I can order a list of files / directories by size:
ls -lS But if I am using du in human readable format:
du --max-depth=1 -h . I get:
128K ./something 3,3M ./more 3,2M ./even-more ... Which is not ordered. Is there any standard tool to order this kind of data? Standard sort does not seem to support this. Do I need to roll my own?
GNU sort has a -h/--human-numeric-sort option and h sort key flag to handle those (it expects 1024-based units (1023 sorts before 1K) which happens to be how GNU du counts as well.
Now note that some precision is lost when you use du -h, so the order may end-up being wrong:
$ du -k a b 1212 a 1208 b $ du -h a b | sort -h 1.2M a 1.2M b As mentioned by @StephenKitt, you can work around it by telling du to give you the full precision and only convert to human format after sorting using for instance GNU numfmt:
$ du --block-size=1 a b | sort -n | numfmt --to=iec 1.2M b 1.2M a (beware that spacing is affected). All of the above assume file names don't contain newline characters.
As for the generic question about ordering by size, zsh globs have a oL glob qualifier for that (note that it's by size, not disk usage).
ls -S could be done (with GNU ls for its -U for unsorted):
ls -ldU -- *(oL) For sorting by size after symlink resolution:
ls -LldU -- *(-oL) wc -c -- *(-oL) 
The GNU variants of ls and sort understand -h (mnemonic: -human). You can either use ls -lSh, ls -sSh or du -h … | sort -h:
du --max-depth=1 -h . | sort -h Keep in mind that ls won't report the size for directory contents.
ls does report the correct size for directories... But a directory’s size isn’t the size of the files it contains.
sorthassort -h(note that your--max-depth=1is GNU-specific already)-hisn’t a standard option fordueither ;-).