How can I display the files in a unix directory sorted by their human readable size, going from largest to smallest?

I tried

du -h | sort -V -k 1 

but it does not seem to work.

  • Can you please clarify if you are expecting the subdirectory sizes to appear in the output too, and also if you are looking for the apparent size of the files or the actual size they use on disk ?
    – jlliagre
    Commented Dec 17, 2011 at 13:42

7 Answers 7


ls(1) /sort:

-S     sort by file size
  • 2
    -S is no longer a valid sort argument at least on ubuntu. The below answer by @alex worked for me. The answer link is superuser.com/a/990437/528836.
    – Prasanna
    Commented Dec 21, 2017 at 5:36
$ ls -lhS

-l     use a long listing format
-h     with -l, print sizes in human readable format (e.g., 1K 234M 2G)
-S     sort by file size

If you have the appropriate sort version you may simply use:

du -h | sort -rh

mine is

$ sort --version
sort (GNU coreutils) 8.12

du -ha | sort -h

du : estimate file disk usage.

-h : for human
-a : all files

sort : sort lines of text.

-h : for human

man du; man sort for more. It works for me on ubuntu v15.


ls -S wasn't an option on the OS for me. The following worked:
ls -l | sort -k 5nr
They "key" was to specify the column to sort (get it, the "key"). Above I'm specifying -k 5nr meaning sort on 5th column which is size (5) evaluated as a number (n) in descending order (n)

Reference sort documentation for more information

  • For a more human readable output, I recommand ls -lh | sort -k 5hr.
    – Lalylulelo
    Commented Dec 18, 2020 at 10:10

I got this to work for me:

ls -l | sort -g -k 5 -r

Which (I just figured-out) is the same as:

ls -lS

Unlike ls -S, this will properly handle sparse files:

ls -lsh | sort -n | sed 's/^[0-9 ]* //'

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