Question

Is there a better way to do this:

perms = product(range(1,7),range(1,7),range(1,7))

so that I can choose how many ranges I use?
I want it to be equivalent to this, but scalable.

def dice(num)
    if num == 1:
        perms = ((i,) for i in range(1,7))
    elif num == 2:
        perms = product(range(1,7),range(1,7))
    elif num == 3:
        perms = product(range(1,7),range(1,7),range(1,7))
    #... and so on

but I know there has to be a better way.
I’m using it for counting dice outcomes.
The actual code

def dice(selection= lambda d: d[2]):   
    perms = itertools.product(range(1,7),range(1,7),range(1,7))      
    return collections.Counter(((selection(sorted(i)) for i in perms)))

where I can call it with a variety of selectors, like sum(d[0:2]) for the sum of the lowest 2 dice or d[1] to get the middle dice.

Answer 1

Yes. Use the repeat keyword argument:

perms = product(range(1, 7), repeat=3)

See the docs for more.