I have a data frame like this
xx<-data.frame(cbind(Sample=c('Ctrl_6h','1+0_6h','1+200_6h','1+5k_6h','Ctrl_5K_6h','ConA_6h'),
IFIT1=c(24,25,24.7,24.5,24.2,24.8)))
grep('[[:digit:]]h',xx$Sample)
yy<-xx$Sample
strsplit(yy,"_")
I have to extract the time information separated by '_' in the sample names,
i tried grep and strsplit, it looks that i am not providing some information
correctly. I would appreciate if someone can point me to the correct way.
Thanks
Sharad
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splitting sample names
6 messages · David Winsemius, Jorge I Velez, Henrique Dallazuanna +1 more
On Aug 18, 2011, at 5:43 PM, 1Rnwb wrote:
I have a data frame like this
xx<-
data
.frame
(cbind
(Sample
=c('Ctrl_6h','1+0_6h','1+200_6h','1+5k_6h','Ctrl_5K_6h','ConA_6h'),
IFIT1=c(24,25,24.7,24.5,24.2,24.8)))
grep('[[:digit:]]h',xx$Sample)
yy<-xx$Sample
strsplit(yy,"_")
You are getting tangled up because of stringsAsFactor + TRUE by
default. (The cbind is not needed and may have been confusing things
as well.)
xx <-
data
.frame
(Sample
=c('Ctrl_6h','1+0_6h','1+200_6h','1+5k_6h','Ctrl_5K_6h','ConA_6h'),
IFIT1=c(24,25,24.7,24.5,24.2,24.8),
stringsAsFactors=FALSE)
sub('(^.+)([[:digit:]]h$)', "\\2", xx$Sample)
1] "6h" "6h" "6h" "6h" "6h" "6h"
yy<-xx$Sample
strsplit(yy,"_")
#--------------
[[1]]
[1] "Ctrl" "6h"
[[2]]
[1] "1+0" "6h"
[[3]]
[1] "1+200" "6h"
[[4]]
[1] "1+5k" "6h"
[[5]]
[1] "Ctrl" "5K" "6h"
[[6]]
[1] "ConA" "6h"
Try instead to use these results to guide you
I have to extract the time information separated by '_' in the sample names, i tried grep and strsplit, it looks that i am not providing some information correctly. I would appreciate if someone can point me to the correct way. Thanks Sharad -- View this message in context: http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3753712.html Sent from the R help mailing list archive at Nabble.com.
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD West Hartford, CT
Hi Sharad, Try xx$newSample <- sapply(with(xx, strsplit(as.character(Sample), "_")), "[", 1) xx HTH, Jorge
On Aug 18, 2011, at 5:43 PM, 1Rnwb wrote:
I have a data frame like this
xx<-data.frame(cbind(Sample=c('Ctrl_6h','1+0_6h','1+200_6h','1+5k_6h','Ctrl_5K_6h','ConA_6h'),
IFIT1=c(24,25,24.7,24.5,24.2,24.8)))
grep('[[:digit:]]h',xx$Sample)
yy<-xx$Sample
strsplit(yy,"_")
I have to extract the time information separated by '_' in the sample names,
i tried grep and strsplit, it looks that i am not providing some information
correctly. I would appreciate if someone can point me to the correct way.
Thanks
Sharad
--
View this message in context: http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3753712.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.
Sorry for the noise. The code should have been sapply(with(xx, strsplit(as.character(Sample), "_")), "[", 2) instead of what I sent previously. HTH, Jorge
On Aug 18, 2011, at 7:30 PM, Jorge I Velez wrote:
Hi Sharad, Try xx$newSample <- sapply(with(xx, strsplit(as.character(Sample), "_")), "[", 1) xx HTH, Jorge On Aug 18, 2011, at 5:43 PM, 1Rnwb wrote:
I have a data frame like this
xx<-data.frame(cbind(Sample=c('Ctrl_6h','1+0_6h','1+200_6h','1+5k_6h','Ctrl_5K_6h','ConA_6h'),
IFIT1=c(24,25,24.7,24.5,24.2,24.8)))
grep('[[:digit:]]h',xx$Sample)
yy<-xx$Sample
strsplit(yy,"_")
I have to extract the time information separated by '_' in the sample names,
i tried grep and strsplit, it looks that i am not providing some information
correctly. I would appreciate if someone can point me to the correct way.
Thanks
Sharad
--
View this message in context: http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3753712.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Try this:
transform(xx, Time = gsub(".*_", "", xx$Sample))
On Thu, Aug 18, 2011 at 6:43 PM, 1Rnwb <sbpurohit at gmail.com> wrote:
I have a data frame like this
xx<-data.frame(cbind(Sample=c('Ctrl_6h','1+0_6h','1+200_6h','1+5k_6h','Ctrl_5K_6h','ConA_6h'),
? ? ? ? ? ? ? ? IFIT1=c(24,25,24.7,24.5,24.2,24.8)))
grep('[[:digit:]]h',xx$Sample)
yy<-xx$Sample
strsplit(yy,"_")
I have to extract the time information separated by '_' in the sample names,
i tried grep and strsplit, it looks that i am not providing some information
correctly. I would appreciate if someone can point me to the correct way.
Thanks
Sharad
--
View this message in context: http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3753712.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.
Henrique Dallazuanna Curitiba-Paran?-Brasil 25? 25' 40" S 49? 16' 22" O
Thanks to all of you for the suggestions and corrections. Sharad -- View this message in context: http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3755297.html Sent from the R help mailing list archive at Nabble.com.