How can I replace patterns using re module in python3

Question

I wanted to replace certain pattern(space between alphabet) multiple times in a line.
Here's my code :

s = re.sub('([a-z]) ([a-z])', '\g<1>_\g<2>', 'series m coupe')

I expected to replace 'series m coupe' to 'series_m_coupe', but what I got is 'series_m coupe'. Even I put count=0, it didn't work...

I guess it's because "m" is 1 syllable. when I put more than 1 syllable like 'series mini coupe', that worked :

s = re.sub('([a-z]) ([a-z])', '\g<1>_\g<2>', 'series mini coupe')
s
'series_mini_coupe'

Wiktor Stribiżew · Accepted Answer · 2016-02-29 08:26:40Z

1

When you use ([a-z]) ([a-z]), the s m is matched in series m coupe and the regex index is after m. So, the regex engine is looking for a second match after that letter, and can't find any.

You need to use a lookahead to match overlapping strings:

s = re.sub('([a-z]) (?=[a-z])', '\g<1>_', 'series m coupe')
                    ^^^     ^

See the regex demo

The (?=[a-z]) lookahead will check if the space is followed with a lowercase ASCII letter, but will not consume it. In the replacement pattern, the \g<2> should be removed as there is no longer the second capturing group.

answered Feb 29, 2016 at 8:26

Wiktor Stribiżew

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1 Comment

sh kim Over a year ago

Wow!! What a smart, wonderful code!! Thanks, I had no idea about "lookahead". It really helpful :)

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