R help - Nov 2016 - Help

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

	[[alternative HTML version deleted]]

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, Andr? Luis Neves <andrluis at
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
>
>

Thank you very much, Sarah!

It worked great in my dataset!

Andre

On Fri, Nov 18, 2016 at 12:31 PM, Sarah Goslee <sarah.goslee at 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, Andr? Luis Neves <andrluis at
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

	[[alternative HTML version deleted]]