Regex Python - How to get the number of occurences that are not consecutives?

Question

Here's my code, I'd like to get the number of combinations based on this pattern : x - y - x , example : "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

It's not working when part of the combinations is imbricated into another one like the first two letters of "rir" in "iri".

Would you know how to get the correct number of those combinations (for the third example) ? Thank you very much in advance

The term imbricated is specific to *botany*. If you want to find some useful information you must look for the term *overlapping*. — Giacomo Alzetta, Jul 09 '19 at 07:45
Thanks , I'm not english so I was searching the term to use :) — GG24, Jul 09 '19 at 07:50

score 0 · Accepted Answer · answered Jul 09 '19 at 07:58

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

Regex Python - How to get the number of occurences that are not consecutives?

1 Answers1