在分隔符之间提取具有一定长度的整数

要允许没有前导或尾随分隔符的第一次和最后一次出现,可以使用负环视框:

(?<![^#])\d{4}(?![^@])

(?<![^#]) 是一个 近的 同义词 (?:^|#) . 这同样适用于负面展望。

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有趣的问题!

这一点可以很容易地通过“向前看”和“向后看”的概念来解决。

输入

pattern = "(?<!\.)(?<=[#@])\d{4}|(?<!\.)\d{4}(?=[@#])"
out = [re.findall(pattern, x) for x in L]
print (out)

[['1759', '1362', '1094', '1234'],
 ['1356'],
 ['1354', '1101', '1108'],
 ['1430', '1431', '1074', '1109'],
 ['1809', '1816', '1076', '1110']]

解释

以上 图案 是由一个 (或操作员)。

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

# 或结束 @ 也是:

[[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']]