但即使在导入os时，我也无法将其称为path.basename。是否可以直接将其称之为basename？

导入os，然后使用os.path.basename

导入os并不意味着你可以在不引用os的情况下使用os.foo。

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

您可以通过以下方式制作自己的产品：

>>> import os
>>> base=os.path.basename('/root/dir/sub/file.ext')
>>> base
'file.ext'
>>> os.path.splitext(base)
('file', '.ext')
>>> os.path.splitext(base)[0]
'file'

重要提示：如果有多个。在文件名中，只删除最后一个。例如：

/root/dir/sub/file.ext.zip -> file.ext

/root/dir/sub/file.ext.tar.gz -> file.ext.tar

请参阅下面的其他答案来解决这个问题。

>>>print（os.path.splitext（os.paath.basename（“/path/to/file/vrun.txt”））[0]）varun

这里/path/to/file/vrun.txt是文件的路径，输出为varun

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

但即使在导入os时，我也无法将其称为path.basename。是否可以直接将其称之为basename？

导入os，然后使用os.path.basename

导入os并不意味着你可以在不引用os的情况下使用os.foo。