fname = str("C:\Windows\paint.exe").split('\\')[-1:][0]

这将返回:paint.exe

更改关于您的路径或操作系统的split函数的sep值。

这是一个仅适用于regex的解决方案，它似乎适用于任何OS上的任何OS路径。

不需要其他模块，也不需要预处理:

import re

def extract_basename(path):
  """Extracts basename of a given path. Should Work with any OS Path on any OS"""
  basename = re.search(r'[^\\/]+(?=[\\/]?$)', path)
  if basename:
    return basename.group(0)


paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c',
         'a/b/../../a/b/c/', 'a/b/../../a/b/c']

print([extract_basename(path) for path in paths])
# ['c', 'c', 'c', 'c', 'c', 'c', 'c']


extra_paths = ['C:\\', 'alone', '/a/space in filename', 'C:\\multi\nline']

print([extract_basename(path) for path in extra_paths])
# ['C:', 'alone', 'space in filename', 'multi\nline']

更新:

If you only want a potential filename, if present (i.e., /a/b/ is a dir and so is c:\windows\), change the regex to: r'[^\\/]+(?![\\/])$' . For the "regex challenged," this changes the positive forward lookahead for some sort of slash to a negative forward lookahead, causing pathnames that end with said slash to return nothing instead of the last sub-directory in the pathname. Of course there is no guarantee that the potential filename actually refers to a file and for that os.path.is_dir() or os.path.is_file() would need to be employed.

这将匹配如下:

/a/b/c/             # nothing, pathname ends with the dir 'c'
c:\windows\         # nothing, pathname ends with the dir 'windows'
c:hello.txt         # matches potential filename 'hello.txt'
~it_s_me/.bashrc    # matches potential filename '.bashrc'
c:\windows\system32 # matches potential filename 'system32', except
                    # that is obviously a dir. os.path.is_dir()
                    # should be used to tell us for sure

正则表达式可以在这里测试。

我从来没有见过双开的路，它们存在吗?python模块os的内置特性在这些方面失败了。所有其他工作，还有你用os.path.normpath()给出的警告:

paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c', 
...     'a/b/../../a/b/c/', 'a/b/../../a/b/c', 'a/./b/c', 'a\b/c']
for path in paths:
    os.path.basename(os.path.normpath(path))

如果您在一个目录中有许多文件，并希望将这些文件名存储到一个列表中。使用下面的代码。

import os as os
import glob as glob
path = 'mypath'
file_list= []
for file in glob.glob(path):
    data_file_list = os.path.basename(file)
    file_list.append(data_file_list)

也许只是我的all in one解决方案，没有重要的一些新的(考虑tempfile创建临时文件:D)

import tempfile
abc = tempfile.NamedTemporaryFile(dir='/tmp/')
abc.name
abc.name.replace("/", " ").split()[-1] 

获取abc.name的值将是这样的字符串:'/tmp/tmpks5oksk7' 所以我可以用空格.replace("/"， " ")替换/，然后调用split()。它会返回一个列表，我得到 列表中最后一个带有[-1]的元素

不需要导入任何模块。

这是工作!

os.path.basename(name)

但是你不能在Linux中通过Windows文件路径获取文件名。Windows。 操作系统。不同操作系统上不同模块的路径加载:

Linux - posixpath Windows - npath

所以你可以用os。路径总是得到正确的结果