How to sort human readable sizeA standard tool to convert a byte-count into human KiB MiB etc; like du, ls1Human-readable ls output under AIX?How to display “human-readable” file sizes in find results?Find, count and sort all audio files. ALAC (M4A) filesSort command inconsistent behaviorSort output of awk except for first line?Find human-readable filesSort function doesn't workFind and sort by file sizeHow do I get rid of human readable format for ls?

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How to sort human readable size


A standard tool to convert a byte-count into human KiB MiB etc; like du, ls1Human-readable ls output under AIX?How to display “human-readable” file sizes in find results?Find, count and sort all audio files. ALAC (M4A) filesSort command inconsistent behaviorSort output of awk except for first line?Find human-readable filesSort function doesn't workFind and sort by file sizeHow do I get rid of human readable format for ls?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;









11

















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










share|improve this question




























  • -k 5 — how does that work?

    – ctrl-alt-delor
    Jun 13 at 19:27











  • @ctrl-alt-delor: I believe the size is in the 5th column of the ls output

    – Jesse_b
    Jun 13 at 19:30






  • 2





    Using du instead of ls could be a good idea.

    – xenoid
    Jun 13 at 19:34











  • ... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes).

    – Stephen Kitt
    Jun 13 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.

    – ctrl-alt-delor
    Jun 13 at 22:10

















11

















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










share|improve this question




























  • -k 5 — how does that work?

    – ctrl-alt-delor
    Jun 13 at 19:27











  • @ctrl-alt-delor: I believe the size is in the 5th column of the ls output

    – Jesse_b
    Jun 13 at 19:30






  • 2





    Using du instead of ls could be a good idea.

    – xenoid
    Jun 13 at 19:34











  • ... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes).

    – Stephen Kitt
    Jun 13 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.

    – ctrl-alt-delor
    Jun 13 at 22:10













11












11








11








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










share|improve this question

















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







find ls sort






share|improve this question
















share|improve this question













share|improve this question




share|improve this question








edited Jun 14 at 2:23







tvo000

















asked Jun 13 at 19:13









tvo000tvo000

587 bronze badges




587 bronze badges















  • -k 5 — how does that work?

    – ctrl-alt-delor
    Jun 13 at 19:27











  • @ctrl-alt-delor: I believe the size is in the 5th column of the ls output

    – Jesse_b
    Jun 13 at 19:30






  • 2





    Using du instead of ls could be a good idea.

    – xenoid
    Jun 13 at 19:34











  • ... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes).

    – Stephen Kitt
    Jun 13 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.

    – ctrl-alt-delor
    Jun 13 at 22:10

















  • -k 5 — how does that work?

    – ctrl-alt-delor
    Jun 13 at 19:27











  • @ctrl-alt-delor: I believe the size is in the 5th column of the ls output

    – Jesse_b
    Jun 13 at 19:30






  • 2





    Using du instead of ls could be a good idea.

    – xenoid
    Jun 13 at 19:34











  • ... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes).

    – Stephen Kitt
    Jun 13 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.

    – ctrl-alt-delor
    Jun 13 at 22:10
















-k 5 — how does that work?

– ctrl-alt-delor
Jun 13 at 19:27





-k 5 — how does that work?

– ctrl-alt-delor
Jun 13 at 19:27













@ctrl-alt-delor: I believe the size is in the 5th column of the ls output

– Jesse_b
Jun 13 at 19:30





@ctrl-alt-delor: I believe the size is in the 5th column of the ls output

– Jesse_b
Jun 13 at 19:30




2




2





Using du instead of ls could be a good idea.

– xenoid
Jun 13 at 19:34





Using du instead of ls could be a good idea.

– xenoid
Jun 13 at 19:34













... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes).

– Stephen Kitt
Jun 13 at 19:35





... or find’s -printf with its %p and %s formatters (followed by a “humanisation” of the sizes).

– Stephen Kitt
Jun 13 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.

– ctrl-alt-delor
Jun 13 at 22:10





@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.

– ctrl-alt-delor
Jun 13 at 22:10










5 Answers
5






active

oldest

votes


















25


















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.






share|improve this answer























  • 5





    Shouldn't parsing the output of ls be avoided?

    – Tomasz
    Jun 13 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 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.

    – D. Ben Knoble
    Jun 17 at 3:00







  • 1





    Also, filenames with whitespace will mangle things

    – D. Ben Knoble
    Jun 17 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.

    – D. Ben Knoble
    Jun 21 at 14:15


















29


















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





share|improve this answer





















  • 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.

    – Toby Speight
    Jun 14 at 10:44






  • 5





    +1 Don't mess around parsing the output of ls.

    – David Richerby
    Jun 14 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 at 16:37












  • This should be the accepted answer.

    – scatter
    Jun 14 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 at 18:58


















5


















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 '%st%sn' "$pathname" "$(zstat +size "$pathname" | numfmt --to=iec)"
done


or, quicker,



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





share|improve this answer



































    4


















    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.






    share|improve this answer



































      -1


















      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.






      share|improve this answer




























      • 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 at 21:53











      • My bad added what it does.

        – Vignesh SP
        Jun 13 at 22:16












      Your Answer








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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      25


















      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.






      share|improve this answer























      • 5





        Shouldn't parsing the output of ls be avoided?

        – Tomasz
        Jun 13 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 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.

        – D. Ben Knoble
        Jun 17 at 3:00







      • 1





        Also, filenames with whitespace will mangle things

        – D. Ben Knoble
        Jun 17 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.

        – D. Ben Knoble
        Jun 21 at 14:15















      25


















      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.






      share|improve this answer























      • 5





        Shouldn't parsing the output of ls be avoided?

        – Tomasz
        Jun 13 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 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.

        – D. Ben Knoble
        Jun 17 at 3:00







      • 1





        Also, filenames with whitespace will mangle things

        – D. Ben Knoble
        Jun 17 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.

        – D. Ben Knoble
        Jun 21 at 14:15













      25














      25










      25









      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.






      share|improve this answer
















      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.







      share|improve this answer















      share|improve this answer




      share|improve this answer








      edited Jun 13 at 19:24

























      answered Jun 13 at 19:23









      ctrl-alt-delorctrl-alt-delor

      14.8k6 gold badges34 silver badges65 bronze badges




      14.8k6 gold badges34 silver badges65 bronze badges










      • 5





        Shouldn't parsing the output of ls be avoided?

        – Tomasz
        Jun 13 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 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.

        – D. Ben Knoble
        Jun 17 at 3:00







      • 1





        Also, filenames with whitespace will mangle things

        – D. Ben Knoble
        Jun 17 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.

        – D. Ben Knoble
        Jun 21 at 14:15












      • 5





        Shouldn't parsing the output of ls be avoided?

        – Tomasz
        Jun 13 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 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.

        – D. Ben Knoble
        Jun 17 at 3:00







      • 1





        Also, filenames with whitespace will mangle things

        – D. Ben Knoble
        Jun 17 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.

        – D. Ben Knoble
        Jun 21 at 14:15







      5




      5





      Shouldn't parsing the output of ls be avoided?

      – Tomasz
      Jun 13 at 19:26





      Shouldn't parsing the output of ls be avoided?

      – Tomasz
      Jun 13 at 19:26




      3




      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 at 22:33






      @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 at 22:33





      1




      1





      @Bloodgain the ls format is not guaranteed to be the same across systems/ls binaries, so parsing it portably is considered impossible.

      – D. Ben Knoble
      Jun 17 at 3:00






      @Bloodgain the ls format is not guaranteed to be the same across systems/ls binaries, so parsing it portably is considered impossible.

      – D. Ben Knoble
      Jun 17 at 3:00





      1




      1





      Also, filenames with whitespace will mangle things

      – D. Ben Knoble
      Jun 17 at 3:06





      Also, filenames with whitespace will mangle things

      – D. Ben Knoble
      Jun 17 at 3:06




      1




      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.

      – D. Ben Knoble
      Jun 21 at 14:15





      @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.

      – D. Ben Knoble
      Jun 21 at 14:15













      29


















      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





      share|improve this answer





















      • 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.

        – Toby Speight
        Jun 14 at 10:44






      • 5





        +1 Don't mess around parsing the output of ls.

        – David Richerby
        Jun 14 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 at 16:37












      • This should be the accepted answer.

        – scatter
        Jun 14 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 at 18:58















      29


















      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





      share|improve this answer





















      • 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.

        – Toby Speight
        Jun 14 at 10:44






      • 5





        +1 Don't mess around parsing the output of ls.

        – David Richerby
        Jun 14 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 at 16:37












      • This should be the accepted answer.

        – scatter
        Jun 14 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 at 18:58













      29














      29










      29









      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





      share|improve this answer














      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






      share|improve this answer













      share|improve this answer




      share|improve this answer










      answered Jun 14 at 3:46









      Mark McKinstryMark McKinstry

      10.1k3 gold badges25 silver badges24 bronze badges




      10.1k3 gold badges25 silver badges24 bronze badges










      • 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.

        – Toby Speight
        Jun 14 at 10:44






      • 5





        +1 Don't mess around parsing the output of ls.

        – David Richerby
        Jun 14 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 at 16:37












      • This should be the accepted answer.

        – scatter
        Jun 14 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 at 18:58












      • 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.

        – Toby Speight
        Jun 14 at 10:44






      • 5





        +1 Don't mess around parsing the output of ls.

        – David Richerby
        Jun 14 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 at 16:37












      • This should be the accepted answer.

        – scatter
        Jun 14 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 at 18:58







      1




      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.

      – Toby Speight
      Jun 14 at 10:44





      -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.

      – Toby Speight
      Jun 14 at 10:44




      5




      5





      +1 Don't mess around parsing the output of ls.

      – David Richerby
      Jun 14 at 10:59






      +1 Don't mess around parsing the output of ls.

      – David Richerby
      Jun 14 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 at 16:37






      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 at 16:37














      This should be the accepted answer.

      – scatter
      Jun 14 at 17:32





      This should be the accepted answer.

      – scatter
      Jun 14 at 17:32




      1




      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 at 18:58





      @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 at 18:58











      5


















      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 '%st%sn' "$pathname" "$(zstat +size "$pathname" | numfmt --to=iec)"
      done


      or, quicker,



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





      share|improve this answer
































        5


















        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 '%st%sn' "$pathname" "$(zstat +size "$pathname" | numfmt --to=iec)"
        done


        or, quicker,



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





        share|improve this answer






























          5














          5










          5









          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 '%st%sn' "$pathname" "$(zstat +size "$pathname" | numfmt --to=iec)"
          done


          or, quicker,



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





          share|improve this answer
















          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 '%st%sn' "$pathname" "$(zstat +size "$pathname" | numfmt --to=iec)"
          done


          or, quicker,



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






          share|improve this answer















          share|improve this answer




          share|improve this answer








          edited Jun 14 at 6:41

























          answered Jun 13 at 20:44









          KusalanandaKusalananda

          170k20 gold badges328 silver badges531 bronze badges




          170k20 gold badges328 silver badges531 bronze badges
























              4


















              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.






              share|improve this answer
































                4


















                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.






                share|improve this answer






























                  4














                  4










                  4









                  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.






                  share|improve this answer
















                  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.







                  share|improve this answer















                  share|improve this answer




                  share|improve this answer








                  edited Jun 13 at 22:18

























                  answered Jun 13 at 19:32









                  Jesse_bJesse_b

                  19.9k3 gold badges47 silver badges90 bronze badges




                  19.9k3 gold badges47 silver badges90 bronze badges
























                      -1


















                      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.






                      share|improve this answer




























                      • 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 at 21:53











                      • My bad added what it does.

                        – Vignesh SP
                        Jun 13 at 22:16















                      -1


















                      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.






                      share|improve this answer




























                      • 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 at 21:53











                      • My bad added what it does.

                        – Vignesh SP
                        Jun 13 at 22:16













                      -1














                      -1










                      -1









                      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.






                      share|improve this answer
















                      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.







                      share|improve this answer















                      share|improve this answer




                      share|improve this answer








                      edited Jun 13 at 22:15

























                      answered Jun 13 at 20:29









                      Vignesh SPVignesh SP

                      2021 silver badge8 bronze badges




                      2021 silver badge8 bronze badges















                      • 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 at 21:53











                      • My bad added what it does.

                        – Vignesh SP
                        Jun 13 at 22:16

















                      • 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 at 21:53











                      • My bad added what it does.

                        – Vignesh SP
                        Jun 13 at 22:16
















                      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 at 21:53





                      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 at 21:53













                      My bad added what it does.

                      – Vignesh SP
                      Jun 13 at 22:16





                      My bad added what it does.

                      – Vignesh SP
                      Jun 13 at 22:16


















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