python - Regex Python - 如何获取不连续出现的次数？

Question

这个问题已经有答案了:

How to use regex to find all overlapping matches (5 个回答)

已关闭 4 年前。

这是我的代码，我想获取基于此模式的组合数量:x - y - x ，例如:"UBU" , "ANA", "INI"

    import re 

    pattern = r"(?P<name>[a-z]).(?P=name)"

    print(len(re.findall(pattern, "hello"))) # should return 0 : OK

    print(len(re.findall(pattern, "mirror"))) # should return 1 because there is "ror" : OK 

    print(len(re.findall(pattern, "irir"))) #should return 2 because there are "iri" and "rir" : return just 1

当部分组合叠瓦式插入另一个组合时，就像“iri”中“rir”的前两个字母一样，它不起作用。

您知道如何获得这些组合的正确数量(对于第三个示例)吗？预先非常感谢您

Answer 1

def pattern_counter(word):
    return sum([1 for index, letter in enumerate(word[:(len(word)-2)]) if letter == word[index+2]])

word1 = "hello"
word2 = "mirror"
word3 = "irir"

print(pattern_counter(word1)) # prints 0
print(pattern_counter(word2)) # prints 1
print(pattern_counter(word3)) # prints 2