您可以将列表推导式与字符串方法一起使用。

>>> s
'foo-bar#baz?qux@127/\\9]'
>>> "".join(x for x in s if x.isalnum())
'foobarbazqux1279'

这是Windows特定路径的另一个答案，使用简单的替换，没有时髦的模块:

import re

def check_for_illegal_char(input_str):
    # remove illegal characters for Windows file names/paths 
    # (illegal filenames are a superset (41) of the illegal path names (36))
    # this is according to windows blacklist obtained with Powershell
    # from: https://stackoverflow.com/questions/1976007/what-characters-are-forbidden-in-windows-and-linux-directory-names/44750843#44750843
    #
    # PS> $enc = [system.Text.Encoding]::UTF8
    # PS> $FileNameInvalidChars = [System.IO.Path]::GetInvalidFileNameChars()
    # PS> $FileNameInvalidChars | foreach { $enc.GetBytes($_) } | Out-File -FilePath InvalidFileCharCodes.txt

    illegal = '\u0022\u003c\u003e\u007c\u0000\u0001\u0002\u0003\u0004\u0005\u0006\u0007\u0008' + \
              '\u0009\u000a\u000b\u000c\u000d\u000e\u000f\u0010\u0011\u0012\u0013\u0014\u0015' + \
              '\u0016\u0017\u0018\u0019\u001a\u001b\u001c\u001d\u001e\u001f\u003a\u002a\u003f\u005c\u002f' 

    output_str, _ = re.subn('['+illegal+']','_', input_str)
    output_str = output_str.replace('\\','_')   # backslash cannot be handled by regex
    output_str = output_str.replace('..','_')   # double dots are illegal too, or at least a bad idea 
    output_str = output_str[:-1] if output_str[-1] == '.' else output_str # can't have end of line '.'

    if output_str != input_str:
        print(f"The name '{input_str}' had invalid characters, "
              f"name was modified to '{output_str}'")

    return output_str

当测试check_for_illegal_char('fas\u0003\u0004good\\..asd.')，我得到:

The name 'fas♥♦good\..asd.' had invalid characters, name was modified to 'fas__good__asd'

为什么不直接用try/except来包装“osopen”，让底层操作系统来判断文件是否有效?

这看起来工作量少得多，而且无论您使用哪种操作系统都是有效的。

你可以看看Django框架(但要考虑到许可!)，看看他们如何从任意文本中创建一个“slug”。段代码是URL和文件名友好的。

Django文本utils定义了一个函数，slugify()，这可能是这种事情的黄金标准。本质上，他们的代码如下。

import unicodedata
import re

def slugify(value, allow_unicode=False):
    """
    Taken from https://github.com/django/django/blob/master/django/utils/text.py
    Convert to ASCII if 'allow_unicode' is False. Convert spaces or repeated
    dashes to single dashes. Remove characters that aren't alphanumerics,
    underscores, or hyphens. Convert to lowercase. Also strip leading and
    trailing whitespace, dashes, and underscores.
    """
    value = str(value)
    if allow_unicode:
        value = unicodedata.normalize('NFKC', value)
    else:
        value = unicodedata.normalize('NFKD', value).encode('ascii', 'ignore').decode('ascii')
    value = re.sub(r'[^\w\s-]', '', value.lower())
    return re.sub(r'[-\s]+', '-', value).strip('-_')

旧版本是:

def slugify(value):
    """
    Normalizes string, converts to lowercase, removes non-alpha characters,
    and converts spaces to hyphens.
    """
    import unicodedata
    value = unicodedata.normalize('NFKD', value).encode('ascii', 'ignore')
    value = unicode(re.sub('[^\w\s-]', '', value).strip().lower())
    value = unicode(re.sub('[-\s]+', '-', value))
    # ...
    return value

还有更多，但我把它省略了，因为它没有解决怠惰，而是逃避。

我喜欢这里的python-slugify方法，但它也剥离点，这是不希望的。所以我优化了上传一个干净的文件名到s3:

pip install python-slugify

示例代码:

s = 'Very / Unsafe / file\nname hähä \n\r .txt'
clean_basename = slugify(os.path.splitext(s)[0])
clean_extension = slugify(os.path.splitext(s)[1][1:])
if clean_extension:
    clean_filename = '{}.{}'.format(clean_basename, clean_extension)
elif clean_basename:
    clean_filename = clean_basename
else:
    clean_filename = 'none' # only unclean characters

输出:

>>> clean_filename
'very-unsafe-file-name-haha.txt'

这是如此的故障安全，它适用于没有扩展名的文件名，甚至只适用于不安全的字符文件名(这里的结果是none)。

一句话:

valid_file_name = re.sub('[^\w_.)( -]', '', any_string)

你也可以用“_”字符让它更具可读性(例如替换斜杠)