如何在python中获取当前.py文件的更高级别目录的路径？

Question

Say there's a directory hierarchy like this: .../A/B/main.py .

main.py is the file being executed, and I need to get the full path of folder A , how can I do that?

Answer 1

Use the os module:

import os
a_path = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))

Answer 2

To get the path of the current directory, you can use:

import os
print os.path.realpath('.')

since . represents the current directory in the file system. So to get the path of the higher level directory, just replace . with .. , which represents parent directory

import os
print os.path.realpath('..')

You can also use the os.getcwd() method to get the current working directory and then get its parent directory with the .. representation.

Answer 3

In Python 3:

from pathlib import Path

mypath = Path().absolute().parent.parent  # each '.parent' goes one level up - vary as required
print(mypath)