Automatically creating directories with file output [duplicate]

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In Python 3.2+, using the APIs requested by the OP, 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")

With the Pathlib module (introduced in Python 3.4), there is an alternate syntax (thanks David258):

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")

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.

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