Checking for a full house

[Tony Meyer]

> def removeAll(element, num2Rem, list):
>     l = list[:]
>     for num in range(0, num2Rem):
>         l.remove(element)
>     return l
> 
> def isfullHouse(roll):
>     for die in range(1,7):
>         if roll.count(die)==3:
>             l = removeAll(die, 3, roll)
>             if l[0]==l[1]:
>                 return 1
>     return 0
> 
> My questions is this: is there a better way to do this?  A way that's
> more natural to python, or just more efficient perhaps?

This is an alternative that doesn't muck about making a second list:

def isfullHouse(roll):
    has_pair = False
    has_triple = False
    for die in xrange(1,7):
        if has_pair and has_triple:
            return True
        count = roll.count(die)
        if count > 3:
            return False # easy out; can't have four/five and full house
        if count == 2:
            has_pair = True
        elif count == 3:
            has_triple = True
    return False

If you want it in one line (and have Python 2.4):

def isfullHouse(roll):
    return len(set(roll)) != 2

If you create a set from the list, it will remove duplicates.  If there are
3,4, or 5 different numbers, then you don't have a triplet.  (I assume that
[1,1,1,1,1] isn't a full house, or this answer is wrong, since that will
have the length be 1).

(These are both untested).

=Tony.Meyer