Automatically creating directories with file output

Question

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)?

Asked By: Phil

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Answer 1

COMMUNITY EDIT:
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)
output_file.write_text("FOOBAR")

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:
    f.write("FOOBAR")

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)):
    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.

Answered By: Krumelur

Automatically creating directories with file output

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