这段代码将返回主脚本的绝对路径。

import os
def whereAmI():
    return os.path.dirname(os.path.realpath(__import__("__main__").__file__))

这甚至可以在一个模块中工作。

考虑一下我的代码:

import os


def readFile(filename):
    filehandle = open(filename)
    print filehandle.read()
    filehandle.close()



fileDir = os.path.dirname(os.path.realpath('__file__'))
print fileDir

#For accessing the file in the same folder
filename = "same.txt"
readFile(filename)

#For accessing the file in a folder contained in the current folder
filename = os.path.join(fileDir, 'Folder1.1/same.txt')
readFile(filename)

#For accessing the file in the parent folder of the current folder
filename = os.path.join(fileDir, '../same.txt')
readFile(filename)

#For accessing the file inside a sibling folder.
filename = os.path.join(fileDir, '../Folder2/same.txt')
filename = os.path.abspath(os.path.realpath(filename))
print filename
readFile(filename)

现在是2018年，Python很久以前就已经进化到__future__了。因此，如何使用Python 3.4附带的惊人的pathlib来完成任务，而不是在os, os中苦苦挣扎。路径，glob, shutil等。

这里有3条路径(可能重复):

Mod_path:简单帮助脚本的路径 Src_path:包含两个等待复制的模板文件。 Cwd:当前目录，这些模板文件的目的地。

问题是:我们没有src_path的完整路径，只知道它到mod_path的相对路径。

现在让我们用惊人的pathlib来解决这个问题:

# Hope you don't be imprisoned by legacy Python code :)
from pathlib import Path

# `cwd`: current directory is straightforward
cwd = Path.cwd()

# `mod_path`: According to the accepted answer and combine with future power
# if we are in the `helper_script.py`
mod_path = Path(__file__).parent
# OR if we are `import helper_script`
mod_path = Path(helper_script.__file__).parent

# `src_path`: with the future power, it's just so straightforward
relative_path_1 = 'same/parent/with/helper/script/'
relative_path_2 = '../../or/any/level/up/'
src_path_1 = (mod_path / relative_path_1).resolve()
src_path_2 = (mod_path / relative_path_2).resolve()

在未来，就这么简单。

此外，我们可以使用pathlib选择、检查和复制/移动这些模板文件:

if src_path != cwd:
    # When we have different types of files in the `src_path`
    for template_path in src_path.glob('*.ini'):
        fname = template_path.name
        target = cwd / fname
        if not target.exists():
            # This is the COPY action
            with target.open(mode='wb') as fd:
                fd.write(template_path.read_bytes())
            # If we want MOVE action, we could use:
            # template_path.replace(target)

一个简单的解决办法是

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

另一个适合我的选择是:

this_dir = os.path.dirname(__file__) 
filename = os.path.realpath("{0}/relative/file.path".format(this_dir))

我认为要在所有系统中使用“ntpath”而不是“os.path”。如今，它在Windows、Linux和Mac OSX上都能很好地工作。

import ntpath
import os
dirname = ntpath.dirname(__file__)
filename = os.path.join(dirname, 'relative/path/to/file/you/want')