I downloaded a lot of images in a directory.
Downloader renamed files which already exist.
I also renamed some of the files manually.

hello.jpg      <-- manually renamed `b(3).jpg`
world.jpg      <-- manually renamed `d.jpg`

How to remove duplicated ones? The result should be:


note: name doesn't matter. I just want uniq files.

10 Answers 10


bash 4.x

declare -A arr
shopt -s globstar

for file in **; do
  [[ -f "$file" ]] || continue
  read cksm _ < <(md5sum "$file")
  if ((arr[$cksm]++)); then 
    echo "rm $file"

This is both recursive and handles any file name. Downside is that it requires version 4.x for the ability to use associative arrays and recursive searching. Remove the echo if you like the results.

gawk version

gawk '
    cmd="md5sum " q FILENAME q
    cmd | getline cksm
    sub(/ .*$/,"",cksm)
      cmd="echo rm " q FILENAME q
  }' q='"' *

Note that this will still break on files that have double-quotes in their name. No real way to get around that with awk. Remove the echo if you like the results.

  • fine, the bash version worked for me, but in my test, with 2 similar folders, it deleted half of duplicates in one folder, and half in the other. why. i would expect deletion of everyone (duplicated) of one folder.
    – Ferroao
    Dec 5 '17 at 17:58
  • @Ferroao Perhaps they were not exact duplicates. If just one bit is off the md5 hash that my script is using to determine duplicity would be completely different. You could add an echo cksm just after the line starting with read if you want to see each file’s hash.
    – SiegeX
    Dec 5 '17 at 20:38
  • no, all "duplicates" (copies) were removed, remaining 1 version, let's say the original. half copies were deleted from one folder, and the other half from the other folder (100% deletion of copies). my 100% is for copies in excess, not of the totality
    – Ferroao
    Dec 5 '17 at 22:40
  • @Ferroao I see. In that case it seems when bash does its recursive path expansion via **, it orders the list in such a way that the two folders are interleaved rather than all of folder 1 then all of folder 2. The script will always leave the first ‘original’ it hits as it iterates through the list. You can echo $file before the read line to see if this is true.
    – SiegeX
    Dec 6 '17 at 0:49
  • Don't reinvent the wheel. This is a kind of dangerous operation and you really don't want to use a program written by someone while browsing stackexchange during a boring meeting.
    – Nobody
    Mar 5 '20 at 11:57

fdupes is the tool of your choice. To find all duplicate files (by content, not by name) in the current directory:

fdupes -r .

To manually confirm deletion of duplicated files:

fdupes -r -d .

To automatically delete all copies but the first of each duplicated file (be warned, this warning, this actually deletes files, as requested):

fdupes -r -f . | grep -v '^$' | xargs rm -v

I'd recommend to manually check files before deletion:

fdupes -rf . | grep -v '^$' > files
... # check files
xargs -a files rm -v
  • Works great, but fails if file names contain spaces. Jun 23 '17 at 12:15
  • 1
    @DanielWolf try with xargs option -d '\n'
    – Jakob
    Jun 27 '17 at 12:13
  • 5
    Also, newer versions of fdupes have the built-in option to delete all but the first in a list of duplicate files: fdupes -rdN . where -r is recursive, -d is delete and -N is no-prompt
    – Rand
    May 15 '19 at 22:38
  • Thank you, This is outstanding because can detect more than 2 duplicates and allows you to select which one of the dups you want to preserve (or all of them).
    – Smeterlink
    Aug 4 '19 at 16:39
  • xargs: unterminated quote
    – Alexey Sh.
    Jan 3 '20 at 5:53

You can try FSLint. It has both command line and GUI interface.


How to test files having unique content?

if diff "$file1" "$file2" > /dev/null; then

How can we get list of files in directory?

files="$( find ${files_dir} -type f )"

We can get any 2 files from that list and check if their names are different and content are same.

# removeDuplicates.sh

if [[ -z "$files_dir" ]]; then
    echo "Error: files dir is undefined"

files="$( find ${files_dir} -type f )"
for file1 in $files; do
    for file2 in $files; do
        # echo "checking $file1 and $file2"
        if [[ "$file1" != "$file2" && -e "$file1" && -e "$file2" ]]; then
            if diff "$file1" "$file2" > /dev/null; then
                echo "$file1 and $file2 are duplicates"
                rm -v "$file2"

For example, we have some dir:

$> ls .tmp -1

So there are only 3 unique files.

Lets run that script:

$> ./removeDuplicates.sh .tmp/
.tmp/text(2) and .tmp/text are duplicates
removed `.tmp/text'
.tmp/all.txt and .tmp/all(2).txt are duplicates
removed `.tmp/all(2).txt'

And we get only 3 files leaved.

$> ls .tmp/ -1

I wrote this tiny script to delete duplicated files


Basically it uses a temporary file (/tmp/list.txt) to create a map of files and their hashes. Later I use that files and the magic of Unix pipes to do the rest.

The script won't delete anything but will print the commands to delete files.

mfilter.sh ./dir | bash

Hope it helps


More concise version of removing duplicated files(just one line)

young@ubuntu-16:~/test$ md5sum `find ./ -type f` | sort -k1 | uniq -w32 -d | xargs rm -fv


#!/usr/bin/env bash
#set -x
#This is small script can find same size of files.

if [[ -z $1 || ! -d $1 ]]
echo "Usage $0 directory_name" ;
 exit $?
echo "current directory is $1"

for i in $(find $dir_name -type f); do
   ls -fl $i
done | awk '{f=""
        if(NF>9)for(i=9;i<=NF;i++)f=f?f" "$i:$i; else f=$9;
        if(a[$5]){ a[$5]=a[$5]"\n"f; b[$5]++;} else a[$5]=f} END{for(x     in b)print a[x] }' | xargs stat -c "%s  %n" #For just list files

find_same_size $1

young@ubuntu-16:~/test$ bash find_same_size.sh tttt/ | awk '{ if($1 !~   /^([[:alpha:]])+/) print $2}' | xargs md5sum | uniq -w32 -d | xargs rm -vf

I found an easier way to perform the same task

for i in `md5sum * | sort -k1 | uniq -w32 -d|awk '{print $2}'`; do
rm -rf $i

This is not what you are asking, but I think someone might find it useful when the checksums are not the same, but the name is similar (with suffix in parentheses). This script removes the files with suffixes as ("digit")

#! /bin/bash
# Warning: globstar excludes hidden directories.
# Turn on recursive globbing (in this script) or exit if the option is not supported:
shopt -s globstar || exit
for f in **
#get only files with parentheses suffix
FILEWITHPAR=$( echo "${f%.*}".$extension | grep -o -P "(.*\([0-9]\)\..*)")
# print file to be possibly deleted
if [ -z "$FILEWITHPAR" ] ;then
echo "$FILEWITHPAR ident"
# identify if a similar file without suffix exists
FILENOPAR=$(echo $FILEWITHPAR | sed -e 's/^\(.*\)([0-9])\(.*\).*/\1\2/')
echo "$FILENOPAR exists?"
if [ -f "$FILENOPAR" ]; then
#delete file with suffix in parentheses
echo ""$FILEWITHPAR" to be deleted"
echo "no"

Most and possibly all of the remaining answers are terribly inefficient by computing the checksum of each and every file in the directory to process.

A potentially orders of magnitude faster approach is to first get the size of each file, which is almost immediate (ls or stat), and then compute and compare the checksums only for the files having a non unique size and only keep one instance of the files sharing both their size and checksum.

Note that even though theoretically hash collisions can occur, there are not enough jpeg files on the entire Internet for a hash collision to reasonably have a chance to happen. Two files sharing both their size and checksum are identical for all intents and purposes.

See: How reliable are SHA1 sum and MD5 sums on very large files?

  • Yes, and you should not take a decision on the checksum only. It denotes a very high probability of content equality, but no more. So, if there is only two files of same size to be compared, it is useless to compute the checksum because in all cases you will need to compare them bit by bit to certify the equality. Jan 14 at 16:19
  • @LaurentSimon Yes, that's what my answer is telling.
    – jlliagre
    Jan 14 at 16:26
  • @jliagre, no, your answer is saying that a checksum is enough to identify duplicates. If you can't tolerate false positives, you have to compare bit by bit, like Laurent pointed out. Feb 27 at 17:12
  • @DominykasMostauskis You are right, I'm telling a checksum is enough to identify duplicates. Comparing bit by bit can be done but is mostly pointless, unless there are hundred of billions of files in that directory. Answer updated.
    – jlliagre
    Feb 28 at 0:01
  • @LaurentSimon Certifying the files are identical by comparing them bit by bit can of course be done but is essentially pointless. An accidental hash collision has certainly much less chance to occur than an accidental bit corruption anyway.
    – jlliagre
    Feb 28 at 12:46

I found a small program that really simplifies this kind of tasks: fdupes.

  • Please add installation instructions and a usage example appropriate for the question.
    – simlev
    Sep 21 '18 at 8:05

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