带扩展名的文件名

filepath = './dir/subdir/filename.ext'
basename = os.path.basename(filepath)
print(basename)
# filename.ext

print(type(basename))
# <class 'str'>

不带扩展名的文件名

basename_without_ext = os.path.splitext(os.path.basename(filepath))[0]
print(basename_without_ext)
# filename

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。

import os
head, tail = os.path.split('path/to/file.exe')

尾巴是你想要的，文件名。

详见python os模块文档

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.

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))

有一个函数返回你想要的结果

import os
print(os.path.basename(your_path))

警告:在POSIX系统上使用os.path.basename()从windows样式的路径(例如。"C:\\my\\file.txt")，则会返回整个路径。

下面的例子来自交互式python shell在Linux主机上运行:

Python 3.8.2 (default, Mar 13 2020, 10:14:16)
[GCC 9.3.0] on Linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> filepath = "C:\\my\\path\\to\\file.txt" # A Windows style file path.
>>> os.path.basename(filepath)
'C:\\my\\path\\to\\file.txt'