Does
do.call('cbind', list_of_dataframes)
do what you want?
S Ellison
> -----Original Message-----
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Wolfgang
> Waser
> Sent: 09 February 2016 09:03
> To: Dénes Tóth; r-help@r-project.org
> Subject: Re: [R] How to extract same columns from identical dataframes in a
> list?
>
> Hi,
>
> sorry if my description was too short / unclear.
>
> > I have a list of 7 data frames, each data frame having 24 rows (hour
> > of the day) and 5 columns (weeks) with a total of 5 x 24 values
>
> [1]
> week1 week2 week3 ...
> 1 x a m ...
> 2 y b n
> 3 z c o
> . . . .
> . . . .
> . . . .
> 24 . . .
>
>
> [2]
> week1 week2 week3 ...
> 1 x2 a2 m2 ...
> 2 y2 b2 n2
> 3 z2 c2 o2
> . . . .
> . . . .
> . . . .
> 24 . . .
>
>
> [3]
> ...
>
> .
> .
> .
>
>
> [7]
> ...
>
>
>
> I now would like to extract e.g. all week2 columns of all data frames in the
> list
> and combine them in a new data frame using cbind.
>
> new data frame
>
> week2 ([1]) week2 ([2]) week2 ([3]) ...
> a a2 .
> b b2 .
> c c2 .
> .
> .
> .
>
> I will then do further row-wise calculations using e.g. apply(x,1,mean), the
> result being a vector of 24 values.
>
>
> I have not found a way to extract specific columns of the data frames in a
> list.
>
>
> As mentioned I can use
>
> sapply(list_of_dataframes,"[",1:24)
>
> which will pick the first 24 values (first column) of each data frame in the
> list
> and arrange them as an array of 24 rows and 7 columns (7 data frames are in
> the list).
> To pick the second column (week2) using sapply I have to use the next 24
> values
> from 25 to 48:
>
> sapply(list_of_dataframes,"[",25:48)
>
>
> It seems that sapply treats the data frames in the list as vectors. I can of
> course
> extract all consecutive weeks using consecutive blocks of
> 24 values, but this seems cumbersome.
>
>
> The question remains, how to select specific columns from data frames in a
> list,
> e.g. all columns 3 of all data frames in the list.
>
>
> Reformatting (unlist(), dim()) in one data frame with one column for each week
> does not help, since I'm not calculating colMeans etc, but row-wise
> calculations
> using apply(x,1,FUN) ("applying a function to margins of an array or matrix").
>
>
>
> Thanks for you help and suggestions!
>
>
> Wolfgang
>
>
>
> On 08/02/16 18:00, Dénes Tóth wrote:
> > Hi,
> >
> > Although you did not provide any reproducible example, it seems you
> > store the same type of values in your data.frames. If this is true, it
> > is much more efficient to store your data in an array:
> >
> > mylist <- list(a = data.frame(week1 = rnorm(24), week2 = rnorm(24)),
> > b = data.frame(week1 = rnorm(24), week2 = rnorm(24)))
> >
> > myarray <- unlist(mylist, use.names = FALSE)
> > dim(myarray) <- c(nrow(mylist$a), ncol(mylist$a), length(mylist))
> > dimnames(myarray) <- list(hour = rownames(mylist$a),
> > week = colnames(mylist$a),
> > other = names(mylist)) # now you can do:
> > mean(myarray[, "week1", "a"])
> >
> > # or:
> > colMeans(myarray)
> >
> >
> > Cheers,
> > Denes
> >
> >
> > On 02/08/2016 02:33 PM, Wolfgang Waser wrote:
> >> Hello,
> >>
> >> I have a list of 7 data frames, each data frame having 24 rows (hour
> >> of the day) and 5 columns (weeks) with a total of 5 x 24 values
> >>
> >> I would like to combine all 7 columns of week 1 (and 2 ...) in a
> >> separate data frame for hourly calculations, e.g.
> >>> apply(new.data.frame,1,mean)
> >>
> >> In some way sapply (lapply) works, but I cannot directly select
> >> columns of the original data frames in the list. As a workaround I
> >> have to select a range of values:
> >>
> >>> sapply(list_of_dataframes,"[",1:24)
> >>
> >> Values 1:24 give the first column, 25:48 the second and so on.
> >>
> >> Is there an easier / more direct way to select for specific columns
> >> instead of selecting a range of values, avoiding loops?
> >>
> >>
> >> Cheers,
> >>
> >> Wolfgang
> >>
> >> ______________________________________________
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> >>
> >
>
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