I want to find files in a certain directory using find command and display the file size as well after finding. Here is what I have come up with so far.
find /my_search_directory -type f -name "abc*" -printf "%f %k KB\n"
%k displays the file size in kbs,I want this to be in gbs. Can anyone help me out with this?
du -BG filename
writes to stdout the size of filename in GB.
You can use -exec to run du for each file found, passing -B GiB to show the size in GiB rather than bytes:
find /my_search_directory -type f -name "abc*" -exec du -B GiB {} \;
You can also use -h instead of -B GiB to make du choose an appropriate unit; this is probably more useful to you because -B rounds upwards. You can also add --apparent-size to show the perceived file size instead of the size taken up on disk.
As an extra tip, sort -h can be used to process the output of du -h:
find /my_search_directory -type f -name "abc*" -exec du h {} \; | sort -h
I've got a bunch of files scattered across folders in a layout, e.g.:
dir1/somefile.gif
dir1/another.mp4
dir2/video/filename.mp4
dir2/some.file
dir2/blahblah.mp4
And I need to find the total disk space used for the MP4 files only. This means it's gotta be recursive somehow.
I've looked at du and fiddling with piping things to grep but can't seem to figure out how to calculate just the MP4 files no matter where they are.
A human readable total disk space output is a must too, preferably in GB, if possible?
Any ideas? Thanks
For individual file size:
find . -name "*.mp4" -print0 | du -sh --files0-from=-
For total disk space in GB:
find . -name "*.mp4" -print0 | du -sb --files0-from=- | awk '{ total += $1} END { print total/1024/1024/1024 }'
You can simply do :
find -name "*.mp4" -exec du -b {} \; | awk 'BEGIN{total=0}{total=total+$1}END{print total}'
The -exec option of find command executes a simple command with {} as the file found by find.
du -b displays the size of the file in bytes.
The awk command initializes a variable at 0 and get the size of each file to display the total at the end of the command.
This will sum all mp4 files size in bytes:
find ./ -name "*.mp4" -printf "%s\n" | paste -sd+ | bc
When I run the command:
find / 2>/dev/null -user root -type f -mmin -1 -exec du -cb {} + | grep total | head -1
I get a rather large number in bytes which is expected.
However, when I run the same command but with human-readable instead of bytes, as in:
find / 2>/dev/null -user root -type f -mmin -1 -exec du -ch {} + | grep total | head -1
I get 0. I also tried removing the head -1 thinking I was grabbing the wrong data, but every print out is 0 total. Why is this? Is there an alternative method to get the total size of all files found using find for both bytes and human-readable print outs?
Use -xdev option to find command to exclude other filesystems.
I don't have an explanation why yet, but I think this is related to tmpfs and devtmpfs filesystems such as /proc.
When I ran your scenario's I had the same results because the -b option adds in the size of /proc/kcore
procfs is a bit of dark magic; no files in it are real. It looks like a filesystem, acts like a filesystem, and is a filesystem. But not one that is stored on disk (or elsewhere).
/proc/kcore specifically is a file which maps directly to every available byte in your virtual memory ... I'm not absolutely clear on the details; the 128TB comes from Linux allocating 47ish bits of the 64bits available for virtual memory.
When I use the -ch argument for du it shows /proc/kcore as 0:
0 /proc/kcore
But when I use the -cb it shows my /proc/kcore as:
140737486266368 /proc/kcore
this is because the -b option:
-b, --bytes
equivalent to '--apparent-size --block-size=1'
and --apparent-size :
--apparent-size
print apparent sizes, rather than disk usage; although the apparent size is
usually smaller, it may be larger due to holes in ('sparse') files, internal
fragmentation, indirect blocks, and the like
References:
/proc kcore file is huge
https://linux.die.net/man/1/du
My linux system is throwing I/O error - disk space full error for any activity. I would like to clear logs, unwanted files with large size.
How to search in file system for files greater than a specific amount?
use the below command to search for files greater than 200MB in desecnding order, use sudo to look for files in descending order
find / -type f -size +200M -exec ls -lh {} \; 2> /dev/null | awk '{ print $NF ": " $5 }' | sort -nrk 2,2
I found that there is no easy to get way the size of a directory in Bash?
I want that when I type ls -<some options>, it can list of all the sum of the file size of directory recursively and files at the same time and sort by size order.
Is that possible?
Simply navigate to directory and run following command:
du -a --max-depth=1 | sort -n
OR add -h for human readable sizes and -r to print bigger directories/files first.
du -a -h --max-depth=1 | sort -hr
Apparently --max-depth option is not in Mac OS X's version of the du command. You can use the following instead.
du -h -d 1 | sort -n
du -s -- * | sort -n
(this willnot show hidden (.dotfiles) files)
Use du -sm for Mb units etc. I always use
du -smc -- * | sort -n
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
Command
du -h --max-depth=0 * | sort -hr
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.
Installation on Ubuntu:
$ sudo apt-get install ncdu
Usage:
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.
Simple and fast:
find . -mindepth 1 -maxdepth 1 -type d | parallel du -s | sort -n
*requires GNU Parallel.
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)
# du -sb /
12684646920 /
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.
Another simple solution.
$ for entry in $(ls); do du -s "$entry"; done | sort -n
the result will look like
2900 tmp
6781 boot
8428 bin
24932 lib64
34436 sbin
90084 var
106676 etc
125216 lib
3313136 usr
4828700 opt
changing "du -s" to "du -sh" will show human readable size, but we won't be able to sort in this method.
you can use the below to list files by size
du -h | sort -hr | more
or
du -h --max-depth=0 * | sort -hr | more
I tend to use du in a simple way.
du -sh */ | sort -n
This provides me with an idea of what directories are consuming the most space. I can then run more precise searches later.
sudo du -hsx 2>/dev/null * | sort -hr | less
4.9G var
2.2G usr
61M root
9.0M etc
6.5M home
824K init
36K run
16K lost+found
4.0K tmp
4.0K srv
4.0K opt
4.0K mnt
4.0K media
4.0K boot
0 sys
0 sbin
0 proc
0 libx32
0 lib64
0 lib32
0 lib
0 dev
0 bin
(END)