Automatically creating directories with file output


Question

Possible Duplicate:
mkdir -p functionality in python

Say I want to make a file:

filename = "/foo/bar/baz.txt"

with open(filename, "w") as f:
    f.write("FOOBAR")

This gives an IOError, since /foo/bar does not exist.

What is the most pythonic way to generate those directories automatically? Is it necessary for me explicitly call os.path.exists and os.mkdir on every single one (i.e., /foo, then /foo/bar)?

1
289
5/27/2017 2:42:58 PM

Accepted Answer

The os.makedirs function does this. Try the following:

import os
import errno

filename = "/foo/bar/baz.txt"
if not os.path.exists(os.path.dirname(filename)):
    try:
        os.makedirs(os.path.dirname(filename))
    except OSError as exc: # Guard against race condition
        if exc.errno != errno.EEXIST:
            raise

with open(filename, "w") as f:
    f.write("FOOBAR")

The reason to add the try-except block is to handle the case when the directory was created between the os.path.exists and the os.makedirs calls, so that to protect us from race conditions.


In Python 3.2+, there is a more elegant way that avoids the race condition above:

filename = "/foo/bar/baz.txt"ยจ
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:
    f.write("FOOBAR")
541
5/27/2017 2:37:31 PM

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