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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