像其他人建议的那样使用os.path.split或os.path.basename并不能在所有情况下工作:如果您在Linux上运行脚本并试图处理经典的windows样式的路径，它将失败。

Windows路径可以使用反斜杠或正斜杠作为路径分隔符。因此，ntpath模块(相当于os. path)在windows上运行时的路径)将适用于所有平台上的所有(1)路径。

import ntpath
ntpath.basename("a/b/c")

当然，如果文件以斜杠结束，basename将为空，所以创建自己的函数来处理它:

def path_leaf(path):
    head, tail = ntpath.split(path)
    return tail or ntpath.basename(head)

验证:

>>> 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']
>>> [path_leaf(path) for path in paths]
['c', 'c', 'c', 'c', 'c', 'c', 'c']

(1) There's one caveat: Linux filenames may contain backslashes. So on linux, r'a/b\c' always refers to the file b\c in the a folder, while on Windows, it always refers to the c file in the b subfolder of the a folder. So when both forward and backward slashes are used in a path, you need to know the associated platform to be able to interpret it correctly. In practice it's usually safe to assume it's a windows path since backslashes are seldom used in Linux filenames, but keep this in mind when you code so you don't create accidental security holes.

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))

我最喜欢的是:

filename = fullname.split(os.sep)[-1]

os.path.split 这是你要找的函数吗

head, tail = os.path.split("/tmp/d/a.dat")

>>> print(tail)
a.dat
>>> print(head)
/tmp/d

这是适用于linux和windows以及标准库

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']

def path_leaf(path):
    return path.strip('/').strip('\\').split('/')[-1].split('\\')[-1]

[path_leaf(path) for path in paths]

结果:

['c', 'c', 'c', 'c', 'c', 'c', 'c']