Hi everyone:
This may be trivial but I just have not been able to figure it out.
Imagine the following dataframe:
a b c d
TRUE TRUE TRUE TRUE
FALSE FALSE FALSE TRUE
FALSE TRUE FALSE FALSE
I would like to create a new dataframe, in which TRUE gets 0 but if
false then add 1 to the cell to the left. So the results for the
example above should be something like:
a b c d
0 0 0 0
1 2 3 0
1 0 1 2
I wonder if you may know?.
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone: Country code: 57
Provider code: 313
Phone 776 2282
From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
conditional Dataframe filling
10 messages · Blaser Nello, arun, Camilo Mora
Here's a possible solution.
dd <- structure(list(a = c(TRUE, FALSE, FALSE),
b = c(TRUE, FALSE, TRUE),
c = c(TRUE, FALSE, FALSE),
d = c(TRUE, TRUE, FALSE)),
.Names = c("a", "b", "c", "d"),
row.names = c(NA, -3L),
class = "data.frame")
ds <- as.data.frame(t(apply(!dd, 1, cumsum)-apply(dd, 1, cumsum)))
ds[as.matrix(dd)] <- 0
ds
Best,
Nello
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Camilo Mora
Sent: Mittwoch, 27. M?rz 2013 09:32
To: r-help at r-project.org
Subject: [R] conditional Dataframe filling
Hi everyone:
This may be trivial but I just have not been able to figure it out.
Imagine the following dataframe:
a b c d
TRUE TRUE TRUE TRUE
FALSE FALSE FALSE TRUE
FALSE TRUE FALSE FALSE
I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like:
a b c d
0 0 0 0
1 2 3 0
1 0 1 2
I wonder if you may know?.
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii Currently available in Colombia
Phone: Country code: 57
Provider code: 313
Phone 776 2282
From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
______________________________________________
R-help at r-project.org mailing list
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.
Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ? b? ? c? ? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ? b? ? c? ? d 0? ? 0? ? 0? ? 0 1? ? 2? ? 3? ? 0 1? ? 0? ? 1? ? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:? Country code: 57 ? ? ? ? Provider code: 313 ? ? ? ? Phone 776 2282 ? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ ______________________________________________ R-help at r-project.org mailing list 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.
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))? #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/ ______________________________________________ R-help at r-project.org mailing list 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.
Dear Arun,
Thank you very much for your help with this.I did not know where to
start looking to solve that problem, so I truly appreciate your input.
The line of code you sent seems to work but it duplicates the results.
Do you know why that may happen?
Below is a larger database, to which I apply your line of code.
Thank you very much again,
Camilo
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone: Country code: 57
Provider code: 313
Phone 776 2282
From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))? #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
______________________________________________ R-help at r-project.org mailing list 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.
Dear Camilo,
How do you want to deal with the NAs?
If I remove the NAs:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(na.omit(dat1))
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
?dat2
#? a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
?dat1
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j??? k??? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE? TRUE
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE TRUE TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE TRUE TRUE FALSE
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 3:27 PM
Subject: Re: [R] conditional Dataframe filling
Dear Arun,
Thank you very? much for your help with this.I did not know where to start looking to solve that problem, so I truly appreciate your input.
The line of code you sent seems to work but it duplicates the results. Do you know why that may happen?
Below is a larger database, to which I apply your line of code.
Thank you very much again,
Camilo
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:? Country code: 57
? ? ? ? Provider code: 313
? ? ? ? Phone 776 2282
? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))? #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
______________________________________________ R-help at r-project.org mailing list 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.
Thanks Arun,
Well that is interesting. My intention was to have a dataframe with
the same number of rows in the original data, and for the rows with
NAs, then return NA (If there are NAs, often the entire row has NAs).
What is interesting is that in your code with NAs, the row that has
NAs gets NAs in the output, which is what I am looking for.
I guess a solution is to subset complete rows and then run your line
of code. Unless there is an alternative, to tell cumsum to leave NAs
as NAs?
Thanks again,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone: Country code: 57
Provider code: 313
Phone 776 2282
From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
Dear Camilo,
How do you want to deal with the NAs?
If I remove the NAs:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(na.omit(dat1))
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
?dat2
#? a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
?dat1
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j??? k??? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE? TRUE
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE TRUE TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE TRUE TRUE FALSE
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 3:27 PM
Subject: Re: [R] conditional Dataframe filling
Dear Arun,
Thank you very? much for your help with this.I did not know where to
start looking to solve that problem, so I truly appreciate your input.
The line of code you sent seems to work but it duplicates the
results. Do you know why that may happen?
Below is a larger database, to which I apply your line of code.
Thank you very much again,
Camilo
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:? Country code: 57
? ? ? ? Provider code: 313
? ? ? ? Phone 776 2282
? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))? #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
______________________________________________ R-help at r-project.org mailing list 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.
Dear Camilo,
You can do this:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y = c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(dat1)
dat2<-dat1
dat2[rowSums(is.na(dat2))==0,]<- t(apply(!dat1[rowSums(is.na(dat1))==0,],1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
dat2
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
Suppose if NAs are there but not for the entire row (if I understand correctly), you wanted to have the whole row NA, right.
datNew<- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE,
NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE),
??? d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE,
??? FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE,
??? NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE
??? ), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE,
??? TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE,
??? NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE
??? )), .Names = c("a", "b", "c", "d", "e", "f", "g", "h", "i",
"j", "k", "l", "m"), row.names = c("w", "x", "y", "z", "u"), class = "data.frame")
datNew
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j???? k???? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE? TRUE? TRUE? TRUE
#x??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA?? NA??? NA??? NA??? NA
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE? TRUE? TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE? TRUE? TRUE FALSE
#u? TRUE? TRUE FALSE FALSE??? NA??? NA? TRUE FALSE??? NA TRUE FALSE FALSE? TRUE
dat2New<- datNew
dat2New[rowSums(is.na(dat2New))==0,]<-t(apply(!datNew[rowSums(is.na(datNew))==0,],1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
dat2New[rowSums(is.na(dat2New))!=0 & rowSums(is.na(dat2New))!=ncol(dat2New),]<-NA
?dat2New
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
#u NA NA NA NA NA NA NA NA NA NA NA NA NA
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 4:10 PM
Subject: Re: [R] conditional Dataframe filling
Thanks Arun,
Well that is interesting. My intention was to have a dataframe with?
the same number of rows in the original data, and for the rows with?
NAs, then return NA (If there are NAs, often the entire row has NAs).?
What is interesting is that in your code with NAs, the row that has?
NAs gets NAs in the output, which is what I am looking for.
I guess a solution is to subset complete rows and then run your line?
of code. Unless there is an alternative, to tell cumsum to leave NAs?
as NAs?
Thanks again,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:? Country code: 57
? ? ? ? ? Provider code: 313
? ? ? ? ? Phone 776 2282
? ? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
Dear Camilo,
How do you want to deal with the NAs?
If I remove the NAs:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =?
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(na.omit(dat1))
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)?
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
?dat2
#? a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
?dat1
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j??? k??? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE TRUE TRUE? TRUE
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE TRUE TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE TRUE TRUE FALSE
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 3:27 PM
Subject: Re: [R] conditional Dataframe filling
Dear Arun,
Thank you very? much for your help with this.I did not know where to?
start looking to solve that problem, so I truly appreciate your input.
The line of code you sent seems to work but it duplicates the?
results. Do you know why that may happen?
Below is a larger database, to which I apply your line of code.
Thank you very much again,
Camilo
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =?
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)?
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:?? Country code: 57
? ? ? ?? Provider code: 313
? ? ? ?? Phone 776 2282
? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x)? unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))?? #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x)? unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if? false then add 1 to the cell to the left. So the results for the? example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
______________________________________________ R-help at r-project.org mailing list 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.
Nice!.
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone: Country code: 57
Provider code: 313
Phone 776 2282
From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
Dear Camilo,
You can do this:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(dat1)
dat2<-dat1
dat2[rowSums(is.na(dat2))==0,]<-
t(apply(!dat1[rowSums(is.na(dat1))==0,],1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
dat2
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
Suppose if NAs are there but not for the entire row (if I understand
correctly), you wanted to have the whole row NA, right.
datNew<- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE,
NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE),
??? d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE,
??? FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE,
??? NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE
??? ), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE,
??? TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE,
??? NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE
??? )), .Names = c("a", "b", "c", "d", "e", "f", "g", "h", "i",
"j", "k", "l", "m"), row.names = c("w", "x", "y", "z", "u"), class =
"data.frame")
datNew
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j????
k???? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE? TRUE?
TRUE? TRUE
#x??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA?? NA???
NA??? NA??? NA
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE? TRUE?
TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE? TRUE?
TRUE FALSE
#u? TRUE? TRUE FALSE FALSE??? NA??? NA? TRUE FALSE??? NA TRUE FALSE
FALSE? TRUE
dat2New<- datNew
dat2New[rowSums(is.na(dat2New))==0,]<-t(apply(!datNew[rowSums(is.na(datNew))==0,],1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
dat2New[rowSums(is.na(dat2New))!=0 &
rowSums(is.na(dat2New))!=ncol(dat2New),]<-NA
?dat2New
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
#u NA NA NA NA NA NA NA NA NA NA NA NA NA
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 4:10 PM
Subject: Re: [R] conditional Dataframe filling
Thanks Arun,
Well that is interesting. My intention was to have a dataframe with
the same number of rows in the original data, and for the rows with
NAs, then return NA (If there are NAs, often the entire row has NAs).
What is interesting is that in your code with NAs, the row that has
NAs gets NAs in the output, which is what I am looking for.
I guess a solution is to subset complete rows and then run your line
of code. Unless there is an alternative, to tell cumsum to leave NAs
as NAs?
Thanks again,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:? Country code: 57
? ? ? ? ? Provider code: 313
? ? ? ? ? Phone 776 2282
? ? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
Dear Camilo,
How do you want to deal with the NAs?
If I remove the NAs:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(na.omit(dat1))
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
?dat2
#? a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
?dat1
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j???
k??? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE TRUE
TRUE? TRUE
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE TRUE
TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE TRUE
TRUE FALSE
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 3:27 PM
Subject: Re: [R] conditional Dataframe filling
Dear Arun,
Thank you very? much for your help with this.I did not know where to
start looking to solve that problem, so I truly appreciate your input.
The line of code you sent seems to work but it duplicates the
results. Do you know why that may happen?
Below is a larger database, to which I apply your line of code.
Thank you very much again,
Camilo
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:?? Country code: 57
? ? ? ?? Provider code: 313
? ? ? ?? Phone 776 2282
? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
______________________________________________ R-help at r-project.org mailing list 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.
Hi Camilo,
No problem.
In case, you wanted to process the partial NA rows, this could help:
datNew<- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE,
NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE),
??? d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE,
??? FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE,
??? NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE
??? ), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE,
??? TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE,
??? NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE
??? )), .Names = c("a", "b", "c", "d", "e", "f", "g", "h", "i",
"j", "k", "l", "m"), row.names = c("w", "x", "y", "z", "u"), class = "data.frame")
dat2New<- datNew
dat2New[rowSums(is.na(dat2New))==0 | rowSums(is.na(dat2New))!=ncol(dat2New),]<- t(apply(!datNew[rowSums(is.na(datNew))==0 | rowSums(is.na(datNew))!=ncol(datNew),],1,function(x) {x[!is.na(x)]<- unlist(lapply(split(x[!is.na(x)],cumsum(c(0,abs(diff(x[!is.na(x)]))))),cumsum));x}))
?dat2New
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
#u? 0? 0? 1? 2 NA NA? 0? 1 NA? 0? 1? 2? 0
?datNew
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j???? k???? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE? TRUE? TRUE? TRUE
#x??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA?? NA??? NA??? NA??? NA
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE? TRUE? TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE? TRUE? TRUE FALSE
#u? TRUE? TRUE FALSE FALSE??? NA??? NA? TRUE FALSE??? NA TRUE FALSE FALSE? TRUE
?
A.K.
----- Original Message -----
From: Camilo Mora <cmora at Dal.Ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 4:49 PM
Subject: Re: [R] conditional Dataframe filling
Nice!.
Thanks,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:? Country code: 57
? ? ? ? ? Provider code: 313
? ? ? ? ? Phone 776 2282
? ? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
Dear Camilo,
You can do this:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =?
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(dat1)
dat2<-dat1
dat2[rowSums(is.na(dat2))==0,]<-?
t(apply(!dat1[rowSums(is.na(dat1))==0,],1,function(x)?
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
dat2
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
Suppose if NAs are there but not for the entire row (if I understand?
correctly), you wanted to have the whole row NA, right.
datNew<- structure(list(a = c(TRUE, NA, FALSE, TRUE, TRUE), b = c(TRUE,
NA, FALSE, TRUE, TRUE), c = c(TRUE, NA, FALSE, TRUE, FALSE),
??? d = c(TRUE, NA, FALSE, TRUE, FALSE), e = c(TRUE, NA, FALSE,
??? FALSE, NA), f = c(FALSE, NA, TRUE, TRUE, NA), g = c(FALSE,
??? NA, TRUE, TRUE, TRUE), h = c(FALSE, NA, FALSE, TRUE, FALSE
??? ), i = c(FALSE, NA, FALSE, FALSE, NA), j = c(TRUE, NA, TRUE,
??? TRUE, TRUE), k = c(TRUE, NA, TRUE, TRUE, FALSE), l = c(TRUE,
??? NA, TRUE, TRUE, FALSE), m = c(TRUE, NA, FALSE, FALSE, TRUE
??? )), .Names = c("a", "b", "c", "d", "e", "f", "g", "h", "i",
"j", "k", "l", "m"), row.names = c("w", "x", "y", "z", "u"), class =?
"data.frame")
datNew
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j?????
k???? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE? TRUE??
TRUE? TRUE
#x??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA??? NA?? NA????
NA??? NA??? NA
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE? TRUE??
TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE? TRUE??
TRUE FALSE
#u? TRUE? TRUE FALSE FALSE??? NA??? NA? TRUE FALSE??? NA TRUE FALSE?
FALSE? TRUE
dat2New<- datNew
dat2New[rowSums(is.na(dat2New))==0,]<-t(apply(!datNew[rowSums(is.na(datNew))==0,],1,function(x)?
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
dat2New[rowSums(is.na(dat2New))!=0 &?
rowSums(is.na(dat2New))!=ncol(dat2New),]<-NA
?dat2New
#?? a? b? c? d? e? f? g? h? i? j? k? l? m
#w? 0? 0? 0? 0? 0? 1? 2? 3? 4? 0? 0? 0? 0
#x NA NA NA NA NA NA NA NA NA NA NA NA NA
#y? 1? 2? 3? 4? 5? 0? 0? 1? 2? 0? 0? 0? 1
#z? 0? 0? 0? 0? 1? 0? 0? 0? 1? 0? 0? 0? 1
#u NA NA NA NA NA NA NA NA NA NA NA NA NA
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 4:10 PM
Subject: Re: [R] conditional Dataframe filling
Thanks Arun,
Well that is interesting. My intention was to have a dataframe with
the same number of rows in the original data, and for the rows with
NAs, then return NA (If there are NAs, often the entire row has NAs).
What is interesting is that in your code with NAs, the row that has
NAs gets NAs in the output, which is what I am looking for.
I guess a solution is to subset complete rows and then run your line
of code. Unless there is an alternative, to tell cumsum to leave NAs
as NAs?
Thanks again,
Camilo
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:?? Country code: 57
? ? ? ? ? Provider code: 313
? ? ? ? ? Phone 776 2282
? ? ? ? ? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
Dear Camilo,
How do you want to deal with the NAs?
If I remove the NAs:
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat1<- as.data.frame(na.omit(dat1))
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
?dat2
#? a b c d e f g h i j k l m
#w 0 0 0 0 0 1 2 3 4 0 0 0 0
#y 1 2 3 4 5 0 0 1 2 0 0 0 1
#z 0 0 0 0 1 0 0 0 1 0 0 0 1
?dat1
#????? a???? b???? c???? d???? e???? f???? g???? h???? i??? j????
k??? l???? m
#w? TRUE? TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE FALSE TRUE TRUE?
TRUE? TRUE
#y FALSE FALSE FALSE FALSE FALSE? TRUE? TRUE FALSE FALSE TRUE TRUE?
TRUE FALSE
#z? TRUE? TRUE? TRUE? TRUE FALSE? TRUE? TRUE? TRUE FALSE TRUE TRUE?
TRUE FALSE
A.K.
----- Original Message -----
From: Camilo Mora <cmora at dal.ca>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Wednesday, March 27, 2013 3:27 PM
Subject: Re: [R] conditional Dataframe filling
Dear Arun,
Thank you very? much for your help with this.I did not know where to
start looking to solve that problem, so I truly appreciate your input.
The line of code you sent seems to work but it duplicates the
results. Do you know why that may happen?
Below is a larger database, to which I apply your line of code.
Thank you very much again,
Camilo
dat1 <- structure(list(
w = c(TRUE,TRUE,TRUE,TRUE,TRUE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,TRUE,TRUE),
x = c(NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA),
y =
c(FALSE,FALSE,FALSE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE,TRUE,TRUE,TRUE,FALSE),
z = c(TRUE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE,FALSE)),
row.names = c(NA, -13L),
class = "data.frame")
dat1<-t(dat1)
colnames(dat1)<-c("a","b","c","d","e","f","g","h","i","j","k", "l","m")
dat2<-dat1
dat2[]<-t(apply(!dat1,1,function(x)
unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum))))
Camilo Mora, Ph.D.
Department of Geography, University of Hawaii
Currently available in Colombia
Phone:?? Country code: 57
? ? ? ?? Provider code: 313
? ? ? ?? Phone 776 2282
? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282
http://www.soc.hawaii.edu/mora/
Quoting arun <smartpink111 at yahoo.com>:
HI, Just a correction: : dat2[]<-t(apply(!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) #should also work A.K. ----- Original Message ----- From: arun <smartpink111 at yahoo.com> To: Camilo Mora <cmora at dal.ca> Cc: R help <r-help at r-project.org> Sent: Wednesday, March 27, 2013 9:09 AM Subject: Re: [R] conditional Dataframe filling Hi, You could try: dat1<- read.table(text=" a??? b??? c??? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE ",sep="",header=TRUE) dat2<-dat1 ?dat2[]<-t(apply(1*!dat1,1,function(x) unlist(lapply(split(x,cumsum(c(0,abs(diff(x))))),cumsum)))) ?dat2 #? a b c d #1 0 0 0 0 #2 1 2 3 0 #3 1 0 1 2 A.K. ----- Original Message ----- From: Camilo Mora <cmora at dal.ca> To: r-help at r-project.org Cc: Sent: Wednesday, March 27, 2013 4:31 AM Subject: [R] conditional Dataframe filling Hi everyone: This may be trivial but I just have not been able to figure it out. Imagine the following dataframe: a? ?? b? ?? c? ?? d TRUE? TRUE? TRUE? TRUE FALSE FALSE FALSE TRUE FALSE? TRUE? FALSE? FALSE I would like to create a new dataframe, in which TRUE gets 0 but if false then add 1 to the cell to the left. So the results for the example above should be something like: a? ?? b? ?? c? ?? d 0? ?? 0? ?? 0? ?? 0 1? ?? 2? ?? 3? ?? 0 1? ?? 0? ?? 1? ?? 2 I wonder if you may know?. Thanks, Camilo Camilo Mora, Ph.D. Department of Geography, University of Hawaii Currently available in Colombia Phone:?? Country code: 57 ? ? ? ?? Provider code: 313 ? ? ? ?? Phone 776 2282 ? ? ? ?? From the USA or Canada you have to dial 011 57 313 776 2282 http://www.soc.hawaii.edu/mora/
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