Python replace string pattern with output of function

Question

I have a string in Python, say The quick @red fox jumps over the @lame brown dog.

I'm trying to replace each of the words that begin with @ with the output of a function that takes the word as an argument.

def my_replace(match):     return match + str(match.index('e'))  #Psuedo-code  string = "The quick @red fox jumps over the @lame brown dog." string.replace('@%match', my_replace(match))  # Result "The quick @red2 fox jumps over the @lame4 brown dog." 

Is there a clever way to do this?

Answer 1

You can pass a function to re.sub. The function will receive a match object as the argument, use .group() to extract the match as a string.

>>> def my_replace(match): ...     match = match.group() ...     return match + str(match.index('e')) ... >>> string = "The quick @red fox jumps over the @lame brown dog." >>> re.sub(r'@\w+', my_replace, string) 'The quick @red2 fox jumps over the @lame4 brown dog.' 

Answer 2

I wasn't aware you could pass a function to a re.sub() either. Riffing on @Janne Karila's answer to solve a problem I had, the approach works for multiple capture groups, too.

import re  def my_replace(match):     match1 = match.group(1)     match2 = match.group(2)     match2 = match2.replace('@', '')     return u"{0:0.{1}f}".format(float(match1), int(match2))  string = 'The first number is 14.2@1, and the second number is 50.6@4.' result = re.sub(r'([0-9]+.[0-9]+)(@[0-9]+)', my_replace, string)  print(result) 

Output:

The first number is 14.2, and the second number is 50.6000.

This simple example requires all capture groups be present (no optional groups).