>>>>>
>>>>> My code
>>>>>
>>>>> sdf <- split(df1,df$ID)
>>>>> lapply(sdf, function(z)
>>>>
>>>> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
>>>>>
>>>>> result:
>>>>> $ a: num [1:3] 2 2 2
>>>>> $ b: num [1:2] 5 5
>>>>>
>>>>> How could I put these two lists back in the split data frame,
>sdf1?
>>>>> Then I could use do.call to reassemble a data frame from the split
>>>>> lists,
>>>>>
>>>>> Thanks,
>>>>> EK
>>>>
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide
>>>> http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>---------------------------------------------------------------------------
>> Jeff Newmiller The ..... ..... Go
>Live...
>> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
>Go...
>> Live: OO#.. Dead: OO#..
>Playing
>> Research Engineer (Solar/Batteries O.O#. #.O#. with
>> /Software/Embedded Controllers) .OO#. .OO#.
>rocks...1k
>>
>---------------------------------------------------------------------------
OPS! Sorry i did indeed posted the code in HTML; should have known better. ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z}) error. unexpected symbol in sdf2 On Mon, Jan 8, 2018 at 11:44 AM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us> wrote:
I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session? -- Sent from my phone. Please excuse my brevity. On January 8, 2018 8:03:45 AM PST, Ek Esawi <esawiek at gmail.com> wrote:
Thank you Jeff. Your code works, as usual , perfectly. I am just
wondering why if i put the whole code in one line, i get an error
message.
sdf2 <- lapply( sdf, function(z){z$Value
<-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
error. unexpected symbol in sdf2
Thanks again
EK
On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us> wrote:
Upon closer examination I see that you are not using the split
version of
df1 as I usually would, so here is a reproducible example: #---- df1 <- read.table( text= "ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA ", header=TRUE, as.is=TRUE ) sdf <- split( df1, df1$ID ) # note the extra [ 1 ] in case you have more than one non-NA value #
per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][ 1 ]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1 ]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that
going
forward in your calculations? That is generally the preferred
approach in R
so you can re-do your calculations easily if you find a mistake
later.
-- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
wrote:
I just came up with a solution right after i posted the question,
but
i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
wrote:
Hi all-- I stumbled on this problem online. I did not like the solution
given
there which was a long UDF. I thought why cannot split and l/s
apply
work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2, which
is
the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame,
sdf1?
Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
---------------------------------------------------------------------------
Jeff Newmiller The ..... ..... Go
Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
---------------------------------------------------------------------------
I don't get exactly that error message,
> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
Error: unexpected symbol in "ifelse(is.na(z$Value),z$Value[!is.na
(z$Value)][1],z$Value)z"
The 'symbol' in "unexpected symbol" refers to a "name" ('z' in this case).
The problem is usually at the end of the code snippet shown in the error
message as in
> f(a)b # missing newline or semicolon?
Error: unexpected symbol in "f(a)b"
or
> f(a b) # missing comma?
Error: unexpected symbol in "f(a b"
You need a newline or semicolon between the end of 'ifelse(...)' and 'z' or
maybe
you don't want the z there at all.
After you fix that it will complain about the unmatched right brace and
parenthesis.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Jan 8, 2018 at 8:55 AM, Ek Esawi <esawiek at gmail.com> wrote:
OPS! Sorry i did indeed posted the code in HTML; should have known better. ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z}) error. unexpected symbol in sdf2 On Mon, Jan 8, 2018 at 11:44 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
I don't know. You seem to be posting in HTML so your code is mangled.
Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session?
-- Sent from my phone. Please excuse my brevity. On January 8, 2018 8:03:45 AM PST, Ek Esawi <esawiek at gmail.com> wrote:
Thank you Jeff. Your code works, as usual , perfectly. I am just
wondering why if i put the whole code in one line, i get an error
message.
sdf2 <- lapply( sdf, function(z){z$Value
<-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
error. unexpected symbol in sdf2
Thanks again
EK
On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller
<jdnewmil at dcn.davis.ca.us> wrote:
Upon closer examination I see that you are not using the split
version of
df1 as I usually would, so here is a reproducible example: #---- df1 <- read.table( text= "ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA ", header=TRUE, as.is=TRUE ) sdf <- split( df1, df1$ID ) # note the extra [ 1 ] in case you have more than one non-NA value #
per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][ 1 ]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1 ]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use that
going
forward in your calculations? That is generally the preferred
approach in R
so you can re-do your calculations easily if you find a mistake
later.
-- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
wrote:
I just came up with a solution right after i posted the question,
but
i figured there must be a better and shorter one.than my solution sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
wrote:
Hi all-- I stumbled on this problem online. I did not like the solution
given
there which was a long UDF. I thought why cannot split and l/s
apply
work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2, which
is
the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame,
sdf1?
Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
----------------------------------------------------------
-----------------
Jeff Newmiller The ..... ..... Go
Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
----------------------------------------------------------
-----------------
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Thank you all. Now everything works. Happy 2018 and beyond EK
On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com> wrote:
Hi all-- I stumbled on this problem online. I did not like the solution given there which was a long UDF. I thought why cannot split and l/s apply work here. My aim is to split the data frame, use l/sapply, make changes on the split lists and combine the split lists to new data frame with the desired changes/output. The data frame shown below has a column named ID which has 2 variables a and b; i want to replace the NAs on the Value column by 2, which is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in the split lists. Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z) ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value)) result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame, sdf1? Then I could use do.call to reassemble a data frame from the split lists, Thanks, EK
Hi Yes, that is why I mentioned "the example". Of course if there are missing values in other columns or data layout is different, na.locf could give undesired result. However it is simple and works with data similar to the example. Cheers Petr
-----Original Message----- From: Jeff Newmiller [mailto:jdnewmil at dcn.davis.ca.us] Sent: Monday, January 8, 2018 4:45 PM To: Eric Berger <ericjberger at gmail.com> Cc: PIKAL Petr <petr.pikal at precheza.cz>; r-help at r-project.org; Ek Esawi <esawiek at gmail.com> Subject: Re: [R] Replace NAs in split lists "Enforce" is overstating it... results will differ if there are no non-NA values for a given ID, and there is a potential further discrepancy if there are multiple non-NA values. But these issues were not identified by the OP, so may not be relevant in their case. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 6:41:33 AM PST, Eric Berger <ericjberger at gmail.com> wrote:
You can enforce these assumptions by sorting on multiple columns, which leads to na.locf(df1[ order(df1$ID,df1$Value), ]) On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote:
Yes, you are right if the IDs are always sequentially-adjacent and
the
first non-NA value appears in the first record for each ID. -- Sent from my phone. Please excuse my brevity. On January 8, 2018 2:29:40 AM PST, PIKAL Petr
<petr.pikal at precheza.cz>
wrote:
Hi With the example, na.locf seems to be the easiest way.
library(zoo)
na.locf(df1)
ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 4 b aa TRUE 5 5 b ab FALSE 5 Cheers Petr
-----Original Message----- From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of
Jeff
Newmiller Sent: Monday, January 8, 2018 9:13 AM To: r-help at r-project.org; Ek Esawi <esawiek at gmail.com> Subject: Re: [R] Replace NAs in split lists Upon closer examination I see that you are not using the split
version of
df1 as I usually would, so here is a reproducible example: #---- df1 <- read.table( text= "ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA ", header=TRUE, as.is=TRUE ) sdf <- split( df1, df1$ID ) # note the extra [ 1 ] in case you have more than one non-NA value
#
per ID
sdf2 <- lapply( sdf
, function( z ) {
z$Value <- ifelse( is.na( z$Value )
, z$Value[ !is.na( z$Value ) ][
1
]
, z$Value
)
z
}
)
df2 <- do.call( rbind, sdf2 )
df2
#> ID ID_2 Firist Value
#> a.1 a aa TRUE 2
#> a.2 a ab FALSE 2
#> a.3 a ac FALSE 2
#> b.4 b aa TRUE 5
#> b.5 b ab FALSE 5
# or using tidyverse methods
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
df3 <- ( df1
%>% group_by( ID )
%>% do({
mutate( .
, Value = ifelse( is.na( Value )
, Value[ !is.na( Value ) ][ 1
]
, Value
)
)
})
%>% ungroup
)
df3
#> # A tibble: 5 x 4
#> ID ID_2 Firist Value
#> <chr> <chr> <lgl> <int>
#> 1 a aa T 2
#> 2 a ab F 2
#> 3 a ac F 2
#> 4 b aa T 5
#> 5 b ab F 5
#----
On Sun, 7 Jan 2018, Jeff Newmiller wrote:
Why do you want to modify df1? Why not just reassemble the parts as a new data frame and use
that
going forward in your calculations? That is generally the
preferred
approach in R so you can re-do your calculations easily if you
find
a
mistake later. -- Sent from my phone. Please excuse my brevity. On January 7, 2018 7:35:59 PM PST, Ek Esawi <esawiek at gmail.com>
wrote:
I just came up with a solution right after i posted the
question,
but
i figured there must be a better and shorter one.than my
solution
sdf1[[1]][1,4]<-lapplyresults[[1]] sdf1[[2]][1,4]<-lapplyresults[[2]] EK On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esawiek at gmail.com>
wrote:
Hi all-- I stumbled on this problem online. I did not like the solution
given
there which was a long UDF. I thought why cannot split and l/s
apply
work here. My aim is to split the data frame, use l/sapply,
make
changes on the split lists and combine the split lists to new
data
frame with the desired changes/output. The data frame shown below has a column named ID which has 2
variables
a and b; i want to replace the NAs on the Value column by 2,
which
is the only numeric entry, for ID=a and by 5 for ID=b. I worked out the solution but could not replace the results in
the
split lists.
Original dataframe , df1 ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA 4 b aa TRUE 5 5 b ab FALSE NA Sdf1 $a ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE NA 3 a ac FALSE NA $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE NA Desired results ID ID_2 Firist Value 1 a aa TRUE 2 2 a ab FALSE 2 3 a ac FALSE 2 $b ID ID_2 Firist Value 4 b aa TRUE 5 5 b ab FALSE 5 My code sdf <- split(df1,df$ID) lapply(sdf, function(z)
ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value))
result: $ a: num [1:3] 2 2 2 $ b: num [1:2] 5 5 How could I put these two lists back in the split data frame,
sdf1?
Then I could use do.call to reassemble a data frame from the
split
lists, Thanks, EK
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
see
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible
code.
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
see
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible
code.
-----------------------------------------------------------
----------------
Jeff Newmiller The ..... .....
Go
Live...
DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#.
Live
Go...
Live: OO#.. Dead: OO#..
Playing
Research Engineer (Solar/Batteries O.O#. #.O#.
with
/Software/Embedded Controllers) .OO#. .OO#.
rocks...1k
______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a
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copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible
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______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code.
________________________________ Tento e-mail a jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient.