R Split character vector of file paths into list by parent directory?

Question

I have a character vector of file paths:

> tail(paths)
[1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz"
[2] "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"
[3] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz"
[4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz"
[5] "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
[6] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

I would like to split this into a list of vectors, by parent folder name, ie:

> tail(desired)
$ "/home/username/data/dir/GCZ98"
[1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz"
[2] "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"
$ "/home/username/data/dir/GCZ98"
[1] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz"
[2] "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz"
[3] "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
[4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

I have tried using split and strsplit with little sucesss, but am struggling to try and find a regular expression which accomplishes my needs.

Thanks for any help

try basename() and dirname() as in this question? http://stackoverflow.com/questions/2548815/find-file-name-from-full-file-path — Beta Barium Borate, Aug 30 '13 at 15:13
I don't this can be done using a regex, you have to find another way. — Ibrahim Najjar, Aug 30 '13 at 15:14
@BetaBariumBorate - Yes awesome! Can't believe i didn't think that. Thanks — rwb, Aug 30 '13 at 15:16

score 4 · Accepted Answer · answered Aug 30 '13 at 15:18

You could combine split and dirname:

path <- c("/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz",
          "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz",
          "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz")

## split by basedir
split(path, dirname(path))

# $`/home/username/data/dir/GCZ98`
# [1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz" "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"
# 
# $`/home/username/data/dir/GCZ99`
# [1] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz" "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz" "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
# [4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

This is correct - however i would like to give @BetaBariumBorate the credit for giving this answer first — rwb, Aug 30 '13 at 15:22
I'm guessing if Beta Barium Borate cared about points s/he'd have answered himself, s/he was shooting a comment because s/he wanted to help guide but didn't want to take the time to write out an answer.. — Tyler Rinker, Aug 30 '13 at 15:36

score 2 · Answer 2 · answered Aug 30 '13 at 15:19

A regex approach:

> split(paths, gsub("(.*)/[^/]+$", "\\1", paths))
$`/home/username/data/dir/GCZ98`
[1] "/home/username/data/dir/GCZ98/GCZ98_1998_12_16.asc.gz"
[2] "/home/username/data/dir/GCZ98/GCZ98_1998_12_20.asc.gz"

$`/home/username/data/dir/GCZ99`
[1] "/home/username/data/dir/GCZ99/GCZ99_1999_12_21.asc.gz"
[2] "/home/username/data/dir/GCZ99/GCZ99_1999_12_23.asc.gz"
[3] "/home/username/data/dir/GCZ99/GCZ99_1999_12_27.asc.gz"
[4] "/home/username/data/dir/GCZ99/GCZ99_1999_12_28.asc.gz"

R Split character vector of file paths into list by parent directory?

2 Answers2