[R] More elegant way of stacking the data

Lee, Chel Hee chl948 at mail.usask.ca
Tue Nov 25 15:42:16 CET 2014


If you do not want to use the loop, a function called 'reshape' may be 
useful:

 > df <- data.frame(a=1:5,b=letters[1:5],c1=1:5,c2=2:6,c3=3:7,c4=4:8)
 > out2 <- reshape(data=df, direction="long", varying=list(3:6), 
times=paste("c",1:4,sep=""))
 > out2
      a b time c1 id
1.c1 1 a   c1  1  1
2.c1 2 b   c1  2  2
3.c1 3 c   c1  3  3
4.c1 4 d   c1  4  4
5.c1 5 e   c1  5  5
1.c2 1 a   c2  2  1
2.c2 2 b   c2  3  2
3.c2 3 c   c2  4  3
4.c2 4 d   c2  5  4
5.c2 5 e   c2  6  5
1.c3 1 a   c3  3  1
2.c3 2 b   c3  4  2
3.c3 3 c   c3  5  3
4.c3 4 d   c3  6  4
5.c3 5 e   c3  7  5
1.c4 1 a   c4  4  1
2.c4 2 b   c4  5  2
3.c4 3 c   c4  6  3
4.c4 4 d   c4  7  4
5.c4 5 e   c4  8  5

I hope this helps.

Chel Hee Lee

On 11/24/2014 5:12 PM, Dimitri Liakhovitski wrote:
> I have the data frame 'df' and my desired solution 'out'.
> I am sure there is a more elegant R-way to do it - without a loop.
>
> df = data.frame(a=1:5,b=letters[1:5],c1=1:5,c2=2:6,c3=3:7,c4=4:8)
> mylist=NULL
> for(i in 1:4){
>    myname<-paste("c",i,sep="")
>    mylist[[i]]<-df[c("a","b",myname)]
>    names(mylist[[i]])<-c("a","b","c")
> }
> out<-do.call(rbind,mylist)
> out
>
> Thank you!
>



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