Fix Python – Automatically creating directories with file output [duplicate]


Asked By – Phil

Say I want to make a file:

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

with open(filename, "w") as f:

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)?

Now we will see solution for issue: Automatically creating directories with file output [duplicate]


As question is closed, @David258’s answer is written as a comment.

from pathlib import Path
output_file = Path("/foo/bar/baz.txt")
output_file.parent.mkdir(exist_ok=True, parents=True)

I leave it to the author of this answer to fold this in or remove it as they see fit.

Original answer starts here:

In Python 3.2+, you can elegantly do the following:

import os

filename = "/foo/bar/baz.txt"
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:

In older python, there is a less elegant way:

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)):
    except OSError as exc: # Guard against race condition
        if exc.errno != errno.EEXIST:

with open(filename, "w") as f:

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.

This question is answered By – Krumelur

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0