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Count files in subdirectories and list the deepest directories

Tags:

find

bash

awk

I want a list of directories with the amount of mp3 files in it; then sorted desc - simply to see, which directories contain the most files.


My command

## Relevant command
find . -mindepth 1 -type f -iname "*.mp3" -printf '%P\0' |
  awk -F/ -vRS='\0' '{n[$1]++}; END{for (i in n) {printf(n[i]" "i" \n")};}' > ./foo.txt

sort -rno ./foo.txt ./foo.txt
## Full command (output improvements only)
find . -mindepth 1 -type f -iname "*.mp3" -printf '%P\0' |
  awk -F/ -vRS='\0' '{n[$1]++}; END{for (i in n) {printf("%03d",n[i]);printf("   ");printf(substr(i,0,60));printf("\n")}; if(length(n)==0) print "NO mp3 found." }' > ./foo.txt
sort -rno ./foo.txt ./foo.txt

Directory structure

./dir_1/fileA.mp3
./dir_2/subdir_1/fileB.mp3
./dir_2/subdir_2/fileC.mp3
./dir_2/subdir_2/fileD.mp3
...

Output

# What I get:
003  dir_2
001  dir_1

# What I want:
002  dir_2/subdir_2
001  dir_2/subdir_1
001  dir_1

The Problem

It only prints the topmost directories, not the deepest possible. It sums up the mp3 count of subdirs.

I cant increase -mindepth because the depth varies.


It would be okay to have both, like this:

003  dir_2
002  dir_2/subdir_2
001  dir_2/subdir_1
001  dir_1

I tried the find -links 2 argument but it only works for -type d not -type f.

like image 537
Jonathan Avatar asked Sep 01 '25 15:09

Jonathan


2 Answers

Except for formatting the count and the error message, awk doesn't seem to be needed.

Since you seem to be using GNU utilities which generally accept a nul delimiter option, if you use %h instead of %P, you can count the directories directly. For example:

find . -type f -iname '*.mp3' -printf '%h\0' |
sort -z |
uniq -zc |
sort -zrn |
tr '\0' '\n'

Or explicitly using gawk for its ability to sort the array itself, and to format the counts:

find . -type f -iname '*.mp3' -printf '%h\0' |
gawk -v RS='\0' '
    {n[$0]++}
    END {
        PROCINFO["sorted_in"]="@val_num_desc"
        if (NR) for (i in n) printf "%03d   %s\n",n[i],i
        else print "NO mp3 found."
    }
'
like image 78
jhnc Avatar answered Sep 04 '25 05:09

jhnc


Setup:

mkdir -p dir_{1..2} dir_2/subdir_{1..2}
touch ./dir_1/fileA.mp3 ./dir_2/subdir_1/fileB.mp3 ./dir_2/subdir_2/file{C,D}.mp3

One awk approach using OP's \0 terminated filenames:

 find . -mindepth 1 -type f -iname "*.mp3" -printf '%P\0' |
 awk -vRS='\0' '
 { match($0,/\/[^/]+$/)                      # find last "/" plus file name
   count[substr($0,1,RSTART-1)]++            # strip off directory name(s) and use an index in count[] array
 }
 END {
   if (NR==0)
      print "NO mp3 found."
   else
      for (dir in count)
          printf "%03d %s\n",count[dir],dir
 }'

This generates:

001 dir_1
001 dir_2/subdir_1
002 dir_2/subdir_2

Piping the output to sort -rn generates:

002 dir_2/subdir_2
001 dir_2/subdir_1
001 dir_1

If we remove all mp3 files and run again this generates:

NO mp3 found.
like image 44
markp-fuso Avatar answered Sep 04 '25 07:09

markp-fuso