Sorting files according to size recursively

Question

I need to find the largest files in a folder.
How do I scan a folder recursively and sort the contents by size?

I have tried using ls -R -S, but this lists the directories as well.
I also tried using find.

Answer 1

You can also do this with just du. Just to be on the safe side I'm using this version of du:

$ du --version
du (GNU coreutils) 8.5

The approach:

$ du -ah <some DIR> | grep -v "/$" | sort -rh

Breakdown of approach

The command du -ah DIR will produce a list of all the files and directories in a given directory DIR. The -h will produce human readable sizes which I prefer. If you don't want them then drop that switch. I'm using the head -6 just to limit the amount of output!

$ du -ah ~/Downloads/ | head -6
4.4M    /home/saml/Downloads/kodak_W820_wireless_frame/W820_W1020_WirelessFrames_exUG_GLB_en.pdf
624K    /home/saml/Downloads/kodak_W820_wireless_frame/easyshare_w820.pdf
4.9M    /home/saml/Downloads/kodak_W820_wireless_frame/W820_W1020WirelessFrameExUG_GLB_en.pdf
9.8M    /home/saml/Downloads/kodak_W820_wireless_frame
8.0K    /home/saml/Downloads/bugs.xls
604K    /home/saml/Downloads/netgear_gs724t/GS7xxT_HIG_5Jan10.pdf

Easy enough to sort it smallest to biggest:

$ du -ah ~/Downloads/ | sort -h | head -6
0   /home/saml/Downloads/apps_archive/monitoring/nagios/nagios-check_sip-1.3/usr/lib64/nagios/plugins/check_ldaps
0   /home/saml/Downloads/data/elasticsearch/nodes/0/indices/logstash-2013.04.06/0/index/write.lock
0   /home/saml/Downloads/data/elasticsearch/nodes/0/indices/logstash-2013.04.06/0/translog/translog-1365292480753
0   /home/saml/Downloads/data/elasticsearch/nodes/0/indices/logstash-2013.04.06/1/index/write.lock
0   /home/saml/Downloads/data/elasticsearch/nodes/0/indices/logstash-2013.04.06/1/translog/translog-1365292480946
0   /home/saml/Downloads/data/elasticsearch/nodes/0/indices/logstash-2013.04.06/2/index/write.lock

Reverse it, biggest to smallest:

$ du -ah ~/Downloads/ | sort -rh | head -6
10G /home/saml/Downloads/
3.8G    /home/saml/Downloads/audible/audio_books
3.8G    /home/saml/Downloads/audible
2.3G    /home/saml/Downloads/apps_archive
1.5G    /home/saml/Downloads/digital_blasphemy/db1440ppng.zip
1.5G    /home/saml/Downloads/digital_blasphemy

Don't show me the directory, just the files:

$ du -ah ~/Downloads/ | grep -v "/$" | sort -rh | head -6 
3.8G    /home/saml/Downloads/audible/audio_books
3.8G    /home/saml/Downloads/audible
2.3G    /home/saml/Downloads/apps_archive
1.5G    /home/saml/Downloads/digital_blasphemy/db1440ppng.zip
1.5G    /home/saml/Downloads/digital_blasphemy
835M    /home/saml/Downloads/apps_archive/cad_cam_cae/salome/Salome-V6_5_0-LGPL-x86_64.run

If you want to exclude all directories from the output, you can use a trick with the presence of a dot character. This assumes that your directory names do not contain dots, and that the files you are looking for do. Then you can filter out the directories with grep -v '\s/[^.]*$':

$ du -ah ~/Downloads/ | grep -v '\s/[^.]*$' | sort -rh | head -2
1.5G    /home/saml/Downloads/digital_blasphemy/db1440ppng.zip
835M    /home/saml/Downloads/apps_archive/cad_cam_cae/salome/Salome-V6_5_0-LGPL-x86_64.run

If you just want the list of smallest to biggest, but the top 6 offending files you can reverse the sort switch, drop (-r), and use tail -6 instead of the head -6.

$ du -ah ~/Downloads/ | grep -v "/$" | sort -h | tail -6
835M    /home/saml/Downloads/apps_archive/cad_cam_cae/salome/Salome-V6_5_0-LGPL-x86_64.run
1.5G    /home/saml/Downloads/digital_blasphemy
1.5G    /home/saml/Downloads/digital_blasphemy/db1440ppng.zip
2.3G    /home/saml/Downloads/apps_archive
3.8G    /home/saml/Downloads/audible
3.8G    /home/saml/Downloads/audible/audio_books

Answer 2

If you want to find all files in the current directory and its sub directories and list them according to their size (without considering their path), and assuming none of the file names contain newline characters, with GNU find, you can do this:

find . -type f -printf "%s\t%p\n" | sort -n

From man find on a GNU system:

   -printf format
          True; print format  on  the  standard  output,
          interpreting  `\'  escapes and `%' directives.
          Field widths and precisions can  be  specified
          as  with the `printf' C function.  Please note
          that many of the  fields  are  printed  as  %s
          rather  than  %d, and this may mean that flags
          don't work as you  might  expect.   This  also
          means  that  the `-' flag does work (it forces
          fields to be  left-aligned).   Unlike  -print,
          -printf  does  not add a newline at the end of
          the string.  The escapes and directives are:

          %p     File's name.
          %s     File's size in bytes.

From man sort:

   -n, --numeric-sort
          compare according to string numerical value

Answer 3

Try the following command:

ls -1Rhs | sed -e "s/^ *//" | grep "^[0-9]" | sort -hr | head -n20

It'll list top-20 biggest files in the current directory recursively.

Note: The option -h for sort is not available on OSX/BSD, so you've to install sort from coreutils (e.g. via brew) and apply the local bin path to PATH, e.g.

export PATH="/usr/local/opt/coreutils/libexec/gnubin:$PATH" # Add a "gnubin" for coreutils.

Alternatively use:

ls -1Rs | sed -e "s/^ *//" | grep "^[0-9]" | sort -nr | head -n20

---

For the biggest directories use du, e.g.:

du -ah . | sort -rh | head -20

or:

du -a . | sort -rn | head -20

Answer 4

This will find all files recursively, and sort them by size. It prints out all file sizes in kb, and rounds down so you may see 0 KB files, but it was close enough for my uses, and works on OSX.

find . -type f -print0 | xargs -0 ls -la | awk '{print int($5/1000) " KB\t" $9}' | sort -n -r -k1

Answer 5

Simple solution for Mac/Linux which skips directories:

find . -type f -exec du -h {} \; | sort -h

Answer 6

With zsh, you'd find the largest file (in terms of apparent size like the size column in ls -l output, not disk usage) with:

ls -ld -- **/*(DOL[1])

For the 6 largest ones:

ls -ld -- **/*(DOL[1,6])

To sort those by file size, you can use ls's -S option. Some ls implementations also have a -U option for ls not to sort the list (as it's already sorted by size by zsh here).

Answer 7

The equivalent in BSD or OSX is

$ du -ah simpl | sort -dr | head -6

Answer 8

This is an incredibly commmon need for a variety of reasons (I like finding the most recent backup in a directory), and is a surprisingly simple task.

I'm going to provide a Linux solution that uses the find, xargs, stat, tail, awk, and sort utilities.

Most people have provided some unique answers, but I prefer mine because it properly handles filenames, and the use case can easily be changed (modify stat, and sort arguments)

I'll also provide a Python solution that should let you use this functionality even on Windows

Linux command line solution

Recursively return entire list of only files from a directory, sorted by file size

find . -type f -print0 | xargs -0 -I{} stat -c '%s %n' {} | sort -n

Same as before, but this time, return the largest file.

# Each utility is split on a new line to help 
# visualize the concept of transforming our data in a stream
find . -type f -print0 | 
xargs -0 -I{} stat -c '%s %n' {} | 
sort -n | 
tail -n 1 |
awk '{print $2}'

Same exact Pattern, but now select the newest file instead of largest

# (Notice only the first argument of stat changed for new functionality!)
find . -type f -print0 | xargs -0 -I{} stat -c '%Y %n' {} | 
sort -n | tail -n 1 | awk '{print $2}'

Explanation:

1. find: Recursively finds all files from current directory, and prints them out with a null character
2. xargs: utility to execute commands using arguments provided from standard input. For every line of output, we want to run the stat utility on that file
3. stat: Stat is an all around awesome command that has so many use cases. I am printing out two columns, the first column being the block size (%s), and the second column being the file name (%n)
4. sort: Sort the results with the numeric switch. Since the first argument is an integer, our results will be sorted properly
5. tail: Only select the last line of output (since the list is sorted, this is the largest file!)
6. awk: Select the second column, which contains the filename, and is the largest file in a recursive directory.

Python solution

#!/usr/bin/env python
import os, sys
files = list()
for dirpath, dirname, filenames in os.walk(sys.argv[1]):
    for filename in filenames:
        realpath = os.path.join(dirpath, filename)
        files.append(realpath)
files_sorted_by_size = sorted(files, key = lambda x: os.stat(x).st_size)
largest_file = files_sorted_by_size[-1]
print(largest_file)

This script takes a little big longer to explain, but essentially if you save that as a script, it will search through the first argument provided on the command line, and return the largest file in that directory. The script does no error checking, but it should give you an idea of how to approach this in Python, which gives you a nice platform independent way of solving this problem.

Answer 9

Try below command with sort option to have folders with size in ascending order

du -sh * | sort -sh

Answer 10

Variant of this answer from a similar question

find . -type f -exec du -ah {} + | sort -rh | more

Answer 11

find -type f -printf '%s %p\n' | numfmt --to=iec | sort -hr | head

• find -type f finds all files under current directory, recursively
• -printf '%s %p\n': for each file it prints file size in bytes and file name, separated by space, with newline
• numfmt --to=iec formats the first field (file size) in human readable format (with K, M, G suffixes) and keep the file name unchanged
• sort -hr sorts in reverse numerically order all lines, understanding human readable suffixes
• head prints only the first 10 lines

Answer 12

Something that works on any platform except AIX and HP-UX is:

find . -ls | sort +6 | tail