Bash: Find folders with less than x files

Solution 1:

• For every subdirectory, print the subdirectory name if there are at most 42 .flac files in the subdirectory. To execute a command on the directories, replace -print by -exec … \;. POSIX compliant.

  find . -type d -exec sh -c 'set -- "$0"/*.flac; [ $# -le 42 ]' {} \; -print
  

  Note that this command won't work to search for directories containing zero .flac files ("$0/*.flac" expands to at least one word). Instead, use

  find . -type d -exec sh -c 'set -- "$0"/*.flac; ! [ -e "$1" ]' {} \; -print
  

• Same algorithm in zsh. **/* expands to all the files in the current directory and its subdirectories recursively. **/*(/) restricts the expansion to directories. {.,**/*}(/) adds the current directory. Finally, (e:…:) restricts the expansion to the matches for which the shell code returns 0.

  echo {.,**/*}(/e:'set -- $REPLY/*.flac(N); ((# <= 42))':)
  

  This can be broken down in two steps for legibility.

  few_flacs () { set -- $REPLY/*.flac(N); ((# <= 42)); }
  echo {.,**/*}(/+few_flacs)
  

_{Changelog:
​• handle x=0 correctly.}

Solution 2:

Replace $MAX with your own limit:

find -name '*.flac' -printf '%h\n' | sort | uniq -c | while read -r n d ; do [ $n -lt $MAX ] && printf '%s\n' "$d" ; done

Note: This will print all the subdirectories with a number of .flac files between 0 and $MAX (both excluded).