错误的unicode字符串是一个意外编码字节的字符串。 例如:
文字:שלום
,Windows-1255编码:\x99\x8c\x85\x8d
,Unicode:u'\u05e9\u05dc\u05d5\u05dd'
,错误的Unicode:u'\x99\x8c\x85\x8d'
在解析MP3文件中的ID3标签时,我有时碰到这样的字符串。我该如何修复这些字符串? (例如,将u'\x99\x8c\x85\x8d'
转换为u'\u05e9\u05dc\u05d5\u05dd'
)
答案 0 :(得分:11)
您可以使用u'\x99\x8c\x85\x8d'
编码将'\x99\x8c\x85\x8d'
转换为latin-1
:
In [9]: x = u'\x99\x8c\x85\x8d'
In [10]: x.encode('latin-1')
Out[10]: '\x99\x8c\x85\x8d'
但是,似乎这不是一个有效的Windows-1255编码字符串。你或许是'\xf9\xec\xe5\xed'
吗?如果是,那么
In [22]: x = u'\xf9\xec\xe5\xed'
In [23]: x.encode('latin-1').decode('cp1255')
Out[23]: u'\u05e9\u05dc\u05d5\u05dd'
将u'\xf9\xec\xe5\xed'
转换为与您发布的所需unicode匹配的u'\u05e9\u05dc\u05d5\u05dd'
。
如果您真的想将u'\x99\x8c\x85\x8d'
转换为u'\u05e9\u05dc\u05d5\u05dd'
,那么这恰好可行:
In [27]: u'\x99\x8c\x85\x8d'.encode('latin-1').decode('cp862')
Out[27]: u'\u05e9\u05dc\u05d5\u05dd'
使用此脚本找到上述编码/解码链:
<强> guess_chain_encodings.py 强>
"""
Usage example: guess_chain_encodings.py "u'баба'" "u'\xe1\xe0\xe1\xe0'"
"""
import six
import argparse
import binascii
import zlib
import utils_string as us
import ast
import collections
import itertools
import random
encodings = us.all_encodings()
Errors = (IOError, UnicodeEncodeError, UnicodeError, LookupError,
TypeError, ValueError, binascii.Error, zlib.error)
def breadth_first_search(text, all = False):
seen = set()
tasks = collections.deque()
tasks.append(([], text))
while tasks:
encs, text = tasks.popleft()
for enc, newtext in candidates(text):
if repr(newtext) not in seen:
if not all:
seen.add(repr(newtext))
newtask = encs+[enc], newtext
tasks.append(newtask)
yield newtask
def candidates(text):
f = text.encode if isinstance(text, six.text_type) else text.decode
results = []
for enc in encodings:
try:
results.append((enc, f(enc)))
except Errors as err:
pass
random.shuffle(results)
for r in results:
yield r
def fmt(encs, text):
encode_decode = itertools.cycle(['encode', 'decode'])
if not isinstance(text, six.text_type):
next(encode_decode)
chain = '.'.join( "{f}('{e}')".format(f = func, e = enc)
for enc, func in zip(encs, encode_decode) )
return '{t!r}.{c}'.format(t = text, c = chain)
def main():
parser = argparse.ArgumentParser()
parser.add_argument('start', type = ast.literal_eval, help = 'starting unicode')
parser.add_argument('stop', type = ast.literal_eval, help = 'ending unicode')
parser.add_argument('--all', '-a', action = 'store_true')
args = parser.parse_args()
min_len = None
for encs, text in breadth_first_search(args.start, args.all):
if min_len is not None and len(encs) > min_len:
break
if type(text) == type(args.stop) and text == args.stop:
print(fmt(encs, args.start))
min_len = len(encs)
if __name__ == '__main__':
main()
运行
% guess_chain_encodings.py "u'\x99\x8c\x85\x8d'" "u'\u05e9\u05dc\u05d5\u05dd'" --all
产量
u'\x99\x8c\x85\x8d'.encode('latin_1').decode('cp862')
u'\x99\x8c\x85\x8d'.encode('charmap').decode('cp862')
u'\x99\x8c\x85\x8d'.encode('rot_13').decode('cp856')
等