I would like a BASH command to list just the count of files in each subdirectory of a directory.

E.g. in directory /tmp there are dir1, dir2, ... I'd like to see :

`dir1` : x files 
`dir2` : x files ...

Assuming you want a recursive count of files only, not directories and other types, something like this should work:

find . -maxdepth 1 -mindepth 1 -type d | while read dir; do
  printf "%-25.25s : " "$dir"
  find "$dir" -type f | wc -l
  • Also, I get "find: warning: you have specified the -maxdepth option after a non-option argument -type, but options are not positional (-maxdepth affects tests specified before it as well as those specified after it). Please specify options before other arguments." – jldupont Sep 14 '12 at 21:38
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    Both answers given so far will give incorrect results in the unlikely case that there are files whose names include newline characters. You can handle that with a find ... -print0 | xargs -0 .... – Scott Sep 14 '12 at 21:57
  • @jldupont: move the depth arguments before the ´-type d´, I've edited the answer. – Thor Sep 14 '12 at 22:58
  • Yes, and let me add the information that this excellent solution will not take any external variables and thus will work with bash alias!! – syntaxerror Nov 25 '14 at 13:07
  • fast, and piping sort -rn -k 2,2 -t$':' you get the DESC list – Andre Figueiredo Apr 18 '18 at 23:05

This task fascinated me so much that I wanted to figure out a solution myself. It doesn't even take a while loop and MAY be faster in execution speed. Needless to say, Thor's efforts helped me a lot to understand things in detail.

So here's mine:

find . -maxdepth 1 -mindepth 1 -type d -exec sh -c 'echo "{} : $(find "{}" -type f | wc -l)" file\(s\)' \;

It looks modest for a reason, for it's way more powerful than it looks. :-)

However, should you intend to include this into your .bash_aliases file, it must look like this:

alias somealias='find . -maxdepth 1 -mindepth 1 -type d -exec sh -c '\''echo "{} : $(find "{}" -type f | wc -l)" file\(s\)'\'' \;'

Note the very tricky handling of nested single quotes. And no, it is not possible to use double quotes for the sh -c argument.

  • 1
    It is slower as it invokes /bin/sh for each directory. You can check this with strace -fc script. Your version makes about 70% more system-calls. +1 for shorter code :-) – Thor Oct 31 '16 at 13:58
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    inspired by this; sorted by filecount: find . -maxdepth 1 -mindepth 1 -type d -exec sh -c 'echo "$(find "{}" -type f | wc -l)" {}' \; | sort -nr – mnagel Mar 21 '17 at 8:44
find . -type f | cut -d"/" -f2 | uniq -c

Lists a folders and files in the current folder with a count of files found beneath. Quick and useful IMO. (files show with count 1).

  • 1
    How about a little explanation of how it is working ? :) – C0deDaedalus Oct 4 '18 at 13:32
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    awesome, thanks! You might want to add | sort -rn to sort subdirs by number of files. – Dennis Golomazov Sep 18 '19 at 8:12
  • Works well except also shows files in the current dir, not just subdirs... – rogerdpack Apr 2 at 18:10

Using find is definitely the way to go if you want to count recursively, but if you just want a count of the files directly under a certain directory:

ls dir1 | wc -l

  • 1
    I don't want to do this for each of the 1000's of directories I've got there... – jldupont Sep 14 '12 at 21:37
  • Then use xargs. ls -d */ | xargs -n1 ls | wc -l (Use the answer you accepted if it already works, though! This is just And Now You Know.) – jrajav Sep 14 '12 at 21:41
  • your proposal didn't show up any results in many seconds whereas the answer I accepted did. – jldupont Sep 15 '12 at 8:03
  • @jrajav this approach absolutely fails for directories with whitespace in them. This is why find is so important. (let alone -print0 and xargs -0, already pointed out by Scott in the other answer) – syntaxerror Nov 25 '14 at 13:03
find . -mindepth 1 -type d -print0 | xargs -0 -I{} sh -c 'printf "%4d : %s\n" "$(find {} -type f | wc -l)" "{}"'

I need often need to count the number of files in my sub-directories and use this command. I prefer the count to appear first.


What I use... This makes an array of all the subdirectories in the one you give as a parameter. Print the subdirectory and the count of that same subdirectory until all the subdirectories are processed.

directories=($(/bin/ls -l $1 | /bin/grep "^d" | /usr/bin/awk -F" " '{print $9}'))

for item in ${directories[*]}
        if [ -d "$1$item" ]; then
            echo "$1$item"
            /bin/ls $1$item | /usr/bin/wc -l

You could use this python code. Boot up the interpreter by running python3 and paste this:

folder_path = '.'
import os, glob
for folder in sorted(glob.glob('{}/*'.format(folder_path))):
    print('{:}: {:>8,}'.format(os.path.split(folder)[-1], len(glob.glob('{}/*'.format(folder)))))

Or a recursive version for nested counts:

import os, glob
def nested_count(folder_path, level=0):
    for folder in sorted(glob.glob('{}/'.format(os.path.join(folder_path, '*')))):
        print('{:}{:}: {:,}'.format('    '*level, os.path.split(os.path.split(folder)[-2])[-1], len(glob.glob(os.path.join(folder, '*')))))
        nested_count(folder, level+1)

Example output:

>>> figures: 5
>>> misc: 1
>>> notebooks: 5
>>>     archive: 65
>>>     html: 12
>>>     py: 12
>>>     src: 14
>>> reports: 1
>>>     content: 6
>>> src: 1
>>>     html_download: 1

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