> Date: Mon, 30 Mar 2015 09:54:39 -0400
> From: Vikram Chhatre <crypticline...@gmail.com>
> To: r-help@r-project.org
> Subject: [R] changing column labels for data frames inside a list
> Message-ID:
> <CAJZnH0=uGay_1VzjVTMMc=fweydkdjxm_tpi4hzo-ardztr...@mail.gmai
> Content-Type: text/plain; charset="UTF-8"
>
> > summary(mygenfreqt)
> Length Class Mode
> dat1.str 59220 -none- numeric
> dat2.str 59220 -none- numeric
> dat3.str 59220 -none- numeric
>
> > head(mylist[[1]])
> 1 2 3 4 5 6 7 8 9 10
> 12
> L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475 0.3
> 0.275
> L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525 0.6
> 0.725
>
> I want to change 1:12 to pop1:pop12
>
> mylist<- lapply(mylist, function(e) colnames(e) <- paste0('pop',1:12))
>
> What this is doing is replacing the data frames with just names
> pop1:pop12. I just want to replace the column labels.
>
> Thanks for any suggestions.
Some readers have already replied, but here is another option that exploits
lapply()'s "..." parameter. First, we make a reproducible example.
(lista <- list(mtcars, mtcars))
Now, we get the unique number of columns of the data frames in the variable
"lista".
(n.cols <- unique(sapply(lista, ncol)))
Finally, we call lapply() and `colnames<-` to change the column names of both
data frames in "lista". See lapply()'s "..." parameter (?lapply).
(lista <- lapply(X = lista, FUN = `colnames<-`, paste0("pop", seq_len(n.cols))))
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