Is calling str.replace() twice the best solution for overlapping matches?

Question

When I execute the following code I expect all ‘ a ‘ to be replaced by ‘ b ‘ yet only non overlapping matches are replaced.

" a a a a a a a a ".replace(' a ', ' b ')
>>>' b a b a b a b a'

So I use the following:

" a a a a a a a a ".replace(' a ', ' b ').replace(' a ', ' b ')
>>>' b b b b b b b b '

Is this a bug or a feature of replace ?

From the docs ALL OCCURENCES are replaced.

str.replace(old, new[, count])

    Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced.

Asked By: thomas

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Source

Answer 1

Most likely your best bet is using regex. Lookbehind/lookahead expressions let you match part of a string surrounded by a specific expression.

import re

s = " a a a a a a a a "

pattern = r'(?<= )a(?= )'

print(re.sub(pattern, "b", s))

Spaces don’t actually become part of the match, so they don’t get replaced.

Answered By: matszwecja

Answer 2

why not just replace only the thing you want to replace that is only ‘a’ and not ‘ a ‘ like this

" a a a a a a a a ".replace('a', 'b')

which gives the output

' b b b b b b b b '

Answered By: Vishnu Balaji

Is calling str.replace() twice the best solution for overlapping matches?

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