os.makedirs 这个功能。请尝试以下操作：

  import os 
 import errno 

 filename =/ foo / bar / baz.txt
如果不是os.path.exists（os.path.dirname（filename））：
 try：
 os.makedirs（os.path.dirname（filename） ）
，除了OSError，如exc：＃防止竞争条件
如果exc.errno！= errno.EEXIST：
打开（文件名，w ）作为f：
 f.write（FOOBAR）
 

try-except block用于处理在 os.path.exists 和 os.makedirs 调用，这样可以保护我们免受竞争的影响。

  filename =/foo/bar/baz.txt\"¨
 os。 makedirs（os.path.dirname（文件名）， （开头的文件名为w）作为f：
 f.write（FOOBAR）
 

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)?

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"¨
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:
    f.write("FOOBAR")