Optional grouping in a simple python regex

Question

All I want to do is search a string for instances of two consecutive digits. If such an instance is found I want to group it, otherwise return none for that particular groups. I thought this would be trivial, but I can't understand where I'm going wrong. In the example below, removing the optional (?) character gets me the numbers, but in strings without numbers, the r evaluates to None, so r.groups() throws an exception.

p = re.compile(r'(\d{2})?')
r = p.search('wqddsel78ffgr')
print r.groups()
>>>(None, )    # why not ('78', )?

# --- update/clarification --- #

Thanks for the answers, but the explanations given are leaving me none-the-wiser. Here's a another go at pin-pointing exactly what it is I don't understand.

pattern = re.compile(r'z.*(A)?')
_string = "aazaa90aabcdefA"
result = pattern.search(_string)
result.group()
>>> zaa90aabcdefA
result.groups()
>>> (None, )

I understand why result.group() produces the result it does, but why doesn't result.groups() produce ('A', )? I thought it worked like this: once the regex hits the z it then matches right to the end of the line using .*. In spite of .* matching everything, the regex engine is aware that it passed over an optional group, and since ? means it will try to match if it can, it should work backwards to try and match. Replacing ? with + does return ('A', ). This suggests that ? won't try and match if it doesn't have to, but this seems to contrast with much of what I've read on the subject (esp. J. Friedl's excellent book).

Answer 1

This works for me:

p = re.compile('\D*(\d{2})?')
r = p.search('wqddsel78ffgr')
print r.groups()  # ('78',)

r = p.search('wqddselffgr')
print r.groups()  # (None,)