If you know that, for instance, every occurrence of '[[' needs to be replaced with ']]', then it is much faster to use regular string methods. If you have the text stored in the variable foo:
foo = foo.replace('[[', '[').replace(']]', ']').replace('->', '')
If you need to recognize the actual pattern, use REs:
regex = re.compile(r'\[\[(.+?) -> (.+?)\]\]')
regex.sub(r'[\1 \2]', foo)
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