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。
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。