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.

  • 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
    echo ${arr[@]}

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

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


  • find "$path" -type f get a recursive listing of all the files on the "$path" folder.
  • sed -e '/.*\/[^\/]*\.[^\/]*$/!s/.*/(none)/' -e 's/.*\.//' regular expressions:
    • /.*\/[^\/]*\.[^\/]*$/!s/.*/(none)/ replace all the files without extension by (none).
    • s/.*\.// get the extension of the remaining files.
  • LC_COLLATE=C sort sort the result, keeping the symbols at the top.
  • uniq -c count the number of repeated entries.
| improve this answer | |

Using Python:

import os
from collections import Counter
from pprint import pprint

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


The output:

Counter({'': 7,
         '.png': 4,
         '.mp3': 3,
         '.jpg': 3,
         '.mkv': 3,
         '.py': 1,
         '.swp': 1,
         '.sh': 1})
| improve this answer | |
  • You probably can get away with list comprehension, like ext = [ f.split('.')[-1] for f in os.listdir('./') ] Thatll make it couple lines shorter and perhaps more Pythonic – Sergiy Kolodyazhnyy Sep 3 '18 at 18:24
  • Thanks for suggestion, I was just trying to write it as clear as I could ... – Ravexina Sep 3 '18 at 18:26
  • 1
    Clarity is the virtue :) Especially when it comes to code and engineering documentation. – Sergiy Kolodyazhnyy Sep 3 '18 at 18:29

If you have GNU awk, you could do something like

printf '%s\0' * | gawk 'BEGIN{RS="\0"; FS="."; OFS="\t"} 
  {a[(NF>1 ? $NF : "(none)")]++} 
  END{for(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.

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:

printf '%s\n' * | mawk 'BEGIN{FS="."; OFS="\t"} {a[(NF>1 ? $NF : "(none)")]++} END{for(i in a) print a[i],i}'
| improve this answer | |

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.

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

$ { printf "file,ext\n"; find -type f -exec sh -c 'f=${1##*/};printf "%s,%s\n" "${1}" "${1##*.}"' sh {} \; ; }  > files.csv
$ sqlite3 <<EOF
> .mode csv
> .import ./files.csv files_tb
> SELECT ext,COUNT(file) FROM files_tb GROUP BY ext;
| improve this answer | |
  • files_tb table I think is being referenced but the table columns are not defined anywhere I can see? – WinEunuuchs2Unix Sep 3 '18 at 20:34
  • @WinEunuuchs2Unix They're defined in csv file itself. That's what the first printf does. And SQLite will default to treating first line of csv file as column names. – Sergiy Kolodyazhnyy Sep 3 '18 at 20:38
  • 1
    Very impressive! +1 – WinEunuuchs2Unix Sep 3 '18 at 21:53

Using PowerShell if that's an option:

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

or shorter, using aliases:

ls -file | group -n Extension
| improve this answer | |
  • 1
    Wow! Great first answer! I didn't even know PowerShell existed for Linux... +1 – Fabby Sep 3 '18 at 22:25
  • 2
    Thanks. It has existed cross-platform and open-source for a while, but there's been a pattern on SO and SU where questions for shell scripting on Windows have often been answered with "Well, install cygwin and use bash, then you can do the following", so I've been hesitant to do the same for Linux SE sites with tools that originated on Windows. But this has been a nice task that shows PowerShell's strengths quite nicely without inviting the old argument about verbosity. – Joey Sep 4 '18 at 7:46

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