听起来你可能还想签出inspect模块。

我有一个脚本，必须在windows环境下工作。 这段代码剪辑是什么我已经完成:

import os,sys
PROJECT_PATH = os.path.abspath(os.path.split(sys.argv[0])[0])

这是个很俗气的决定。但它不需要外部库，这是我的情况下最重要的事情。

我写了一个函数，它考虑到eclipse调试器和单元测试。 它返回您启动的第一个脚本的文件夹。你可以选择指定__file__变量，但最主要的是你不必在所有调用层次结构中共享这个变量。

也许你可以处理一些我没见过的特殊情况，但对我来说是可以的。

import inspect, os
def getRootDirectory(_file_=None):
    """
    Get the directory of the root execution file
    Can help: http://stackoverflow.com/questions/50499/how-do-i-get-the-path-and-name-of-the-file-that-is-currently-executing
    For eclipse user with unittest or debugger, the function search for the correct folder in the stack
    You can pass __file__ (with 4 underscores) if you want the caller directory
    """
    # If we don't have the __file__ :
    if _file_ is None:
        # We get the last :
        rootFile = inspect.stack()[-1][1]
        folder = os.path.abspath(rootFile)
        # If we use unittest :
        if ("/pysrc" in folder) & ("org.python.pydev" in folder):
            previous = None
            # We search from left to right the case.py :
            for el in inspect.stack():
                currentFile = os.path.abspath(el[1])
                if ("unittest/case.py" in currentFile) | ("org.python.pydev" in currentFile):
                    break
                previous = currentFile
            folder = previous
        # We return the folder :
        return os.path.dirname(folder)
    else:
        # We return the folder according to specified __file__ :
        return os.path.dirname(os.path.realpath(_file_))

下面返回当前主脚本所在的路径。我用Linux、Win10、IPython和Jupyter Lab进行了测试。我需要一个解决方案，工作于本地Jupyter笔记本电脑以及。

import builtins
import os
import sys

def current_dir():
    if "get_ipython" in globals() or "get_ipython" in dir(builtins):
        # os.getcwd() is PROBABLY the dir that hosts the active notebook script.
        # See also https://github.com/ipython/ipython/issues/10123
        return os.getcwd()
    else:
        return os.path.abspath(os.path.dirname(sys.argv[0]))

print(__file__)
print(__import__("pathlib").Path(__file__).parent)

你可以使用inspect.stack()

import inspect,os
inspect.stack()[0]  => (<frame object at 0x00AC2AC0>, 'g:\\Python\\Test\\_GetCurrentProgram.py', 15, '<module>', ['print inspect.stack()[0]\n'], 0)
os.path.abspath (inspect.stack()[0][1]) => 'g:\\Python\\Test\\_GetCurrentProgram.py'