python:获取目录两个级别 [英] python: get directory two levels up

问题描述

好吧...我不知道模块x在哪里，但是我知道我需要把目录的路径上移两个级别.

Ok...I dont know where module x is, but I know that I need to get the path to the directory two levels up.

因此，有没有一种更优雅的方法:

So, is there a more elegant way to do:

import os
two_up = os.path.dirname(os.path.dirname(__file__))

欢迎使用适用于Python 2和3的解决方案！

Solutions for both Python 2 and 3 are welcome!

推荐答案

您可以使用 pathlib .不幸的是，这仅在Python 3.4的stdlib中可用.如果您使用的是旧版本，则必须在此处处从PyPI安装副本.使用pip应该很容易做到.

You can use pathlib. Unfortunately this is only available in the stdlib for Python 3.4. If you have an older version you'll have to install a copy from PyPI here. This should be easy to do using pip.

from pathlib import Path

p = Path(__file__).parents[1]

print(p)
# /absolute/path/to/two/levels/up

这使用parents序列，该序列提供对父目录的访问并选择第二个目录.

This uses the parents sequence which provides access to the parent directories and chooses the 2nd one up.

请注意，在这种情况下，p将是某种形式的Path对象，具有自己的方法.如果您需要路径为字符串，则可以在其上调用str.

Note that p in this case will be some form of Path object, with their own methods. If you need the paths as string then you can call str on them.

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