Get parent of current directory from Python script

问题:

I want to get the parent of current directory from Python script. For example I launch the script from /home/kristina/desire-directory/scripts the desire path in this case is /home/kristina/desire-directory

I know sys.path[0] from sys. But I don't want to parse sys.path[0] resulting string. Is there any another way to get parent of current directory in Python?

回答1:

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'

回答2:

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

回答3:

This worked for me (I am on Ubuntu):

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

回答4:

import os
current_file = os.path.abspath(os.path.dirname(__file__))
parent_of_parent_dir = os.path.join(current_file, '../../')

回答5:

'..' returns parent of current directory.

import os
os.chdir('..')

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

回答6:

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

回答7:

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])