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正則表達式查找<>

[英]Regular Expression to find <>

我有一個字符串

"Absolutely<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>\r\n"

我只想將 <> 部分與字符串分開。 我試過<.*>但它回來了

<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>

我不要他們之間的話。 我想要輸出,

 ["<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64>", "<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>"]

任何幫助。 我被困在 python 中。

你需要一個負面的前瞻 此模式匹配,直到找到第一個>后面沒有<

import re

text = "Absolutely<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>\r\n"

pattern = "<.*?>(?!<)"

print re.findall(pattern, text)
#['<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64>', '<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>']

您可以使用<.*?>(?!<)而不是<.*>

這是你可以做的

s = "Absolutely<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E64> Friendship goals exceeded here!! Sydney, Melbourne, Connecticut &amp; South Carolina<U+653C><U+3E64><U+613C><U+3E30><U+623C><U+3E64><U+653C><U+3E64><U+623C><U+3E31><U+383C><U+3E61>\r\n"

result = re.findall('<.*?>(?!<)',s)

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