Python Regex：匹配由一个其他单词完全分隔的任何重复单词

Question

I encountered this problem where I need to use regex to find repeated words separated by another word.

So if:

"all in all" will return: "all"

"good good good" will return: Null (Same word not another word)

I have tried:

p = re.compile(r'(\b\w+\b)\s\w+\s\1')
m = p.findall('all in all day in and day out bit by bit good good good')

print(m)

This returns ['all', 'bit', 'good'] , but I only want it to return ['all','bit'] .

Thanks in advance!

Answer 1

You just need to add a negative lookahead for the word immediately following the initial capture group to ensure your regex can't match (for example) good good :

import re

p = re.compile(r'(\b\w+\b)(?!\s\1\b)\s\w+\s\1\b')
m = p.findall('all in all day in and day out bit by bit good good good')

print(m)

Output:

['all', 'bit']

If you want to include overlapping matches, make the entire regex a positive lookahead (thanks @ggorlen):

p = re.compile(r'(?=(\b\w+\b)(?!\s\1\b)\s\w+\s\1\b)')
m = p.findall('foo bar foo bar foo')

Output:

['foo', 'bar', 'foo']

If you also need to remove duplicate matches, convert to a set and back to a list :

p = re.compile(r'(?=(\b\w+\b)(?!\s\1\b)\s\w+\s\1\b)')
m = list(set(p.findall('foo bar foo bar foo')))
print(m)

Output:

['foo', 'bar']

Answer 2

No need for regex; normal programming constructs can handle this sort of problem just fine. Write a loop and add a conditional:

s = 'all in all day in and day out bit by bit good good good'

words = s.split()
result = []

for i in range(len(words) - 2):
    if words[i] == words[i+2] and words[i] != words[i+1]:
        result.append(words[i])

print(result) # ['all', 'bit']