>>> sys.version
2.6.5 (r265:79096, Mar 19 2010, 21:48:26) [MSC v.1500 32 bit
(Intel)]
>>> import re
>>> re.match("[-+]?[0-9]+[.]?[0-9]*([eE][-+]?[0-9]+)?", "1.23e-4").group()
1.23e-4
>>> re.search("[-+]?[0-9]+[.]?[0-9]*([eE][-+]?[0-9]+)?", "1.23e-4").group()
1.23e-4
>>> for e in re.finditer("[-+]?[0-9]+[.]?[0-9]*([eE][-+]?[0-9]+)?", "1.23e-4"):
        print e.group()
        
1.23e-4

but

>>> re.findall("[-+]?[0-9]+[.]?[0-9]*([eE][-+]?[0-9]+)?", "1.23e-4")
['e-4']
>>> 

It seems re.findall() does not like patterns containing parentheses.

>>> re.findall("[-+]?[0-9]+[.]?[0-9]*[eE][-+]?[0-9]+", "1.23e-4")
['1.23e-4']
>>> 
>>> re.findall("a(b)?", "ab")
['b']
>>> re.findall("ab?", "ab")
['ab']
>>>

Seems consistent to me:

.match, .search and .finditer return a MatchObject whose .group() return the *entire matched string*. If you use .group(1) you'll get similar results to .findall() which returns a list of (possibly of tuples) of the captured groupings.

This is the findall equivalent (using a non-capturing group):

>>> re.findall('a(?:b)?', 'ab')
['ab']

What Jon said is correct, .group() is equivalent to .group(0) and returns the whole match. re.findall returns all the groups captured by each set of () as a list of strings (if there is 0 or 1 group) or a list of tuples (if there are more than 1).