Creating data frames from each row of a list element

Question

assume, that I've a list of data frames and every data frame inside the list has the same structure:

$final01_1
    k    h      value
85 12 h=12 0.02867449
86 11 h=23 0.02885711
87 10 h=15 0.02868290
88  9 h=14 0.02872784
89  8 h=13 0.02864886
90  7 h=19 0.02869149
.   . ..    .....

$final02_1
    k    h      value
85 12 h=12 0.02013402
86 11 h=17 0.02012700
87 10 h=17 0.01972729
88  9 h=26 0.02025178
89  8 h=13 0.02035556
90  7 h=13 0.02022585

$final03_1
k    h      value
85 12 h=55 0.04254915
86 11 h=56 0.04283464
87 10 h=24 0.04244404
88  9 h=22 0.04239187
89  8 h=21 0.04318766
90  7 h=21 0.04510606

and so on. I want to create (in this case) 6 new data frames crated by the rows of each $final. That means the first data frame should contain all first rows of each data frame like this:

$foo
85 12 h=12 0.02867449
85 12 h=12 0.02013402
85 12 h=55 0.04254915

and this for all rows. I thinks something with lapply would be the solution, but I don't get an efficient solution.

Hope, that you can help me

akrun · Accepted Answer · 2018-04-20 11:58:26Z

2

We can loop through the rownames and then extract the rows

lst2 <- lapply(rownames(lst[[1]]), function(rn)
           do.call(rbind, lapply(lst, function(y) y[rn,])))

Note: A data.frame cannot have the same row names so, the duplicate row names are made into a unique row name with make.unique

This can also be done with split. Loop through the list with map_df, create a row names column with (rownames_to_column) and get a single dataset, then split by the 'rn' column so that we get a list of data.frames having the same 'rn'

library(tidyverse)
lst %>%
    map_df(rownames_to_column, 'rn') %>%
    split(.$rn)

-output

#$`85`
#   rn  k    h      value
#1  85 12 h=12 0.02867449
#7  85 12 h=12 0.02013402
#13 85 12 h=55 0.04254915

#$`86`
#   rn  k    h      value
#2  86 11 h=23 0.02885711
#8  86 11 h=17 0.02012700
#14 86 11 h=56 0.04283464

#$`87`
#   rn  k    h      value
#3  87 10 h=15 0.02868290
#9  87 10 h=17 0.01972729
#15 87 10 h=24 0.04244404

#$`88`
#   rn k    h      value
#4  88 9 h=14 0.02872784
#10 88 9 h=26 0.02025178
#16 88 9 h=22 0.04239187

#$`89`
#   rn k    h      value
#5  89 8 h=13 0.02864886
#11 89 8 h=13 0.02035556
#17 89 8 h=21 0.04318766

#$`90`
#   rn k    h      value
#6  90 7 h=19 0.02869149
#12 90 7 h=13 0.02022585
#18 90 7 h=21 0.04510606

As mentioned above, it is better to have unique row names. So, we will leave the 'rn' as a column

edited Apr 20, 2018 at 11:58

answered Apr 20, 2018 at 11:41

akrun

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2 Comments

rook1996 Over a year ago

Thanks for answering so fast. It seems to be false because one column isn't correctly. The column h contains factors. The result would be h factor,1 factor2, factor,1

akrun Over a year ago

@rook1996 I updated the answer from changing sapply to lapply. Can you please check

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