带字符串的正则表达式:出现两次的字母对

如何使用Python的正则表达式在字符串中找到两次出现的字母对？

我想遍历一个字符串列表,找到那些具有重复字母对的字符串,并将它们放入列表中.字母不需要相同,只需要重复,但字母可以相同.

例如: xxhgfhdeifhjfrikfoixx- 这个有xx两次,所以我想保留这个字符串 kwofhdbugktrkdidhdnbk- 这个也会保留,因为hd重复

我得到的最好的是找到对: ([a-z][a-z])\1|([a-z])\2

我需要找到哪些字符串有重复的对.

正则表达式

(\w{2}).*?(\1)

https://regex101.com/r/yB3nX6/1

可视化

码

迭代所有比赛

for match in re.finditer(r"(\w{2}).*?(\1)", subject, re.IGNORECASE):
    # match start: match.start()
    # match end (exclusive): match.end()
    # matched text: match.group()

获取字符串中所有正则表达式匹配的数组

result = re.findall(r"(\w{2}).*?(\1)", subject, re.IGNORECASE)

人类可读

# (\w{2}).*?(\1)
# 
# Options: Case insensitive; Exact spacing; Dot doesn’t match line breaks; ^$ don’t match at line breaks; Regex syntax only
# 
# Match the regex below and capture its match into backreference number 1 «(\w{2})»
#    Match a single character that is a “word character” (Unicode; any letter or ideograph, any number, underscore) «\w{2}»
#       Exactly 2 times «{2}»
# Match any single character that is NOT a line break character (line feed) «.*?»
#    Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
# Match the regex below and capture its match into backreference number 2 «(\1)»
#    Match the same text that was most recently matched by capturing group number 1 (case insensitive; fail if the group did not participate in the match so far) «\1»

笔记

您可以切换出\w的[a-z],如果你想具体谈谈只接受a-z字符.