Thanks for the useful reproducible example.
Here's one of the various ways this can be done:
> lapply(seq_along(mylist), function(i)setNames(mylist[[i]], c("CaZyme",
> names(mylist)[i])))
[[1]]
CaZyme A
1 1 3
2 2 3
[[2]]
CaZyme B
1 1 3
2 2 3
[[3]]
CaZyme C
1 1 4
2 2 5
On Fri, Nov 18, 2016 at 2:02 PM, André Luis Neves <andrl...@ualberta.ca> wrote:
> Dear,
>
> I have the following list (mylist), in which I need to pass to the column 2
> the name of the list itself.
> So, running the follwing commands:
>
> A= data.frame(1:2,3)
> B= data.frame(1:4,3)
> C= data.frame(c(1:2),c(4:5))
> mylist=list(A=A,B=A,C=C)
> lapply(mylist, setNames, paste(c("CaZyme")))
>
> The output would be:
>> lapply(mylist, setNames, paste(c("CaZyme")))
> $A
> CaZyme NA
> 1 1 3
> 2 2 3
>
> $B
> CaZyme NA
> 1 1 3
> 2 2 3
>
> $C
> CaZyme NA
> 1 1 4
> 2 2 5
> 3 3 6
>
> My question is:
> How could I name the second column with the name of each dataframe of the
> list, such that NA would be substitute for A, B and C, respectively.
>
> Thank you very much,
>
> Andre
>
>
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
