filesystem - Python 获取文件路径及文件目录

Question

我正在学习Python，不过遇到一些问题，想请教OS模块中的os.path.dirname(__file__)和os.path.abspath(__file__)运行os.path.dirname(__file__)时候，为什么返回的是空白呢？是不是因为他运行的是相对路径？？？如...

迷茫 · Answer

建議你可以稍微瀏覽一下 Python doc: os.path, 你就會明白囉:

我放上跟你問題相關的幾個條目:

os.path.abspath(path)
- Return a normalized absolutized version of the pathname path. On most platforms, this is equivalent to calling the function normpath() as follows: normpath(join(os.getcwd(), path)).
os.path.normpath(path)
- Normalize a pathname by collapsing redundant separators and up-level references so that A//B, A/B/, A/./B and A/foo/../B all become A/B. This string manipulation may change the meaning of a path that contains symbolic links. On Windows, it converts forward slashes to backward slashes. To normalize case, use normcase().
os.path.dirname(path)
- Return the directory name of pathname path. This is the first element of the pair returned by passing path to the function split().
os.path.split(path)
- Split the pathname path into a pair, (head, tail) where tail is the last pathname component and head is everything leading up to that. The tail part will never contain a slash; if path ends in a slash, tail will be empty. If there is no slash in path, head will be empty. If path is empty, both head and tail are empty. Trailing slashes are stripped from head unless it is the root (one or more slashes only). In all cases, join(head, tail) returns a path to the same location as path (but the strings may differ). Also see the functions dirname() and basename().

我們做以下觀察:

test.py

import os

print(__file__)
print(os.path.dirname(__file__))
print(os.path.abspath(__file__))
print(os.path.abspath(os.path.dirname(__file__)))
print(os.path.dirname(os.path.abspath(__file__)))

運行:

$ pwd
/home/dokelung
$ python test.py

結果:

test.py

/home/dokelung/test.py
/home/dokelung
/home/dokelung

首先 __file__ 的值其實就是在命令列上 invoke Python 時給的 script 名稱:

$ python test.py          # 此時 __file__ 是 test.py
$ python ../test.py       # 此時 __file__ 是 ../test.py
$ python hello/../test.py # 此時 __file__ 是 hello/../test.py

在這裡, 因為 __file__ 的值為 test.py, 所以 print(__file__) 的結果是 test.py 的值為

, 所以 print(__file__) 的結果是

也就不意外了。os.path.dirname(__file__)之所以得出空白(空字串), 是因為 __file__ 就只是一個單純的名稱(非路徑) 且 dirname 也只是很單純的利用 os.path.split()

接著, os.path.dirname(__file__)之所以得出空白(空字串), 是因為

就只是一個單純的名稱(非路徑) 且 dirname 也只是很單純的利用 os.path.split() 來切割這個名稱(這當然沒甚麼好切的, 連路徑分割符都沒有):head 是空字串, 所以 dirname

>>> import os
>>> os.path.split('test.py')
('', 'test.py')
>>> os.path.split('test.py')[0]
''

我分會發現切出來的

的結果是空白。abspath 動用了 os.getcwd()

所以即便給定的是單純的名稱, 也能返回路徑:os.path.abspath(os.path.dirname(__file__)) 的結果就等於是 os.getcwd() 的結果去接上 dirname

>>> os.getcwd()
'/home/dokelung'

>>> os.path.join(os.getcwd(), 'test.py')
'/home/dokelung/test.py'

>>> os.path.normpath(os.path.join(os.getcwd(), 'test.py'))
'/home/dokelung/test.py'

而

得到的空字串:os.path.dirname(os.path.abspath(__file__))

>>> os.path.dirname('test.py')
''

>>> os.path.join(os.getcwd(), os.path.dirname('test.py'))
'/home/dokelung/'

最後,

的結果是這麼來的:

>>> os.path.abspath('test.py')
'/home/dokelung/test.py'

>>> os.path.split(os.path.abspath('test.py'))
('/home/dokelung', 'test.py')

>>> os.path.split(os.path.abspath('test.py'))[0]
'/home/dokelung'

🎜希望講到這裡有讓你明白!🎜

Conclusion

Now briefly answer your question

Why is dirname blank?
- Because you gave a simple name when running, so __file__ is a simple name and not a path
How can I know whether the file in the brackets is run with a relative path or an absolute path?
- It’s very simple, it depends on how you run Python
Why is this file run with a relative path when I run the example script below?
- Because $ python 1.py you gave the relative path yourself
os.path.abspath(os.path.dirname(__file__)) 和 os.path.dirname(os.path.abspath(__file__)) Are the properties the same?
- Basically the same

Questions I answered: Python-QA