我有一个字符串,我想用它作为文件名,所以我想用Python删除文件名中不允许的所有字符。

我宁愿严格一点,所以假设我想只保留字母、数字和一小组其他字符,如“_-.()”。”。最优雅的解决方案是什么?

文件名需要在多个操作系统(Windows, Linux和Mac OS)上有效——它是我库中的一个MP3文件,以歌曲标题为文件名,并在3台机器之间共享和备份。


当前回答

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'

其他回答

就像S.Lott回答的那样,你可以看看Django框架如何将字符串转换为有效的文件名。

最新和更新的版本在utils/text.py中,并定义了"get_valid_filename",如下所示:

def get_valid_filename(s):
    s = str(s).strip().replace(' ', '_')
    return re.sub(r'(?u)[^-\w.]', '', s)

(见https://github.com/django/django/blob/master/django/utils/text.py)

给,这应该涵盖了所有的基础。它为您处理所有类型的问题,包括(但不限于)字符替换。

适用于Windows、*nix和几乎所有其他文件系统。只允许打印字符。

def txt2filename(txt, chr_set='normal'):
    """Converts txt to a valid Windows/*nix filename with printable characters only.

    args:
        txt: The str to convert.
        chr_set: 'normal', 'universal', or 'inclusive'.
            'universal':    ' -.0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
            'normal':       Every printable character exept those disallowed on Windows/*nix.
            'extended':     All 'normal' characters plus the extended character ASCII codes 128-255
    """

    FILLER = '-'

    # Step 1: Remove excluded characters.
    if chr_set == 'universal':
        # Lookups in a set are O(n) vs O(n * x) for a str.
        printables = set(' -.0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz')
    else:
        if chr_set == 'normal':
            max_chr = 127
        elif chr_set == 'extended':
            max_chr = 256
        else:
            raise ValueError(f'The chr_set argument may be normal, extended or universal; not {chr_set=}')
        EXCLUDED_CHRS = set(r'<>:"/\|?*')               # Illegal characters in Windows filenames.
        EXCLUDED_CHRS.update(chr(127))                  # DEL (non-printable).
        printables = set(chr(x)
                         for x in range(32, max_chr)
                         if chr(x) not in EXCLUDED_CHRS)
    result = ''.join(x if x in printables else FILLER   # Allow printable characters only.
                     for x in txt)

    # Step 2: Device names, '.', and '..' are invalid filenames in Windows.
    DEVICE_NAMES = 'CON,PRN,AUX,NUL,COM1,COM2,COM3,COM4,' \
                   'COM5,COM6,COM7,COM8,COM9,LPT1,LPT2,' \
                   'LPT3,LPT4,LPT5,LPT6,LPT7,LPT8,LPT9,' \
                   'CONIN$,CONOUT$,..,.'.split()        # This list is an O(n) operation.
    if result in DEVICE_NAMES:
        result = f'-{result}-'

    # Step 3: Maximum length of filename is 255 bytes in Windows and Linux (other *nix flavors may allow longer names).
    result = result[:255]

    # Step 4: Windows does not allow filenames to end with '.' or ' ' or begin with ' '.
    result = re.sub(r'^[. ]', FILLER, result)
    result = re.sub(r' $', FILLER, result)

    return result

这个解决方案不需要外部库。它也替代了不可打印的文件名,因为它们并不总是容易处理。

仍然没有找到一个好的库来生成有效的文件名。注意,在德语、挪威语或法语等语言中,文件名中的特殊字符非常常见,完全可以接受。所以我最终有了自己的图书馆:

# util/files.py

CHAR_MAX_LEN = 31
CHAR_REPLACE = '_'

ILLEGAL_CHARS = [
    '#',  # pound
    '%',  # percent
    '&',  # ampersand
    '{',  # left curly bracket
    '}',  # right curly bracket
    '\\',  # back slash
    '<',  # left angle bracket
    '>',  # right angle bracket
    '*',  # asterisk
    '?',  # question mark
    '/',  # forward slash
    ' ',  # blank spaces
    '$',  # dollar sign
    '!',  # exclamation point
    "'",  # single quotes
    '"',  # double quotes
    ':',  # colon
    '@',  # at sign
    '+',  # plus sign
    '`',  # backtick
    '|',  # pipe
    '=',  # equal sign
]


def generate_filename(
        name, char_replace=CHAR_REPLACE, length=CHAR_MAX_LEN, 
        illegal=ILLEGAL_CHARS, replace_dot=False):
    ''' return clean filename '''
    # init
    _elem = name.split('.')
    extension = _elem[-1].strip()
    _length = length - len(extension) - 1
    label = '.'.join(_elem[:-1]).strip()[:_length]
    filename = ''
    
    # replace '.' ?
    if replace_dot:
        label = label.replace('.', char_replace)
    
    # clean
    for char in label + '.' + extension:
        if char in illegal:
            char = char_replace
        filename += char      
    
    return filename

generate_虚构(“nucgae zutaaer .0.1 docx”,replace_dot=False)

nucgae_zutaäer..0.1.docx

generate_虚构(“nucgae zutaaer .0.1 docx”,replace_dot=True)

nucgae_zutaäer__0_1.docx

This whitelist approach (ie, allowing only the chars present in valid_chars) will work if there aren't limits on the formatting of the files or combination of valid chars that are illegal (like ".."), for example, what you say would allow a filename named " . txt" which I think is not valid on Windows. As this is the most simple approach I'd try to remove whitespace from the valid_chars and prepend a known valid string in case of error, any other approach will have to know about what is allowed where to cope with Windows file naming limitations and thus be a lot more complex.

>>> import string
>>> valid_chars = "-_.() %s%s" % (string.ascii_letters, string.digits)
>>> valid_chars
'-_.() abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
>>> filename = "This Is a (valid) - filename%$&$ .txt"
>>> ''.join(c for c in filename if c in valid_chars)
'This Is a (valid) - filename .txt'

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。