Thank you very much, Sarah!
It worked great in my dataset!
Andre
On Fri, Nov 18, 2016 at 12:31 PM, Sarah Goslee <sarah.gos...@gmail.com>
wrote:
> 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
> >
> >
>
--
Andre
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