Find the number of files for each extension in a directory

I want to count the number of files for each extension in a directory as well as the files without extension.

I have tried a few options, but I haven't found a working solution yet:

find "$folder" -type f | sed 's/.*.//' | sort | uniq -c is an option but doesn't work if there is no file extension. I need to know how many files do not have an extension.

find "$folder" -type f | sed 's/.*.//' | sort | uniq -c

I have also tried a find loop into an array and then sum the results, but at this time that code throws an undeclared variable error, but only outside of the loop:

declare -a arr
arr=()
echo $arr[@]


This throws an undeclared variable, as well as once the find loop completes.

5 Answers
5

find "$path" -type f | sed -e '/.*/[^/]*.[^/]*$/!s/.*/(none)/' -e 's/.*.//' | LC_COLLATE=C sort | uniq -c


Explanation:

find "$path" -type f

"$path"

sed -e '/.*/[^/]*.[^/]*$/!s/.*/(none)/' -e 's/.*.//'

/.*/[^/]*.[^/]*$/!s/.*/(none)/

s/.*.//

LC_COLLATE=C sort

uniq -c

Using Python:

import os
from collections import Counter
from pprint import pprint

lst =
for file in os.listdir('./'):
name, ext = os.path.splitext(file)
lst.append(ext)

pprint(Counter(lst))


The output:

Counter('': 7,
'.png': 4,
'.mp3': 3,
'.jpg': 3,
'.mkv': 3,
'.py': 1,
'.swp': 1,
'.sh': 1)


ext = [ f.split('.')[-1] for f in os.listdir('./') ]

If you have GNU awk, you could do something like

printf '%s' * | gawk 'BEGINRS=""; FS="."; OFS="t"
a[(NF>1 ? $NF : "(none)")]++
ENDfor(i in a) print a[i],i
'


i.e. construct / increment an associative array keyed on the last . separated field, or some arbitrary fixed string such as (none) if there is no extension.

.

(none)

mawk doesn't seem to allow a null-byte record separator - you could use mawk with the default newline separator if you are confident that you don't need to deal with newlines in your file names:

mawk

mawk

printf '%sn' * | mawk 'BEGINFS="."; OFS="t" a[(NF>1 ? $NF : "(none)")]++ ENDfor(i in a) print a[i],i'


With basic /bin/sh or even bash the task can be a little difficult, but as you can see in other answers the tools that can work on aggregate data can deal with such task particularly easy. One such tool would be sqlite database.

/bin/sh

bash

sqlite

The very simple process to use sqlite database would be to create a .csv file with two fields: file name and extension. Later sqlite can use simple aggregate statement COUNT() with GROUP BY ext to perform counting of files based on extension field

sqlite

.csv

sqlite

COUNT()

GROUP BY ext

$ printf "file,extn"; find -type f -exec sh -c 'f=$1##*/;printf "%s,%sn" "$1" "$1##*."' sh ; ; > files.csv
$ sqlite3 <<EOF
> .mode csv
> .import ./files.csv files_tb
> SELECT ext,COUNT(file) FROM files_tb GROUP BY ext;
> EOF
csv,1
mp3,6
txt,1
wav,27


files_tb

printf

Using PowerShell if that's an option:

Get-ChildItem -File | Group-Object Extension -NoElement


or shorter, using aliases:

ls -file | group -n Extension


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