https://docs.python.org/3/library/os.path.html

在python 3中，pathlib“pathlib模块提供高级路径对象。”所以

>>> from pathlib import Path

>>> p = Path("/a/b/c.txt")
>>> p.with_suffix('')
WindowsPath('/a/b/c')
>>> p.stem
'c'

使用pathlib.Path.stem是正确的方法，但这里有一个丑陋的解决方案，它比基于pathlib的方法更有效。

您有一个文件路径，其字段由正斜杠/分隔，斜杠不能出现在文件名中，因此您将文件路径拆分为/，最后一个字段是文件名。

扩展名始终是通过按点分割文件名创建的列表的最后一个元素。，因此，如果反转文件名并按点拆分一次，则第二个元素的反转是不带扩展名的文件名。

name = path.split('/')[-1][::-1].split('.', 1)[1][::-1]

性能：

Python 3.9.10 (tags/v3.9.10:f2f3f53, Jan 17 2022, 15:14:21) [MSC v.1929 64 bit (AMD64)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.28.0 -- An enhanced Interactive Python. Type '?' for help.

In [1]: from pathlib import Path

In [2]: file = 'D:/ffmpeg/ffmpeg.exe'

In [3]: Path(file).stem
Out[3]: 'ffmpeg'

In [4]: file.split('/')[-1][::-1].split('.', 1)[1][::-1]
Out[4]: 'ffmpeg'

In [5]: %timeit Path(file).stem
6.15 µs ± 433 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [6]: %timeit file.split('/')[-1][::-1].split('.', 1)[1][::-1]
671 ns ± 37.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [7]:

import os
filename, file_extension =os.path.splitext(os.path.basename('/d1/d2/example.cs'))

文件名为“example”文件扩展名为“.cs”

'

我们可以做一些简单的拆分/弹出魔术，如图所示(https://stackoverflow.com/a/424006/1250044)，以提取文件名（考虑windows和POSIX的差异）。

def getFileNameWithoutExtension(path):
  return path.split('\\').pop().split('/').pop().rsplit('.', 1)[0]

getFileNameWithoutExtension('/path/to/file-0.0.1.ext')
# => file-0.0.1

getFileNameWithoutExtension('\\path\\to\\file-0.0.1.ext')
# => file-0.0.1

# use pathlib. the below works with compound filetypes and normal ones
source_file = 'spaces.tar.gz.zip.rar.7z'
source_path = pathlib.Path(source_file)
source_path.name.replace(''.join(source_path.suffixes), '')
>>> 'spaces'

尽管上面描述了许多工作实现，我还是添加了这个^，因为它只使用pathlib，并且适用于复合文件类型和普通文件类型

使用Pathlib回答几个场景

使用Pathlib，当只有一个扩展名（或没有扩展名）时，获取文件名很简单，但处理多个扩展名的一般情况可能会很困难。

零或一扩展

from pathlib import Path

pth = Path('./thefile.tar')

fn = pth.stem

print(fn)      # thefile


# Explanation:
# the `stem` attribute returns only the base filename, stripping
# any leading path if present, and strips the extension after
# the last `.`, if present.


# Further tests

eg_paths = ['thefile',
            'thefile.tar',
            './thefile',
            './thefile.tar',
            '../../thefile.tar',
            '.././thefile.tar',
            'rel/pa.th/to/thefile',
            '/abs/path/to/thefile.tar']

for p in eg_paths:
    print(Path(p).stem)  # prints thefile every time

两个或更少的扩展

from pathlib import Path

pth = Path('./thefile.tar.gz')

fn = pth.with_suffix('').stem

print(fn)      # thefile


# Explanation:
# Using the `.with_suffix('')` trick returns a Path object after
# stripping one extension, and then we can simply use `.stem`.


# Further tests

eg_paths += ['./thefile.tar.gz',
             '/abs/pa.th/to/thefile.tar.gz']

for p in eg_paths:
    print(Path(p).with_suffix('').stem)  # prints thefile every time

任意数量的扩展名（0、1或更多）

from pathlib import Path

pth = Path('./thefile.tar.gz.bz.7zip')

fn = pth.name
if len(pth.suffixes) > 0:
    s = pth.suffixes[0]
    fn = fn.rsplit(s)[0]

# or, equivalently

fn = pth.name
for s in pth.suffixes:
    fn = fn.rsplit(s)[0]
    break

# or simply run the full loop

fn = pth.name
for _ in pth.suffixes:
    fn = fn.rsplit('.')[0]

# In any case:

print(fn)     # thefile


# Explanation
#
# pth.name     -> 'thefile.tar.gz.bz.7zip'
# pth.suffixes -> ['.tar', '.gz', '.bz', '.7zip']
#
# If there may be more than two extensions, we can test for
# that case with an if statement, or simply attempt the loop
# and break after rsplitting on the first extension instance.
# Alternatively, we may even run the full loop and strip one 
# extension with every pass.


# Further tests

eg_paths += ['./thefile.tar.gz.bz.7zip',
             '/abs/pa.th/to/thefile.tar.gz.bz.7zip']

for p in eg_paths:
    pth = Path(p)
    fn = pth.name
    for s in pth.suffixes:
        fn = fn.rsplit(s)[0]
        break

    print(fn)  # prints thefile every time

已知第一个扩展的特殊情况

例如，如果扩展名可以是.tar、.tar.gz、.tar/gz.bz等；您可以简单地rsplit已知的扩展并获取第一个元素：

pth = Path('foo/bar/baz.baz/thefile.tar.gz')

fn = pth.name.rsplit('.tar')[0]

print(fn)      # thefile