[R] 'which' statement for recode?
Hi, Is there a way to easily convert the list of course terms into sequential integers in the dataframe (see code below)? eg. 199801 = 1; 199808=2 I know I can use recode but shouldn't which work? Thanks in advance! sc = data.frame(c(200208, 200701, 201201)) names(sc) = c(TERM) TermList = c(NA, 199801, 199808, 199901, 199908, 21, 28, 200101, 200108, 200201, 200208, 200301, 200308, 200401, 200408, 200501, 200508, 200601, 200608, 200701, 200708, 200801, 200808, 200901, 200908, 201001, 201008, 201101, 201108, 201201, 201308, 201401, 201408) which(TermList==sc$TERM) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] which LETTERS?
I'm confused: Thanks in advance. which(LETTERS == c(A)) [1] 1 which(LETTERS == c(A,B)) [1] 1 2 which(LETTERS == c(A,B,C)) [1] 1 2 3 Warning message: In LETTERS == c(A, B, C) : longer object length is not a multiple of shorter object length Charles Stangor Professor Dept of Psychology University of Maryland Academic Achievement Research Group http://www.charlesstangor.com/AARG [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Can't understand syntax
OK, I need help!! I've been searching, but I don't understand the logic of some this dataframe addressing syntax. What is this type of code called? test [[v3]] [is.na(test[[v2]])] -10 #choose column v3 where column v2 is == 4 and replace with 10 and where is it documented? The code below works for what I want to do (find the non-missing value in a row), but why? test - read.table(text= v1 v2 v3 result 3 NA NA NA NA 3 NA NA NA NA 3 NA , header=TRUE) test [[result]] [!(is.na(test[[v1]]))] - test [[v1]] [!(is.na (test[[v1]]))] test [[result]] [!(is.na(test[[v2]]))] - test [[v2]] [!(is.na (test[[v2]]))] test [[result]] [!(is.na(test[[v3]]))] - test [[v3]] [!(is.na (test[[v3]]))] thanks! On Fri, Jul 13, 2012 at 6:41 AM, Rui Barradas ruipbarra...@sapo.pt wrote: Hello, Check the structure of what you have, df and newdf. You will see that in df dateTime is of class POSIXlt and in newDf newDateTime is of class POSIXct. Solution: [...] df$dateTime - strptime(df$dateTime,%m/%d/%Y %H:%M) df$dateTime - as.POSIXct(df$dateTime) [...] Hope this helps, Rui Barradas Em 13-07-2012 10:24, vioravis escreveu: I have the following dataframe with the first column being of type datetime: dateTime - c(10/01/2005 0:00, 10/01/2005 0:20, 10/01/2005 0:40, 10/01/2005 1:00, 10/01/2005 1:20) var1 - c(1,2,3,4,5) var2 - c(10,20,30,40,50) df - data.frame(dateTime = dateTime, var1 = var1, var2 = var2) df$dateTime - strptime(df$dateTime,%m/%d/%Y %H:%M) I want to create 10 minute interval data as follows: minTime - min(df$dateTime) maxTime - max(df$dateTime) newTime - seq(minTime,maxTime,600) newDf - data.frame(newDateTime = newTime) newDf - merge(newDf,df,by.x = newDateTime,by.y = dateTime,all.x = TRUE) The objective here is to create a data frame with values from df for the datetime in df and NA for the missing ones. However, I am getting the following data frame with both Var1 and Var2 having all NAs. newDf newDateTime var1 var2 1 2005-10-01 00:00:00 NA NA 2 2005-10-01 00:10:00 NA NA 3 2005-10-01 00:20:00 NA NA 4 2005-10-01 00:30:00 NA NA 5 2005-10-01 00:40:00 NA NA 6 2005-10-01 00:50:00 NA NA 7 2005-10-01 01:00:00 NA NA 8 2005-10-01 01:10:00 NA NA 9 2005-10-01 01:20:00 NA NA Can someone help me on how to do the merge based on the two datetime columns? Thank you. Ravi -- View this message in context: http://r.789695.n4.nabble.com/** Merging-on-Datetime-Column-**tp4636417.htmlhttp://r.789695.n4.nabble.com/Merging-on-Datetime-Column-tp4636417.html Sent from the R help mailing list archive at Nabble.com. __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Charles Stangor Professor and Associate Chair [[alternative HTML version deleted]] __ R-help@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.
Re: [R] Substitute list value
thanks! On Fri, Jul 13, 2012 at 4:35 AM, Jessica Streicher j.streic...@micromata.de wrote: two things: - R always counts from 1, not from 0 - listmembers are accessed by using [[ ]] , not [ ] try t1[t==ll[[1]], v] - 99 greetings Jessi On 11.07.2012, at 15:47, Charles Stangor wrote: I can't seem to determine how to get the name of a list member to substitute: ll - list(a1 = a,a2 = b) t1[t==ll[0], v] - 99 why doesn't this substitute to: t1[t==a, v] - 99 Thank you! -- Charles Stangor Professor [[alternative HTML version deleted]] __ R-help@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. -- Charles Stangor Professor and Associate Chair [[alternative HTML version deleted]] __ R-help@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.
[R] which() in subset()
Why does the subset not work in the which() version below? Thank you v1 - subset(t1, version_1==as.character(100-1) | version_1==as.character(100-2)) a-c(100-1, 100-2) v1 - subset(t1, which(a==as.character(version_1)) != 0) [[alternative HTML version deleted]] __ R-help@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.
Re: [R] which() in subset()
Thanks Dennis and Mike... I'm getting it!!! Sent from my Android Rui Barradas ruipbarra...@sapo.pt wrote: Hello, To know why, just evaluate the condition, with 't1$' before 'version_1': which(as.character(t1$version_1) %in% a) != 0 [1] TRUE TRUE It allways evaluates to TRUE, therefore, subset() returns all rows. See if this isn't simpler than both of your forms. v2 - subset(t1, version_1 %in% a) v2 id version_1 1 1 100-1 2 2 100-2 The trick is to use %in% when doing multiple comparisons. With the vector with length equal to the number of observations on the left hand side. Hope this helps, Rui Barradas Em 13-07-2012 12:12, Charles Stangor escreveu: Why does the subset not work in the which() version below? Thank you v1 - subset(t1, version_1==as.character(100-1) | version_1==as.character(100-2)) a-c(100-1, 100-2) v1 - subset(t1, which(a==as.character(version_1)) != 0) [[alternative HTML version deleted]] __ R-help@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. __ R-help@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.
[R] Substitute list value
I can't seem to determine how to get the name of a list member to substitute: ll - list(a1 = a,a2 = b) t1[t==ll[0], v] - 99 why doesn't this substitute to: t1[t==a, v] - 99 Thank you! -- Charles Stangor Professor [[alternative HTML version deleted]] __ R-help@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.
[R] Substitute list value
I can't seem to determine how to get the name of a list member to substitute: ll - list(a1 = a,a2 = b) t1[t==ll[0], v] - 99 why doesn't this substitute to: t1[t==a, v] - 99 Thank you! [[alternative HTML version deleted]] __ R-help@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.
[R] Substitute list value
I can't seem to determine how to get the name of a list member to substitute as a variable name: ll - list(a1 = a,a2 = b) t1[t==ll[1], v] - 99 why doesn't this substitute to: t1[t==a, v] - 99 Thank you! -- Charles Stangor Professor -- Charles Stangor Professor [[alternative HTML version deleted]] __ R-help@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.
Re: [R] Help with loop
I think I just learned this myself:
Don't put the $ extension in the bracket :
df1$cola[is.na(df1$cola)]-
df2$cola
Instead substitute using brackets within the brackets:
df1[cola]is.na(df1[cola])]-
df2[cola]
then the cola s can be substituted.
Maybe this will help
On Wed, Jul 11, 2012 at 10:11 AM, paulalou pl...@medschl.cam.ac.uk wrote:
Hi,
I have two dataframes:
The first, df1, contains some missing data:
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
The second, df2, contains the following:
cola colb colc cold cole
1 1.4 0.8 0.02 1.6 0.6
I'm wanting all missing data in df1$cola to be replaced by the value of
df2$cola. Then the missing data in df1$colb to be replaced with the
corresponding value in df2$colb etc.
I can get this to work column by column with single input lines but as my
original dataset is a lot larger I'm wanting a create a loop but can't work
out how.
The single line command is:
df1$cola[is.na(df1$cola)]-df2$cola
I've tried a replace function within a loop but get error messages:
list-colnames(df1)
for (i in list) {
r-replace(df1$i,df1$i[is.na(df1$i)],df2$i)
}
with error messages of:
Warning messages:
1: In is.na(mymat$snp) :
is.na() applied to non-(list or vector) of type 'NULL'
Can anyone help me with this?
Thanks
--
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[R] sapply question
Why does this sapply code change df3 but not df1?
Thanks
df1 - read.table(text=
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
, header=TRUE)
df2 -df1*2
df1
df2
df3 -sapply(names(df1),function(x) {df1[[x]]- df2[[x]]})
df1
df3
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[R] Analyzing subsets
Say I want to perform transformations on row subsets of my dataframe. Do I have to break the dataframe into subsets, perform the analysis on each subset, and rbind() them together again? ..or is there another way. Thank you! [[alternative HTML version deleted]] __ R-help@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.
