create file of particular size in python

Question

I want to create a file of particular size (say, 1GiB). The content is not important since I will fill stuff into it.

What I am doing is:

f = open(\"         


        

Answer 1

WARNING This solution gives the result that you might not expect. See UPD ...

1 Create new file.

2 seek to size-1 byte.

3 write 1 byte.

4 profit :)

f = open('newfile',"wb")
f.seek(1073741824-1)
f.write(b"\0")
f.close()
import os
os.stat("newfile").st_size

1073741824

UPD: Seek and truncate both create sparse files on my system (Linux + ReiserFS). They have size as needed but don't consume space on storage device in fact. So this can not be proper solution for fast space allocation. I have just created 100Gib file having only 25Gib free and still have 25Gib free in result.

Minor Update: Added b prefix to f.write("\0") for Py3 compatibility.

Answer 2

The question has been answered before. Not sure whether the solution is cross platform, but it works in Windows (NTFS file system) flawlessly.

with open("file.to.create", "wb") as out:
    out.truncate(1024 * 1024 * 1024)

This answer uses seek and write:

with open("file.to.create", "wb") as out:
    out.seek((1024 * 1024 * 1024) - 1)
    out.write('\0')