自动创建带有文件输出的目录 [英] Automatically creating directories with file output

问题描述

可能的重复:
mkdir -p python 中的功能

说我想制作一个文件:

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

with open(filename, "w") as f:
    f.write("FOOBAR")

这给出了一个 IOError，因为 /foo/bar 不存在.

This gives an IOError, since /foo/bar does not exist.

自动生成这些目录的最pythonic 的方法是什么?我是否有必要在每一个(即/foo，然后是/foo/bar)上显式调用 os.path.exists 和 os.mkdir ?

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)?

推荐答案

os.makedirs 函数就是这样做的.请尝试以下操作:

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")

添加try-except 块的原因是为了处理在os.path.exists 和os 之间创建目录的情况.makedirs 调用，以便保护我们免受竞争条件的影响.

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.

在 Python 3.2+ 中，有一种更优雅的方式 避免了上面的竞争条件:

In Python 3.2+, there is a more elegant way that avoids the race condition above:

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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