How to count number of files in each directory?

Question

I am able to list all the directories by

find ./ -type d

I attempted to list the contents of each directory and count the number of files i

Answer 1

Easy way to recursively find files of a given type. In this case, .jpg files for all folders in current directory:

find . -name *.jpg -print | wc -l

Answer 2

You could arrange to find all the files, remove the file names, leaving you a line containing just the directory name for each file, and then count the number of times each directory appears:

find . -type f |
sed 's%/[^/]*$%%' |
sort |
uniq -c

The only gotcha in this is if you have any file names or directory names containing a newline character, which is fairly unlikely. If you really have to worry about newlines in file names or directory names, I suggest you find them, and fix them so they don't contain newlines (and quietly persuade the guilty party of the error of their ways).

---

If you're interested in the count of the files in each sub-directory of the current directory, counting any files in any sub-directories along with the files in the immediate sub-directory, then I'd adapt the sed command to print only the top-level directory:

find . -type f |
sed -e 's%^\(\./[^/]*/\).*$%\1%' -e 's%^\.\/[^/]*$%./%' |
sort |
uniq -c

The first pattern captures the start of the name, the dot, the slash, the name up to the next slash and the slash, and replaces the line with just the first part, so:

./dir1/dir2/file1

is replaced by

./dir1/

The second replace captures the files directly in the current directory; they don't have a slash at the end, and those are replace by ./. The sort and count then works on just the number of names.

Answer 3

A super fast miracle command, which recursively traverses files to count the number of images in a directory and organize the output by image extension:

find . -type f | sed -e 's/.*\.//' | sort | uniq -c | sort -n | grep -Ei '(tiff|bmp|jpeg|jpg|png|gif)$'

Credits: https://unix.stackexchange.com/a/386135/354980

Answer 4

find . -type f -printf '%h\n' | sort | uniq -c

gives for example:

  5 .
  4 ./aln
  5 ./aln/iq
  4 ./bs
  4 ./ft
  6 ./hot

Answer 5

This can also be done with looping over ls instead of find

for f in */; do echo "$f -> $(ls $f | wc -l)"; done

Explanation:

for f in */; - loop over all directories

do echo "$f -> - print out each directory name

$(ls $f | wc -l) - call ls for this directory and count lines

Answer 6

I combined @glenn jackman's answer and @pcarvalho's answer(in comment list, there is something wrong with pcarvalho's answer because the extra style control function of character '`'(backtick)).

My script can accept path as an augument and sort the directory list as ls -l, also it can handles the problem of "space in file name".

#!/bin/bash
OLD_IFS="$IFS"
IFS=$'\n'
for dir in $(find $1 -maxdepth 1 -type d | sort); 
do
    files=("$dir"/*)
    printf "%5d,%s\n" "${#files[@]}" "$dir"
done
FS="$OLD_IFS"

My first answer in stackoverflow, and I hope it can help someone ^_^