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extracting columns with same partial name
4 messages · Henrique Dallazuanna, Peter Ehlers, Martin Striz
Try this:
DF[,grep("BOUTLENGTHTOT", l)]
On Thu, Jan 14, 2010 at 4:43 PM, Martin Striz <mstriz at gmail.com> wrote:
Hi folks, I'm new to the list.
I have a data file with 256 columns. ?Here's just a subset of names(data):
?[1] "MOUSE" ? ? ? ? ? ? ? "BASEDATE1" ? ? ? ? ? "PERCENTSLEEPTOT1"
?[4] "PERCENTSLEEPNIGHT1" ?"PERCENTSLEEPDAY1" ? ?"BOUTLENGTHTOT1"
?[7] "BOUTLENGTHNITE1" ? ? "BOUTLENGTHDAY1" ? ? ?"BOUTTHRESTOTP1"
?[10] "BOUTTHRESHNITEP1" ? ?"BOUTTHRESHDAYP1" ? ? "BOUTTHRESTOTN1"
?[13] "BOUTTHRESHNITEN1" ? ?"BOUTTHRESHDAYN1" ? ? "ACTONSET1"
?[16] "PEAKACT1" ? ? ? ? ? ?"BASEDATE2" ? ? ? ? ? "PERCENTSLEEPTOT2"
?[19] "PERCENTSLEEPNIGHT2" ?"PERCENTSLEEPDAY2" ? ?"BOUTLENGTHTOT2"
?[22] "BOUTLENGTHNITE2" ? ? "BOUTLENGTHDAY2" ? ? ?"BOUTTHRESTOTP2"
?[25] "BOUTTHRESHNITEP2" ? ?"BOUTTHRESHDAYP2" ? ? "BOUTTHRESTOTN2"
?[28] "BOUTTHRESHNITEN2" ? ?"BOUTTHRESHDAYN2" ? ? "ACTONSET2"
?[31] "PEAKACT2" ? ? ? ? ? ?"BASEDATE3" ? ? ? ? ? "PERCENTSLEEPTOT3"
?[34] "PERCENTSLEEPNIGHT3" ?"PERCENTSLEEPDAY3" ? ?"BOUTLENGTHTOT3"
You'll notice that certain names repeat, like BOUTLENGTHTOT1,
BOUTLENGTHTOT2, etc. ?These represent each day of data recording, and they
go up to 17.
I want to extract, for example, BOUTLENGTHTOT1 - 17. ?I'm new to R, and the
only ways that I know how to this are:
newData <- data[, c("BOUTLENGTHTOT1", "BOUTLENGTHTOT2", ....)]
or
newData <- data.frame(col1 = data$BOUTLENGTHTOT1, col2 =
data$BOUTLENGTHTOT2, .....)
Both of which take a long time for 17 columns. ?Is there a way that I can
use a wildcard to grab all BOUTLENGTHTOT* columns?
Thanks,
Martin
? ? ? ?[[alternative HTML version deleted]]
______________________________________________ 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
Try
dat[names(dat) %in% paste("BOUTLENGTHTOT", 1:17, sep="")]
-Peter Ehlers
Martin Striz wrote:
Hi folks, I'm new to the list.
I have a data file with 256 columns. Here's just a subset of names(data):
[1] "MOUSE" "BASEDATE1" "PERCENTSLEEPTOT1"
[4] "PERCENTSLEEPNIGHT1" "PERCENTSLEEPDAY1" "BOUTLENGTHTOT1"
[7] "BOUTLENGTHNITE1" "BOUTLENGTHDAY1" "BOUTTHRESTOTP1"
[10] "BOUTTHRESHNITEP1" "BOUTTHRESHDAYP1" "BOUTTHRESTOTN1"
[13] "BOUTTHRESHNITEN1" "BOUTTHRESHDAYN1" "ACTONSET1"
[16] "PEAKACT1" "BASEDATE2" "PERCENTSLEEPTOT2"
[19] "PERCENTSLEEPNIGHT2" "PERCENTSLEEPDAY2" "BOUTLENGTHTOT2"
[22] "BOUTLENGTHNITE2" "BOUTLENGTHDAY2" "BOUTTHRESTOTP2"
[25] "BOUTTHRESHNITEP2" "BOUTTHRESHDAYP2" "BOUTTHRESTOTN2"
[28] "BOUTTHRESHNITEN2" "BOUTTHRESHDAYN2" "ACTONSET2"
[31] "PEAKACT2" "BASEDATE3" "PERCENTSLEEPTOT3"
[34] "PERCENTSLEEPNIGHT3" "PERCENTSLEEPDAY3" "BOUTLENGTHTOT3"
You'll notice that certain names repeat, like BOUTLENGTHTOT1,
BOUTLENGTHTOT2, etc. These represent each day of data recording, and they
go up to 17.
I want to extract, for example, BOUTLENGTHTOT1 - 17. I'm new to R, and the
only ways that I know how to this are:
newData <- data[, c("BOUTLENGTHTOT1", "BOUTLENGTHTOT2", ....)]
or
newData <- data.frame(col1 = data$BOUTLENGTHTOT1, col2 =
data$BOUTLENGTHTOT2, .....)
Both of which take a long time for 17 columns. Is there a way that I can
use a wildcard to grab all BOUTLENGTHTOT* columns?
Thanks,
Martin
[[alternative HTML version deleted]]
______________________________________________ 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.
Peter Ehlers University of Calgary 403.202.3921
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