Hi,

I received several really good suggestions, but the feeling seems to be to use a for loop. I did this at first, but my curiosity took over -- hence, my question. The loop, coupled with a list of relevant columns, seems best.

Thanks all,

Walt

________________________

Walter R. Paczkowski, Ph.D.
Data Analytics Corp.
44 Hamilton Lane
Plainsboro, NJ 08536
________________________
(V) 609-936-8999
(F) 609-936-3733
w...@dataanalyticscorp.com
www.dataanalyticscorp.com
_____________________________________________________


On 9/8/2013 10:06 AM, John Fox wrote:
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 <smartpink...@yahoo.com> 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. <w...@dataanalyticscorp.com>
To: R help <r-help@r-project.org>
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
w...@dataanalyticscorp.com
www.dataanalyticscorp.com

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and provide commented, minimal, self-contained, reproducible code.


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------------------------------------------------
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.


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