给出一个字符串列表,如:
L = ['1759@1@83@0#[email protected]@[email protected]#1094@[email protected]@14.4',
'[email protected]@[email protected]',
'[email protected]@[email protected]#1101@2@40@0#1108@2@30@0',
'1430@[email protected]@2.15#1431@[email protected]@60.29#1074@[email protected]@58.8#1109',
'1809@[email protected]@292.66#1816@[email protected]@95.44#1076@[email protected]@1110.61']
我需要在分隔符#
或@
之间提取长度为4的所有整数,并提取第一个和最后一个整数。没有花车。
我的解决方案有点过于复杂 - 用空格替换然后应用this解决方案:
pat = r'(?<!\S)\d{4}(?!\S)'
out = [re.findall(pat, re.sub('[#@]', ' ', x)) for x in L]
print (out)
"""
[['1759', '1362', '1094'],
['1356'],
['1354', '1101', '1108'],
['1430', '1431', '1074', '1109'],
['1809', '1816', '1076']]
"""
是否可以更改正则表达式,因为不使用re.sub
必须替换?还有另一种性能更好的解决方案吗?
要允许没有前导或尾随分隔符的第一次和最后一次出现,您可以使用负面外观:
(?<![^#])\d{4}(?![^@])
(?<![^#])
是(?:^|#)
的同义词。这同样适用于负前瞻。
看到现场demo here
有趣的问题!
这可以通过前瞻和后瞻的概念轻松解决。
INPUT
pattern = "(?<!\.)(?<=[#@])\d{4}|(?<!\.)\d{4}(?=[@#])"
out = [re.findall(pattern, x) for x in L]
print (out)
OUTPUT
[['1759', '1362', '1094', '1234'],
['1356'],
['1354', '1101', '1108'],
['1430', '1431', '1074', '1109'],
['1809', '1816', '1076', '1110']]
说明
上述图案是由|分隔的两个单独图案的组合(OR运算符)。
pattern_1 = "(?<!\.)(?<=[#@])\d{4}"
\d{4} --- Extract exactly 4 digits
(?<!\.) --- The 4 digits must not be preceded by a period(.) NEGATIVE LOOKBEHIND
(?<=[#@]) --- The 4 digits must be preceded by a hashtag(#) or at(@) POSITIVE LOOKBEHIND
pattern_2 = "(?<!\.)\d{4}(?=[@#])"
\d{4} --- Extract exactly 4 digits
(?<!\.) --- The 4 digits must not be preceded by a period(.) NEGATIVE LOOKBEHIND
(?=[@#] --- The 4 digits must be followed by a hashtag(#) or at(@) POSITIVE LOOKAHEAD
为了更好地理解这些概念,click here
如果你考虑长度为4而没有起始#
或结束@
的整数,那么这里是一个复杂的列表理解而不使用正则表达式:
[[n for o in p for n in o] for p in [[[m for m in k.split("@") if m.isdigit() and str(int(m))==m and len(m) ==4] for k in j.split("#")] for j in L]]
输出:
[['1759', '1362', '1094'], ['1356'], ['1354', '1101', '1108'], ['1430', '1431', '1074', '1109'], ['1809', '1816', '1076']]