[R] How to extract same columns from identical dataframes in a list?

Ulrik Stervbo ulrik.stervbo at gmail.com
Mon Feb 8 16:33:54 CET 2016


Hi Wolfgang,

I'm not sure exactly what you want, but the ldply in the package plyr can
help you make a data.frame from a list of data.frames:

library(plyr)

dfa <- data.frame(cola = LETTERS[1:5], colb = c(1:5))
dfb <- data.frame(cola = LETTERS[1:5], colb = c(1:5))

df.lst <- list(dfa.name = dfa, dfb.name = dfb)

# If you want to use column number
ldply(df.lst, function(cur.df){return(cur.df[, 2])})

# If the column name is always the same
ldply(df.lst, function(cur.df){return(cur.df$colb)})

# Use the entire data.frame
ldply(df.lst, function(cur.df){return(cur.df)})

# The latter can also be done with do.call
do.call(rbind, df.lst)

Hope this helps,
Ulrik

On Mon, 8 Feb 2016 at 16:07 Wolfgang Waser <waser at frankenfoerder-fg.de>
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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