Python factorization

Question

I\'d just like to know the best way of listing all integer factors of a number, given a dictionary of its prime factors and their exponents.
For example if we have {2:3

Answer 1

Well, not only you have 3 loops, but this approach won't work if you have more than 3 factors :)

One possible way:

def genfactors(fdict):    
    factors = set([1])

    for factor, count in fdict.iteritems():
        for ignore in range(count):
            factors.update([n*factor for n in factors])
            # that line could also be:
            # factors.update(map(lambda e: e*factor, factors))

    return factors

factors = {2:3, 3:2, 5:1}

for factor in genfactors(factors):
    print factor

Also, you can avoid duplicating some work in the inner loop: if your working set is (1,3), and want to apply to 2^3 factors, we were doing:

• (1,3) U (1,3)*2 = (1,2,3,6)
• (1,2,3,6) U (1,2,3,6)*2 = (1,2,3,4,6,12)
• (1,2,3,4,6,12) U (1,2,3,4,6,12)*2 = (1,2,3,4,6,8,12,24)

See how many duplicates we have in the second sets?

But we can do instead:

• (1,3) + (1,3)*2 = (1,2,3,6)
• (1,2,3,6) + ((1,3)*2)*2 = (1,2,3,4,6,12)
• (1,2,3,4,6,12) + (((1,3)*2)*2)*2 = (1,2,3,4,6,8,12,24)

The solution looks even nicer without the sets:

def genfactors(fdict):
    factors = [1]

    for factor, count in fdict.iteritems():
        newfactors = factors
        for ignore in range(count):
            newfactors = map(lambda e: e*factor, newfactors)
            factors += newfactors

    return factors