附加到在字符串[Python]中找到的每个正则表达式匹配项

Question

Meet string 's':

AA Well, This is amazing. AA It is. BB Wow. AA As I said.

Here's the code:

s = AA Well, This is amazing. AA It is. BB Wow. AA As I said.
separate_words_tuple = s.split()
i = 0
for word in separate_words_tuple:
    check_word = word
    if re.search("[A-Z]+", str(check_word)):
        match = re.search("[A-Z]+", str(check_word))
        if (len(match.group(0)) != 1):
            actor_name = match.group(0) + "_" + str(i)
            print(re.sub(r"[A-Z]+", actor_name, s))
            i += 1

Here's what 's' looks like with that code:

AA_3 AA_3ell, AA_3his is amazing. AA_3 AA_3t is. AA_3 AA_3ow. AA_3 AA_3s AA_3 said.

Here's what I want 's' to look like:

AA_0 Well, This is amazing. AA_1 It is. BB_2 Wow. AA_3 As I said.

Answer 1

Use function's static property to count all matched items. incCaps() is a replacement callback.

s = 'AA Well, This is amazing. AA It is. BB Wow. AA As I said.'

def incCaps(m):
    replaced = m.group(0) + '_' + str(incCaps.counter)
    incCaps.counter += 1
    return replaced

incCaps.counter = 0
s = re.sub(r'\b[A-Z]{2,}\b', incCaps, s)

print(s)

The output:

AA_0 Well, This is amazing. AA_1 It is. BB_2 Wow. AA_3 As I said.

Answer 2

@RomanPerekhrest is right - re.sub provides us with a great opportunity to pass a function as an argument. But having a global function with a state is a code smell - this function isn't testable or even reusable.

I would suggest two similar ways of refactoring:

1. Having match_counter global for reuse:

    def match_counter(): counter = -1 def inc_caps(m): counter += 1 return '{}_{}'.format(m.group(0), counter) return inc_caps s = re.sub(r'\\b[AZ]{2,}\\b', match_counter(), s) 

2. or making the whole thing a single unit of functionality:

    def replaced_by_counts(regex, string): counter = -1 def inc_caps(m): counter += 1 return '{}_{}'.format(m.group(0), counter) return regex.sub(inc_caps, string)