Why does re.findall() find more matches than re.sub()?

Question

Consider the following:

>>> import re
>>> a = "first:second"
>>> re.findall("[^:]*", a)
['first', '', 'second', '']
>>> re.sub("[^:]*", r"(\g<0>)", a)
'(first):(second)'

re.sub()'s behavior makes more sense initially, but I can also understand re.findall()'s behavior. After all, you can match an empty string between first and : that consists only of non-colon characters (exactly zero of them), but why isn't re.sub() behaving the same way?

Shouldn't the result of the last command be (first)():(second)()?

Answer 1

You use the * which allows empty matches:

'first'   -> matched
':'       -> not in the character class but, as the pattern can be empty due 
             to the *, an empty string is matched -->''
'second'  -> matched
'$'       -> can contain an empty string before,
             an empty string is matched -->''

Quoting the documentation for re.findall():

Empty matches are included in the result unless they touch the beginning of another match.

The reason you don't see empty matches in sub results is explained in the documentation for re.sub():

Empty matches for the pattern are replaced only when not adjacent to a previous match.

Try this:

re.sub('(?:Choucroute garnie)*', '#', 'ornithorynque') 

And now this:

print re.sub('(?:nithorynque)*', '#', 'ornithorynque')

There is no consecutive #