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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T15:30:38+00:00 2026-05-15T15:30:38+00:00

I’m creating sparse files in python as follows: >>> f = open(‘testfile’, ‘ab’) >>>

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I’m creating sparse files in python as follows:

>>> f = open('testfile', 'ab')
>>> f.truncate(1024000)
>>> f.close()

when the file is done, it takes up 0 disk space, but its inode size is set to my truncated value (1000K):

igor47@piglet:~/test$ ls -lh testfile 
-rw-r--r-- 1 igor47 igor47 1000K 2010-07-09 04:02 testfile
igor47@piglet:~/test$ du -hs testfile 
0   testfile

How do I get the file’s real space usage (allocated size) inside python? The stat call returns the file’s apparent size, and I have no idea how else to get the real usage other than to read the entire file (it may become quite large)

>>> os.stat('testfile').st_size
1024000
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T15:30:39+00:00Added an answer on May 15, 2026 at 3:30 pm
    >>> os.stat('testfile').st_blocks*512
    0
    

    Tadaa 🙂

    st_blocks is the number of 512-byte blocks actually allocated to the file. Note that st_blocks is not guaranteed to be present in all operating systems, but those that support sparse files generally do.

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