How to list the size of each file and directory and sort by descending size in Bash?
I found that there is no easy to get way the size of a directory in Bash? I want that when I type ls —
What exactly do you mean by the «size» of a directory? The number of files under it (recursively or not)? The sum of the sizes of the files under it (recursively or not)? The disk size of the directory itself? (A directory is implemented as a special file containing file names and other information.)
@KeithThompson @KitHo du command estimates file space usage so you cannot use it if you want to get the exact size.
@ztank1013: Depending on what you mean by «the exact size», du (at least the GNU coreutils version) probably has an option to provide the information.
12 Answers 12
Simply navigate to directory and run following command:
OR add -h for human readable sizes and -r to print bigger directories/files first.
du -a -h --max-depth=1 | sort -hr du -h requires sort -h too, to ensure that, say 981M sorts before 1.3G ; with sort -n only the numbers would be taken into account and they’d be the wrong way round.
This doesn’t list the size of the individual files within the current directory, only the size of its subdirectories and the total size of the current directory. How would you include individual files in the output as well (to answer OP’s question)?
@ErikTrautman to list the files also you need to add -a and use —all instead of —max-depth=1 like so du -a -h —all | sort -h
Apparently —max-depth option is not in Mac OS X’s version of the du command. You can use the following instead.
Unfortunately this does not show the files, but only the folder sizes. -a does not work with -d either.
To show files and folders, I combined 2 commands: l -hp | grep -v / && du -h -d 1 , which shows the normal file size from ls for files, but uses du for directories.
(this willnot show hidden (.dotfiles) files)
Use du -sm for Mb units etc. I always use
because the total line ( -c ) will end up at the bottom for obvious reasons 🙂
PS:
- See comments for handling dotfiles
- I frequently use e.g. ‘du -smc /home// | sort -n |tail’ to get a feel of where exactly the large bits are sitting
du —max-depth=1|sort -n or find . -mindepth 1 -maxdepth 1|xargs du -s|sort -n for including dotfiles too.
@arnaud576875 find . -mindepth 1 -maxdepth 1 -print0 | xargs -0 du -s | sort -n if some of the found paths could contain spaces.
This is a great variant to get a human readable view of the biggest: sudo du -smch * | sort -h | tail
Command
Output
3,5M asdf.6000.gz 3,4M asdf.4000.gz 3,2M asdf.2000.gz 2,5M xyz.PT.gz 136K xyz.6000.gz 116K xyz.6000p.gz 88K test.4000.gz 76K test.4000p.gz 44K test.2000.gz 8,0K desc.common.tcl 8,0K wer.2000p.gz 8,0K wer.2000.gz 4,0K ttree.3 Explanation
- du displays «disk usage»
- h is for «human readable» (both, in sort and in du)
- max-depth=0 means du will not show sizes of subfolders (remove that if you want to show all sizes of every file in every sub-, subsub-, . folder)
- r is for «reverse» (biggest file first)
ncdu
When I came to this question, I wanted to clean up my file system. The command line tool ncdu is way better suited to this task.
Just type ncdu [path] in the command line. After a few seconds for analyzing the path, you will see something like this:
$ ncdu 1.11 ~ Use the arrow keys to navigate, press ? for help --- / --------------------------------------------------------- . 96,1 GiB [##########] /home . 17,7 GiB [# ] /usr . 4,5 GiB [ ] /var 1,1 GiB [ ] /lib 732,1 MiB [ ] /opt . 275,6 MiB [ ] /boot 198,0 MiB [ ] /storage . 153,5 MiB [ ] /run . 16,6 MiB [ ] /etc 13,5 MiB [ ] /bin 11,3 MiB [ ] /sbin . 8,8 MiB [ ] /tmp . 2,2 MiB [ ] /dev ! 16,0 KiB [ ] /lost+found 8,0 KiB [ ] /media 8,0 KiB [ ] /snap 4,0 KiB [ ] /lib64 e 4,0 KiB [ ] /srv ! 4,0 KiB [ ] /root e 4,0 KiB [ ] /mnt e 4,0 KiB [ ] /cdrom . 0,0 B [ ] /proc . 0,0 B [ ] /sys @ 0,0 B [ ] initrd.img.old @ 0,0 B [ ] initrd.img @ 0,0 B [ ] vmlinuz.old @ 0,0 B [ ] vmlinuz Delete the currently highlighted element with d , exit with CTRL + c
ls -S sorts by size. Then, to show the size too, ls -lS gives a long ( -l ), sorted by size ( -S ) display. I usually add -h too, to make things easier to read, so, ls -lhS .
Ah, sorry, that was not clear from your post. You want du , seems someone has posted it. @sehe: Depends on your definition of real — it is showing the amount of space the directory is using to store itself. (It’s just not also adding in the size of the subentries.) It’s not a random number, and it’s not always 4KiB.
find . -mindepth 1 -maxdepth 1 -type d | parallel du -s | sort -n I think I might have figured out what you want to do. This will give a sorted list of all the files and all the directories, sorted by file size and size of the content in the directories.
(find . -depth 1 -type f -exec ls -s <> \;; find . -depth 1 -type d -exec du -s <> \;) | sort -n [enhanced version]
This is going to be much faster and precise than the initial version below and will output the sum of all the file size of current directory:
echo `find . -type f -exec stat -c %s <> \; | tr '\n' '+' | sed 's/+$//g'` | bc the stat -c %s command on a file will return its size in bytes. The tr command here is used to overcome xargs command limitations (apparently piping to xargs is splitting results on more lines, breaking the logic of my command). Hence tr is taking care of replacing line feed with + (plus) sign. sed has the only goal to remove the last + sign from the resulting string to avoid complains from the final bc (basic calculator) command that, as usual, does the math.
Performances: I tested it on several directories and over ~150.000 files top (the current number of files of my fedora 15 box) having what I believe it is an amazing result:
# time echo `find / -type f -exec stat -c %s <> \; | tr '\n' '+' | sed 's/+$//g'` | bc 12671767700 real 2m19.164s user 0m2.039s sys 0m14.850s Just in case you want to make a comparison with the du -sb / command, it will output an estimated disk usage in bytes ( -b option)
As I was expecting it is a little larger than my command calculation because the du utility returns allocated space of each file and not the actual consumed space.
[initial version]
You cannot use du command if you need to know the exact sum size of your folder because (as per man page citation) du estimates file space usage. Hence it will lead you to a wrong result, an approximation (maybe close to the sum size but most likely greater than the actual size you are looking for).
I think there might be different ways to answer your question but this is mine:
ls -l $(find . -type f | xargs) | cut -d" " -f5 | xargs | sed 's/\ /+/g'| bc It finds all files under . directory (change . with whatever directory you like), also hidden files are included and (using xargs ) outputs their names in a single line, then produces a detailed list using ls -l . This (sometimes) huge output is piped towards cut command and only the fifth field ( -f5 ), which is the file size in bytes is taken and again piped against xargs which produces again a single line of sizes separated by blanks. Now take place a sed magic which replaces each blank space with a plus ( + ) sign and finally bc (basic calculator) does the math.
It might need additional tuning and you may have ls command complaining about arguments list too long.
How to List All Files Ordered by Size in Linux
In one of our several articles about listing files using the popular ls command, we covered how to list and sort files by last modification time (date and time) in Linux. In this short handy article, we will present a number of useful ls command options to list all of the files in a certain directory and sort them by file size in Linux.
To list all files in a directory, open a terminal window and run the following command. Note that when ls invoked without any arguments, it will list the files in the current working directory.
In the following command the -l flag means long listing and -a tells ls to list all files including (.) or hidden files. To avoid showing the . and .. files, use the -A option instead of -a .
$ ls -la OR $ ls -la /var/www/html/admin_portal/
To list all files and sort them by size, use the -S option. By default, it displays output in descending order (biggest to smallest in size).
$ ls -laS /var/www/html/admin_portal/
You can output the file sizes in human-readable format by adding the -h option as shown.
$ ls -laSh /var/www/html/admin_portal/
And to sort in reverse order, add the -r flag as follows.
$ ls -laShr /var/www/html/admin_portal/
Besides, you can list subdirectories recursively using the -R option.
$ ls -laShR /var/www/html/admin_portal/
You will also find the following related articles useful:
If you any other way to list the files ordered by sizes in Linux, do share with us or do you have questions or thoughts to share about this guide? If yes, reach us via the feedback form below.
Linux Command , how to find files by size larger than x?
I’m trying to find files with size larger than «x» , ex: 32 bytes But what I found in ls —help was only ls -S , that just sort by size and not satisfied my demand I’m new to Linux , I’ve tried Google but I don’t know what keywords should I use to find answer , could somebody give me advice , thank you !
Your question is not about programing but about linux command that would be better in serverfault.com. If you want to ask something about shell script please read this first. stackoverflow.com/help/mcve
4 Answers 4
Try to use the power of find utility.
Find files greater than 32 bytes:
Of course you could also ask for files smaller than 32 bytes:
And files with exact 32 bytes:
For files with 32 bytes or more:
Or not smaller than 32 bytes (the same as above, but written another way):
And files between 32 and 64 bytes:
And you can also apply other common suffixes:
If you get find: illegal option — i error, specify the folder to search in after find . Add a dot to find files in the current folder: find . -size +32k -size -64M
You can change 32 for the size you want. In this case, I used your example.
Both answers you two provided me are excellent 😀 But I think I will give @rslemos the best answer tag since he’s «1 minute earier» , but still thanks to you !
Explanation: Use unix command find Using -size oprator
The find utility recursively descends the directory tree for each path listed, evaluating an expression (composed of the ‘primaries’ and ‘operands’) in terms of each file in the tree.
Solution: According to the documentation
-size n[ckMGTP] True if the file's size, rounded up, in 512-byte blocks is n. If n is fol- lowed by a c, than the primary is true if the file's size is n bytes (charac- ters). Similarly if n is followed by a scale indicator than the file's size is compared to n scaled as: k kilobytes (1024 bytes) M megabytes (1024 kilobytes) G gigabytes (1024 megabytes) T terabytes (1024 gigabytes) P petabytes (1024 terabytes) Usage: perform find operation with -size flag and threshold measurement arguments: more than(+)/less than(-), number(n) and measurement type (k/M/G/T/P).
Formula: find -size [+/-]
1.Greater Than — Find all files in my current directory (.) that greater than 500 kilobyte
2.Less Than — Find all files in my current directory (.) that less than 100 megabyte.
3.Range — Find specific file (test) in my current directory (.) that greater than 500 kilobyte less than 100 megabyte (500k-1000k)
find . -name "test" -size +500k -size -100M 4.Complex Range with Regex Find all .mkv files in all filesystem (root /) that are greater than 1 gigabyte and created this month, and present info of them.
find / -name "*.mkv" -size +1G -type f -ctime -4w | xargs ls -la