在Windows系统上，我也使用了驱动程序名称前缀，例如：

>>> s = 'c:\\temp\\akarmi.txt'
>>> print(os.path.splitext(s)[0])
c:\temp\akarmi

因为我不需要驱动器号或目录名，所以我使用：

>>> print(os.path.splitext(os.path.basename(s))[0])
akarmi

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

以下情况如何？

import pathlib
filename = '/path/to/dir/stem.ext.tar.gz'
pathlib.Path(filename).name[:-len(''.join(pathlib.Path(filename).suffixes))]
# -> 'stem'

或者这个等价物？

pathlib.Path(filename).name[:-sum(map(len, pathlib.Path(filename).suffixes))]

import os
path = "a/b/c/abc.txt"
print os.path.splitext(os.path.basename(path))[0]

如果扩展名中有多个点，os.path.splitext（）将无法工作。

例如，images.tar.gz

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> print os.path.splitext(file_name)[0]
images.tar

您只需找到basename中第一个点的索引，然后对basename进行切片，即可获得不带扩展名的文件名。

>>> import os
>>> file_path = '/home/dc/images.tar.gz'
>>> file_name = os.path.basename(file_path)
>>> index_of_dot = file_name.index('.')
>>> file_name_without_extension = file_name[:index_of_dot]
>>> print file_name_without_extension
images

在Windows系统上，我也使用了驱动程序名称前缀，例如：

>>> s = 'c:\\temp\\akarmi.txt'
>>> print(os.path.splitext(s)[0])
c:\temp\akarmi

因为我不需要驱动器号或目录名，所以我使用：

>>> print(os.path.splitext(os.path.basename(s))[0])
akarmi