[R] Split a dataframe by rownames and/or colnames
David Winsemius
dwinsemius at comcast.net
Fri Feb 20 20:36:15 CET 2015
On Feb 20, 2015, at 9:33 AM, Tim Richter-Heitmann wrote:
> Dear List,
>
> Consider this example
>
> df <- data.frame(matrix(rnorm(9*9), ncol=9))
> names(df) <- c("c_1", "d_1", "e_1", "a_p", "b_p", "c_p", "1_o1", "2_o1", "3_o1")
> row.names(df) <- names(df)
>
>
> indx <- gsub(".*_", "", names(df))
>
> I can split the dataframe by the index that is given in the column.names after the underscore "_".
>
> list2env(
> setNames(
> lapply(split(colnames(df), indx), function(x) df[x]),
> paste('df', sort(unique(indx)), sep="_")),
> envir=.GlobalEnv)
>
> However, i changed my mind and want to do it now by rownames. Exchanging colnames with rownames does not work, it gives the exact same output (9 rows x 3 columns). I could do
> as.data.frame(t(df_x),
> but maybe that is not elegant.
> What would be the solution for splitting the dataframe by rows?
The split.data.frame method seems to work perfectly well with a rownames-derived index argument:
> split(df, sub(".+_","", rownames(df) ) )
$`1`
c_1 d_1 e_1 a_p b_p c_p 1_o1 2_o1 3_o1
c_1 -0.11 -0.04 1.33 -0.87 -0.16 -0.25 -0.75 0.34 0.14
d_1 -0.62 -0.94 0.80 -0.78 -0.70 0.74 0.11 1.44 -0.33
e_1 0.98 -0.83 0.48 0.19 -0.32 -1.01 1.28 1.04 -2.16
$o1
c_1 d_1 e_1 a_p b_p c_p 1_o1 2_o1 3_o1
1_o1 -0.93 -0.02 0.69 -0.67 1.04 1.04 -1.50 -0.36 0.50
2_o1 0.02 -0.16 -0.09 -1.50 -0.02 -1.04 1.07 -0.45 1.56
3_o1 -1.42 0.88 -0.05 0.85 -1.35 0.21 1.35 0.92 -0.76
$p
c_1 d_1 e_1 a_p b_p c_p 1_o1 2_o1 3_o1
a_p -1.35 0.91 -0.58 -0.63 0.94 -1.13 0.71 0.25 0.82
b_p -0.25 -0.73 -0.41 -1.71 1.28 0.19 -0.35 1.74 -0.93
c_p -0.01 -1.11 -0.12 0.58 1.51 0.03 -0.99 -0.23 -0.03
>
> Thank you very much!
>
> --
> Tim Richter-Heitmann
>
--
David Winsemius
Alameda, CA, USA
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