从 Python 脚本获取工作目录的父代

我想从 Python 脚本中找到工作目录的父母。例如,我从 /home/kristina/desire-directory/scripts启动脚本,在这种情况下,需要的路径是 /home/kristina/desire-directory

我知道 sys.path[0]sys。但是我不想解析 sys.path[0]产生的字符串。有没有别的方法可以让工作目录的父母用 Python?

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Using os.path

To get the parent directory of the directory containing the script (regardless of the current working directory), you'll need to use __file__.

Inside the script use os.path.abspath(__file__) to obtain the absolute path of the script, and call os.path.dirname twice:

from os.path import dirname, abspath
d = dirname(dirname(abspath(__file__))) # /home/kristina/desire-directory

Basically, you can walk up the directory tree by calling os.path.dirname as many times as needed. Example:

In [4]: from os.path import dirname


In [5]: dirname('/home/kristina/desire-directory/scripts/script.py')
Out[5]: '/home/kristina/desire-directory/scripts'


In [6]: dirname(dirname('/home/kristina/desire-directory/scripts/script.py'))
Out[6]: '/home/kristina/desire-directory'

If you want to get the parent directory of the current working directory, use os.getcwd:

import os
d = os.path.dirname(os.getcwd())

Using pathlib

You could also use the pathlib module (available in Python 3.4 or newer).

Each pathlib.Path instance have the parent attribute referring to the parent directory, as well as the parents attribute, which is a list of ancestors of the path. Path.resolve may be used to obtain the absolute path. It also resolves all symlinks, but you may use Path.absolute instead if that isn't a desired behaviour.

Path(__file__) and Path() represent the script path and the current working directory respectively, therefore in order to get the parent directory of the script directory (regardless of the current working directory) you would use

from pathlib import Path
# `path.parents[1]` is the same as `path.parent.parent`
d = Path(__file__).resolve().parents[1] # Path('/home/kristina/desire-directory')

and to get the parent directory of the current working directory

from pathlib import Path
d = Path().resolve().parent

Note that d is a Path instance, which isn't always handy. You can convert it to str easily when you need it:

In [15]: str(d)
Out[15]: '/home/kristina/desire-directory'
import os
current_file = os.path.abspath(os.path.dirname(__file__))
parent_of_parent_dir = os.path.join(current_file, '../../')
from os.path import dirname
from os.path import abspath


def get_file_parent_dir_path():
"""return the path of the parent directory of current file's directory """
current_dir_path = dirname(abspath(__file__))
path_sep = os.path.sep
components = current_dir_path.split(path_sep)
return path_sep.join(components[:-1])

This worked for me (I am on Ubuntu):

import os
os.path.dirname(os.getcwd())

Use Path.parent from the pathlib module:

from pathlib import Path


# ...


Path(__file__).parent

You can use multiple calls to parent to go further in the path:

Path(__file__).parent.parent

'..' returns parent of current directory.

import os
os.chdir('..')

Now your current directory will be /home/kristina/desire-directory.

You can simply use../your_script_name.py For example suppose the path to your python script is trading system/trading strategies/ts1.py. To refer to volume.csv located in trading system/data/. You simply need to refer to it as ../data/volume.csv

import os def parent_directory(): # Create a relative path to the parent # of the current working directory path = os.getcwd() parent = os.path.dirname(path)

relative_parent = os.path.join(path, parent)


# Return the absolute path of the parent directory
return relative_parent

print(parent_directory())

import os
import sys
from os.path import dirname, abspath


d = dirname(dirname(abspath(__file__)))
print(d)
path1 = os.path.dirname(os.path.realpath(sys.argv[0]))
print(path1)
path = os.path.split(os.path.realpath(__file__))[0]
print(path)

For me, this is what worked:

from os import path
path.dirname(path.dirname(__file__))

You get the current directory using the current file as a reference, and then call the path.dirname again to get the parent directory.