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I have a directory which has files with random names. What I want to do is rename the files with file1, file2 and so on.The lexicographically smaller file name should be numbered with smaller number. How can I do that?

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Also look at pyrenamer. Depending on your distro, it may be in your standard repositories. – Paddy Landau Jul 1 '13 at 11:23
up vote 7 down vote accepted

A simple bash script should do it.

#!/usr/bin/env bash

for i in *; do
    mv "${i}" file${count}.`echo "${i}" | awk -F. '{print $2}'`

To help with a line-by-line explanation:

  1. Location of the bash shell this is executing under, determined by your environmental variable.
  2. Set the variable "count" to 0
  3. Creates a for-loop, iterating over the output of the command ls (sorted alphabetically by default).
  4. Moves (i.e. renames) the file "i" (current one in the loop) to file#.ext. The "#" is the current number in "count" and the stuff past the "." is a quick way to get the current file extension (so this could work in a folder of various file extensions).
  5. Increment the counter variable

loop, loop, loop

  1. done!


  1. This executes in whatever directory you are executing it from. So as a script, you'd want to add a command line argument. Instead I would open a terminal, navigate to the directory to do this on, and execute the following (exactly):

    for i in *; do mv "${i}" "file${count}".`echo "${i}" | 
      awk -F. '{print $2}'`; ((++count)); done
  2. This is assuming that you only have file names such as "file.txt" not ""

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Two things: Use * instead of `ls`, and quote ${i}: "${i}". – Christoffer Hammarström Jul 1 '13 at 7:01
As @ChristofferHammarström says, use * not ls, because your code will handle files with spaces incorrectly, and because the output from ls is not guaranteed always to be the same on different systems. Also quote your use of the variables, as "${i}" and "${2}", because multiple, leading and trailing spaces will be affected. A minor point: In the calculation, you can omit the parameter-substitution, as count=$(( count + 1 )) or, even better, just use (( ++count )) by itself. – Paddy Landau Jul 1 '13 at 11:19
@PaddyLandau - Thanks for the tips (and explanation)! Updated. – nerdwaller Jul 1 '13 at 13:03
Your script assumes all files have extensions which is not necessarily true. – terdon Jul 1 '13 at 13:55
@terdon - Good point, in which case the OP can simply drop everything between ${count} and the following ; (exclusive). – nerdwaller Jul 1 '13 at 14:28

You can use this python script:

import os
#enter the path to the folder: relative or absolute
direc = raw_input('Enter the path to directory: ')
files = os.listdir('.')
def func(x):
    return map(str.lower,os.path.splitext(x))
files.sort(key = func)
for i,x in enumerate(files, 1):
    name, ext = os.path.splitext(x)
    os.rename(x, 'file{}{}'.format(i,ext))
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I would suggest making your script respect extensions (an easy update), I'll +1 when you do that since otherwise their files may be like "wtf mate?" – nerdwaller Jul 1 '13 at 4:18
@nerdwaller solution updated. – Ashwini Chaudhary Jul 1 '13 at 4:54

A possibility (similar to nerdwaller's solution, but without subshells and forks to ):


shopt -s nullglob

for i in *; do
    mv -nv -- "$i" "file$((++count))"

Switches for mv:

  • -n for no clobber: won't overwrite an otherwise existing file. Remove it if you actually want to overwrite files (might not be available depending on your version of mv),
  • -v for verbose: tell what it's doing. Highly optional (but I like to have these lines on my terminal to impress my colleagues),
  • -- for end of options: without this, and if a filename starts with a hyphen, that would confuse mv as it would try to interpret it as an option. Might not be available depending on your version of mv. Use it if available!

I'm also using the shell optional behavior nullglob so that if there are no files, then you won't get any errors (the loop will not get executed as * would expand to nothing in this case).

If you need to handle extensions too (as in nerdwaller's version), you don't need , it can all be done in :


shopt -s nullglob

for i in *; do
    mv -nv -- "$i" "file$((++count)).${i##*.}"

Observe that the numbering will not have any leading zeros, so that you'll get files named so:


and this might screw up the order of the files in listings. If you need leading zeros too:


shopt -s nullglob

files=( * )

for ((n=${#files[@]};n;n/=10)); do ((++lz)); done

for i in "${files[@]}"; do
    printf -v n "%0${lz}d" $((++count))
    mv -nv -- "$i" "file$n"

And if you want it to keep the original file's extension, replace the line

mv -nv -- "$i" "file$n"


mv -nv -- "$i" "file$n.${i##*.}"


  • Because this uses globbing, it might be quite slow if you have a huge number of files in your directory (globbing can take some time).
  • This will rename everything, including directories, but not hidden files.


  • I don't have any solutions regarding speed in case of huge number of files in .
  • If you need to discard directories from this renaming procedure, and only rename files, add [[ -f $i ]] || continue just after the for i in *; do statement.
  • If you need to also rename hidden files, add shopt -s dotglob just after the shopt -s nullglob statement.

If you read down to here, I guess you'll be able to craft something that matches your very own need from the examples I gave you, hopefully using good practice (provided you choose to solve your problem using of course).

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I wanted to implement the same when I was a student and was studying the Linux Shell Scripts. then I created the below script.. Have a look..

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