Dear Walt and A.K.,
One shouldn't reflexively avoid loops in R. In this case, it seems to me
clearer to use a loop, and it's no less "efficient" (especially, I would guess,
when one takes into account the time to figure out how to do the computation).
I get
> system.time({
+
res<-do.call(rbind,lapply(split(colnames(dat1),((seq_len(ncol(dat1))-1)%/%21)+1),function(x)
{x1<- dat1[,x]; colnames(x1)<- paste("V",1:21);x1}))
+ row.names(res)<- 1:nrow(res)
+ })
user system elapsed
0.02 0.00 0.02
> dim(res)
[1] 1170 21
> system.time({
+ res2 <- as.data.frame(matrix(0, 1170, 21))
+ for (i in 1:9){
+ res2[((i - 1)*130 + 1):(i*130), ] <- dat1[, ((i - 1)*21 + 1):(i*21)]
+ }
+ })
user system elapsed
0.02 0.00 0.01
> dim(res2)
[1] 1170 21
> all(res == res2)
[1] TRUE
Best,
John
On Sun, 8 Sep 2013 06:29:42 -0700 (PDT)
arun <[email protected]> wrote:
>
>
> Hi,
>
> You could try:
> set.seed(48)
> dat1<- as.data.frame(matrix(sample(1:40,189*130,replace=TRUE),ncol=189))
> res<-do.call(rbind,lapply(split(colnames(dat1),((seq_len(ncol(dat1))-1)%/%21)+1),function(x)
> {x1<- dat1[,x]; colnames(x1)<- paste("V",1:21);x1}))
> row.names(res)<- 1:nrow(res)
> dim(res)
> #[1] 1170 21
> A.K.
>
>
>
> ----- Original Message -----
> From: Data Analytics Corp. <[email protected]>
> To: R help <[email protected]>
> Cc:
> Sent: Saturday, September 7, 2013 11:33 PM
> Subject: [R] melting a data frame
>
> Hi,
>
> Suppose I have a data frame with 189 columns. The columns are actually 9
> blocks of 21 columns each, each block representing measures on each of 9
> products. There are 130 rows. Suppose I extract the first block of 21
> columns and make them into a separate data frame. I then want to take the
> second block of 21 columns and rbind it to the first; then the third set of
> 21 and rbind it to the first two; etc. The final data frame should have 1170
> (= 9 * 130) rows and 21 columns. Is there an easy way to melt the blocks
> comparable to using the melt function in the plyr package (which is why I'm
> referring to what I want to do as "melting")? It seems that there should be
> a simple way to do this. I used a for loop which worked, but I want to see
> if there's a more efficient way.
>
> Thanks,
>
> Walt
>
> ________________________
>
> Walter R. Paczkowski, Ph.D.
> Data Analytics Corp.
> 44 Hamilton Lane
> Plainsboro, NJ 08536
> ________________________
> (V) 609-936-8999
> (F) 609-936-3733
> [email protected]
> www.dataanalyticscorp.com
>
> ______________________________________________
> [email protected] 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.
>
>
> ______________________________________________
> [email protected] mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
------------------------------------------------
John Fox
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
