Even though I work with R since a year or so I still struggle with simple problems. I hope someone can help me with this. Been trying for days and am a little frustrated now. I have a data frame somewhat like the one bellow: dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4)) I want to know how many NAs I have in week 3 and in week 4. -- View this message in context: http://r.789695.n4.nabble.com/count-NAs-per-week-tp4643351.html Sent from the R help mailing list archive at Nabble.com.
count NAs per week
8 messages · R. Michael Weylandt, William Dunlap, arun +2 more
On Mon, Sep 17, 2012 at 11:03 AM, Tagmarie <Ramgad82 at gmx.net> wrote:
Even though I work with R since a year or so I still struggle with simple problems. I hope someone can help me with this. Been trying for days and am a little frustrated now. I have a data frame somewhat like the one bellow: dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4)) I want to know how many NAs I have in week 3 and in week 4.
Thanks for the reproducible example: there are many ways to do this (aggregate, tapply, ave, etc.) but they are all based on the paradigm of: break up your data by "Week" --> apply the function "function(x) sum(is.na(x))" --> recombine. (See, inter alia, the JSS paper on the plyr package) I'm not at a computer with R right now so this is a little untested, but one way might be: with(dattrial, tapply(a, Week, function(x) sum(is.na(x)))) Cheers, Michael
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______________________________________________ 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, Try this: dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4)) ?aggregate(dattrial$a,list(dattrial$Week),function(x) sum(is.na(x))) #? Group.1 x #1?????? 3 1 #2?????? 4 0 #or ?ddply(dattrial,.(Week),summarize, sum(is.na(a))) #? Week ..1 #1??? 3?? 1 #2??? 4?? 0 #or list1<-split(dattrial,dattrial$Week) ?unlist(lapply(lapply(list1,`[`, 1),function(x) sum(is.na(x)))) #3 4 #1 0 A.K. ----- Original Message ----- From: Tagmarie <Ramgad82 at gmx.net> To: r-help at r-project.org Cc: Sent: Monday, September 17, 2012 6:03 AM Subject: [R] count NAs per week Even though I work with R since a year or so I still struggle with simple problems. I hope someone can help me with this. Been trying for days and am a little frustrated now. I have a data frame somewhat like the one bellow: dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4)) I want to know how many NAs I have in week 3 and in week 4. -- View this message in context: http://r.789695.n4.nabble.com/count-NAs-per-week-tp4643351.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ 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.
Thank you Michael, that worked perfectly!
Now I wonder, if it is possible to break my data further apart and put it
together again.
Assume I include a column for an ID in the data frame like this:
dattrial2<-data.frame(a=c(1,NA,NA,NA,2,3), Week=c(3,3,3,4,4,4),
AnimalID=c("Ernie","Bert", "Ernie", "Bert", "Bert", "Ernie"))
Is it possible to get two different lists in the output, one for Ernie and
one for Bert? Or do I have to do it seperately for each animal?
Thank you again! I learn a lot by doing and by people helping me. Thank you
for the hint with the paper.
--
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with(dattrial, table(A_is_missing=is.na(a), Week))
Week
A_is_missing 3 4
FALSE 2 3
TRUE 1 0
with(dattrial, table(A_is_missing=is.na(a), Week))["TRUE", ] # extract the missing="TRUE" row, note quotes
3 4 1 0 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com
-----Original Message----- From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Tagmarie Sent: Monday, September 17, 2012 3:03 AM To: r-help at r-project.org Subject: [R] count NAs per week Even though I work with R since a year or so I still struggle with simple problems. I hope someone can help me with this. Been trying for days and am a little frustrated now. I have a data frame somewhat like the one bellow: dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4)) I want to know how many NAs I have in week 3 and in week 4. -- View this message in context: http://r.789695.n4.nabble.com/count-NAs-per-week- tp4643351.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ 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, May be this is what you wanted. ?res1<-aggregate(dattrial2$a,list(dattrial2$Week,dattrial2$AnimalID),function(x) sum(is.na(x)))
split(res1,res1$Group.2)
#$Bert
#? Group.1 Group.2 x
#1?????? 3??? Bert 1
#2?????? 4??? Bert 1
#$Ernie
?# Group.1 Group.2 x
#3?????? 3?? Ernie 1
#4?????? 4?? Ernie 0
A.K.
----- Original Message -----
From: Tagmarie <Ramgad82 at gmx.net>
To: r-help at r-project.org
Cc:
Sent: Monday, September 17, 2012 9:07 AM
Subject: Re: [R] count NAs per week
Thank you Michael, that worked perfectly!
Now I wonder, if it is possible to break my data further apart and put it
together again.
Assume I include a column for an ID in the data frame like this:
dattrial2<-data.frame(a=c(1,NA,NA,NA,2,3), Week=c(3,3,3,4,4,4),
AnimalID=c("Ernie","Bert", "Ernie", "Bert", "Bert", "Ernie"))
Is it possible to get two different lists in the output, one for Ernie and
one for Bert? Or do I have to do it seperately for each animal?
Thank you again! I learn a lot by doing and by people helping me. Thank you
for the hint with the paper.
--
View this message in context: http://r.789695.n4.nabble.com/count-NAs-per-week-tp4643351p4643371.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
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.
On Sep 17, 2012, at 6:07 AM, Tagmarie wrote:
Thank you Michael, that worked perfectly!
Now I wonder, if it is possible to break my data further apart and put it
together again.
Assume I include a column for an ID in the data frame like this:
dattrial2<-data.frame(a=c(1,NA,NA,NA,2,3), Week=c(3,3,3,4,4,4),
AnimalID=c("Ernie","Bert", "Ernie", "Bert", "Bert", "Ernie"))
Is it possible to get two different lists in the output, one for Ernie and
one for Bert? Or do I have to do it seperately for each animal?
split(dattrial2, dattrial2[['AnimalID']])
Thank you again! I learn a lot by doing and by people helping me. Thank you for the hint with the paper. -- View this message in context: http://r.789695.n4.nabble.com/count-NAs-per-week-tp4643351p4643371.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ 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.
David Winsemius, MD Alameda, CA, USA
Thank you Arun, that was exactly what I was looking for. This is the second time you helped me in this mailing list. Thanks! And also thanks to all of the others of you whou helped me. It is amazing how many possibilities there are to solve my problem (which isn't a prob now anymore). Have a great day, Tagmarie -- View this message in context: http://r.789695.n4.nabble.com/count-NAs-per-week-tp4643351p4643486.html Sent from the R help mailing list archive at Nabble.com.