4 messages · Dimitri Liakhovitski, David Winsemius, Chel Hee Lee
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!
Dimitri Liakhovitski
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
Observe the wonders of recycling in dataframes. If you wanted to drop the last column from this it would be the same modulo changing one column nmae. cbind( df[1:2], stack(df[-c(1:2)] ) ) # a b values ind 1 1 a 1 c1 2 2 b 2 c1 3 3 c 3 c1 4 4 d 4 c1 5 5 e 5 c1 6 1 a 2 c2 7 2 b 3 c2 8 3 c 4 c2 9 4 d 5 c2 10 5 e 6 c2 11 1 a 3 c3 12 2 b 4 c3 13 3 c 5 c3 14 4 d 6 c3 15 5 e 7 c3 16 1 a 4 c4 17 2 b 5 c4 18 3 c 6 c4 19 4 d 7 c4 20 5 e 8 c4
David Winsemius Alameda, CA, USA
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
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!
Thanks a lot, guys!
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!
Dimitri Liakhovitski