这可以在没有模块的情况下完成。

def get_path():
    return (__file__.replace(f"<your script name>.py", ""))
print(get_path())

这可以在没有模块的情况下完成。

def get_path():
    return (__file__.replace(f"<your script name>.py", ""))
print(get_path())

也适用于__file__不可用的情况(jupyter笔记本)

import sys
from pathlib import Path
path_file = Path(sys.path[0])
print(path_file)

还使用了pathlib，这是python 3中处理路径的面向对象的方式。

试试这个:

import os
dir_path = os.path.dirname(os.path.realpath(__file__))

python中有用的路径属性:

from pathlib import Path

#Returns the path of the current directory
mypath = Path().absolute()
print('Absolute path : {}'.format(mypath))

#if you want to go to any other file inside the subdirectories of the directory path got from above method
filePath = mypath/'data'/'fuel_econ.csv'
print('File path : {}'.format(filePath))

#To check if file present in that directory or Not
isfileExist = filePath.exists()
print('isfileExist : {}'.format(isfileExist))

#To check if the path is a directory or a File
isadirectory = filePath.is_dir()
print('isadirectory : {}'.format(isadirectory))

#To get the extension of the file
fileExtension = mypath/'data'/'fuel_econ.csv'
print('File extension : {}'.format(filePath.suffix))

输出: 绝对路径是放置python文件的路径

绝对路径:D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib和seaborn Part2

文件路径:D:\Study\Machine Learning\Jupitor Notebook\JupytorNotebookTest2\Udacity_Scripts\Matplotlib and seaborn Part2\data\ fuel_econc .csv

isfileExist: True

isadirectory: False

文件扩展名:.csv

Python 2和3

你还可以简单地做:

from os import sep
print(__file__.rsplit(sep, 1)[0] + sep)

输出如下:

C:\my_folder\sub_folder\