Automatically creating directories with file output [duplicate]

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)?

Answer 1

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")  

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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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