Question

Situation:

• text: a string
• R: a regex that matches part of the string. This might be expensive to calculate.

I want to both delete the R-matches from the text, and see what they actually contain. Currently, I do this like:

import re ab_re = re.compile('[ab]') text='abcdedfe falijbijie bbbb laifsjelifjl' ab_re.findall(text) # ['a', 'b', 'a', 'b', 'b', 'b', 'b', 'b', 'a'] ab_re.sub('',text) # 'cdedfe flijijie  lifsjelifjl' 

This runs the regex twice, near as I can tell. Is there a technique to do it all on pass, perhaps using re.split? It seems like with split based solutions I’d need to do the regex at least twice as well.

Answer 1

import re  r = re.compile('[ab]') text = 'abcdedfe falijbijie bbbb laifsjelifjl'  matches = [] replaced = [] pos = 0 for m in r.finditer(text):     matches.append(m.group(0))     replaced.append(text[pos:m.start()])     pos = m.end() replaced.append(text[pos:])  print matches print ''.join(replaced) 

Outputs:

['a', 'b', 'a', 'b', 'b', 'b', 'b', 'b', 'a'] cdedfe flijijie  lifsjelifjl