Find all files, create CSV with one row per subdirectory and file names in collumns

Question

I have a directory with subdirectories and files structured like this:

01/fileA
01/fileB
01/fileC
02/fileD
02/fileE
03/fileF
03/fileG
03/fileH
04/fileI

I'd like to get a CSV that looks like this:

01, fileA, fileB, fileC
02, fileD, fileE
03, fileF, fileG, fileH
04, fileI

In other words, I want to generate a CSV with one row per subdirectory, with files listed as columns.

Is it possible to do this from the Linux command line?

Answer 1

That can be done in a number of ways. One simple method could be this

for d in *
do  echo -n "$d, "
    ls -m $d
done

Answer 2

It's probably overkill, but using GNU datamash

find 0? -type f | sort -t/ | datamash -t\/ groupby 1 collapse 2 | sed 's/\//,/'
01,fileA,fileB,fileC
02,fileD,fileE
03,fileF,fileG,fileH
04,fileI

Or with a perl hash of arrays

find 0? -type f | perl -F/ -alne '
  push @{$dirs{$F[0]}}, $F[1]; 
  END{
    for $d (sort keys %dirs) {print join ",", $d, sort @{$dirs{$d}}}
  }'
01,fileA,fileB,fileC
02,fileD,fileE
03,fileF,fileG,fileH
04,fileI

or with GNU awk

find 0? -type f | sort -t/ | gawk -F/ '
  {dirs[$1] = dirs[$1] "," $2} 
  END {
    n = asorti(dirs,sdirs); 
    for(i=1;i<=n;i++) print sdirs[i] "" dirs[sdirs[i]]
}'
01,fileA,fileB,fileC
02,fileD,fileE
03,fileF,fileG,fileH
04,fileI

With GNU awk > 4.0 you can simplify the array traversal to

  END {
    PROCINFO["sorted_in"] = "@ind_num_asc";
    for (d in dirs) print d "" dirs[d];
  }'

Answer 3

Here is another solution

find * -type d -printf "\n%p, " -exec ls -w0 -m {} \; |
    sed -e '/^$/d' -e 's/, *$//'

Output

01, fileA, fileB, fileC
02, fileD, fileE
03, fileF, fileG, fileH
04, fileI