I'm basically looking for files then sorting by the size. The script works if I don't sort the size by human readable. But I want the size to be human readable. How can I sort sizes that are human readable?

For example:

 ls -l | sort -k 5 -n | awk '{print $9 " " $5}'

This works as expected, I got the size of my files in bytes ascending:

1.txt 1
test.txt 3
bash.sh* 573
DocGeneration.txt 1131
andres_stuff.txt 1465
Branches.xlsx 15087
foo 23735
bar 60566
2016_stuff.pdf 996850

Now, I want the size to be human readable, so I added an -h parameter to ls, and now some files are out of order:

 ls -lh | sort -k 5 -n | awk '{print $9 " " $5}'
1.txt 1
DocGeneration.txt 1.2K
andres_stuff.txt 1.5K
test.txt 3
Branches.xlsx 15K
foo 24K
bar 60K
bash.sh* 573
2016_stuff.pdf 974K

  • -k 5 — how does that work? Jun 13, 2019 at 19:27
  • @ctrl-alt-delor: I believe the size is in the 5th column of the ls output
    – jesse_b
    Jun 13, 2019 at 19:30
  • 2
    Using du instead of ls could be a good idea.
    – xenoid
    Jun 13, 2019 at 19:34
  • ... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes). Jun 13, 2019 at 19:35
  • @Jesse_b my error, I just assumed that the data in the question (marked as this is what I got) was the sorted input.I was wrong. Jun 13, 2019 at 22:10

5 Answers 5


Try sort -h k2

-h, --human-numeric-sort compare human readable numbers (e.g., 2K 1G)

It is part of gnu sort, BSD sort, and others.

  • 5
    Shouldn't parsing the output of ls be avoided?
    – user147505
    Jun 13, 2019 at 19:26
  • 3
    @Tomasz Not always. If it provides the output you need, piping it to another formatting operation is not particularly dangerous. What you should not do is loop over the output of ls, and instead use file globbing directly. Globbing alone won't work here. That said, I would probably prefer du for this.
    – Bloodgain
    Jun 14, 2019 at 22:33
  • 1
    @Bloodgain the ls format is not guaranteed to be the same across systems/ls binaries, so parsing it portably is considered impossible. Jun 17, 2019 at 3:00
  • 1
    Also, filenames with whitespace will mangle things Jun 17, 2019 at 3:06
  • 1
    @Bloodgain : files=(); for f in *; do [[ -L "$f" ]] && files+=("$f"); done; echo ${#files[@]} (I might have the is a symlink test switch wrong). If you don’t care about symlinks, files=(*); echo ${#files[@]}, which becomes portable if you use set and not arrays. Jun 21, 2019 at 14:15

ls has this functionality built in, use the -S option and sort in reverse order: ls -lShr

       -r, --reverse
              reverse order while sorting

       -S     sort by file size, largest first
  • 1
    -h is not a standard ls option, but must be usable if OP already has it. The rest are standard, and it's certainly the answer I would have written. Jun 14, 2019 at 10:44
  • 6
    +1 Don't mess around parsing the output of ls. Jun 14, 2019 at 10:59
  • This is the best answer, but it should include the info in @Toby's comment: -S might not be available for your ls. FWIW, -S is supported even with Emacs's libraryls-lisp.el, which is used when the OS has no ls. It works in Emacs on MS Windows, for example.
    – Drew
    Jun 14, 2019 at 16:37
  • This should be the accepted answer.
    – scatter
    Jun 14, 2019 at 17:32
  • 1
    @Drew: Toby's comment says that -h may not be universally available, but OP is already using it anyway. -S really should be universally available, because it's in the POSIX link that Toby provides. However, quite a few non-POSIX toolkits do exist out there.
    – Kevin
    Jun 16, 2019 at 18:58

Since no specific shell was mentioned, here's how to do the whole thing in the zsh shell:

ls -lhf **/*(.Lk-1024oL)

The ** glob pattern matches like * but across / in pathnames, i.e. like a recursive search would do.

The ls command would enable human readable sizes with -h, and long list output format with -l. The -f option disables sorting, so ls would just list the files in the order they are given.

This order is arranged by the **/*(.Lk-1024oL) filename globbing pattern so that the smaller files are listed first. The **/* bit matches every file and directory in this directory and below, but the (...) modifies the glob's behaviour (it's a "glob qualifier").

It's the oL at the end that orders (o) the names by file size (L, "length").

The . at the start makes the glob only match regular files (no directories).

The Lk-1024 bit selects files whose size is less than 1024 KB ("length in KB less than 1024").

If zsh is not your primary interactive shell, then you could use

zsh -c 'ls -lf **/*(.Lk-1024oL)'

Use setopt GLOB_DOTS (or zsh -o GLOB_DOTS -c ...) to also match hidden names. ... or just add D to the glob qualifier string.

Expanding on the above, assuming that you'd want a 2-column output with pathnames and human readable sizes, and also assuming that you have numfmt from GNU coreutils,

zmodload -F zsh/stat b:zstat

for pathname in **/*(.Lk-1024oL); do
    printf '%s\t%s\n' "$pathname" "$(zstat +size "$pathname" | numfmt --to=iec)"

or, quicker,

paste <( printf '%s\n' **/*(.Lk-1024oL) ) \
      <( zstat -N +size **/*(.Lk-1024oL) | numfmt --to=iec )

If your sort does not have the -h option you could use an (albeit very long) awk command like the following:

find . -type f -size -1024k -exec ls -al {} \; | sort -k 5 -n | awk '{if ($5 > 1099511627776) {print $9,$5/1024/1024/1024/1024"T"} else if ($5 > 1073741824) {print $9,$5/1024/1024/1024"G"} else if ($5 > 1048576) {print $9,$5/1024/1024"M"} else if ($5 > 1024) {print $9,$5/1024"K"} else {print $9,$5"B"}}' | column -t

This will sort your output in bytes and then convert them to their human readable size afterward.


Would this work?

ls -l | awk '{if ($5<=1024) {print}}' | sort -k 5 -n | awk '{print $9"\t"substr($5/1024,1,3)"k"} '| column -t

The first awk exp will look for the files lesser than 1M and the second one will take the byte size from the result and convert it to the KB and prints the first 3 elements to give a human-readable size.

  • That does not really solve OPs question - it only looks in the current directory and will only print regular files. Also will compare against 1Kb instead of 1MB. Finally we are after answers with some explanation about why the code works.
    – grochmal
    Jun 13, 2019 at 21:53
  • My bad added what it does.
    – Vignesh SP
    Jun 13, 2019 at 22:16

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