1

Is there a way to do substitution on a group?

Say I am trying to insert a link into text, based on custom formatting. So, given something like this:

This is a random text. This should be a [[link somewhere]]. And some more text at the end.

I want to end up with 

This is a random text. This should be a <a href="/link_somewhere">link somewhere</a>. And some more text at the end.

I know that '\[\[(.*?)\]\]' will match stuff within square brackets as group 1, but then I want to do another substitution on group 1, so that I can replace space with _.

Is that doable in a single re.sub regex expression?

Goro
  • 9,919
  • 22
  • 74
  • 108

2 Answers2

3

You can use a function as a replacement instead of string.

>>> import re
>>> def as_link(match):
...     link = match.group(1)
...     return '<a href="{}">{}</a>'.format(link.replace(' ', '_'), link)
...
>>> text = 'This is a random text. This should be a [[link somewhere]]. And some more text at the end.'
>>> re.sub(r'\[\[(.*?)\]\]', as_link, text)
'This is a random text. This should be a <a href="link_somewhere">link somewhere</a>. And some more text at the end.'
falsetru
  • 357,413
  • 63
  • 732
  • 636
1

You could do something like this.

import re

pattern = re.compile(r'\[\[([^]]+)\]\]')

def convert(text): 
    def replace(match):
        link = match.group(1)
        return '<a href="{}">{}</a>'.format(link.replace(' ', '_'), link)
    return pattern.sub(replace, text)

s = 'This is a random text. This should be a [[link somewhere]]. .....'
convert(s)

See working demo

hwnd
  • 69,796
  • 4
  • 95
  • 132