How to make Project Euler 32 in Python faster?

We shall say that an n-digit number is pandigital if it makes use of
all the digits 1 to n exactly once; for example, the 5-digit number,
15234, is 1 through 5 pandigital.


The product 7254 is unusual, as the identity, 39 × 186 = 7254,
containing multiplicand, multiplier, and product is 1 through 9
pandigital.


Find the sum of all products whose multiplicand/multiplier/product
identity can be written as a 1 through 9 pandigital.


HINT: Some products can be obtained in more than one way so be sure to
only include it once in your sum.

I need help on making this code faster. Right now I'm using double nested for-loops; is there a way I can change that?

from timeit import default_timer as timer

start = timer()
products = set()
for a in range(12, 100):
    for b in range(102, 1000): # to get 9-digit: 2-digit * 3-digit
        product = a * b
        digits = list(str(a) + str(b) + str(product))
        digits = [int(x) for x in digits]
        if sorted(digits, key = int) == range(1, 10):
            products.add(product)

for a in range(1, 10):
    for b in range(1023, 10000): # other option is 1-digit * 4-digit
        product = a * b
        digits = list(str(a) + str(b) + str(product))
        digits = [int(x) for x in digits]
        if sorted(digits, key = int) == range(1, 10):
            products.add(product)

ans = sum(list(products))
elapsed_time = (timer() - start) * 1000 # s --> ms

print "Found %d in %r ms." % (ans, elapsed_time)

Since you know that all of the non-zero digits must be present in the answer, you can skip most of the numbers in the range. That is, it's pointless to do a test multiplication of 22*3333 because those both repeat digits.

So what is needed is an efficient way to run through all of the permutations of digits quickly, and fortunately, Python's itertools provides exactly that.

import itertools
start = timer()
products = set()
def makeint(n):
    return int(''.join(map(str,n)))

target = range(1,10)
d = itertools.permutations(target, 5)
for a in d:
    product = makeint(a[:2])*makeint(a[2:])
    if sorted(list(a)+map(int,str(product))) == target:
        products.add(product)
    product = makeint(a[:1])*makeint(a[1:])
    if sorted(list(a)+map(int,str(product))) == target:
        products.add(product)

ans = sum(list(products))

The way this works is to pick off permutations 5 digits at a time and then group as 2 and 3 or as 1, 4 as you have done. There are other refinements one can make, such as skipping every number which ends in a 5 (because 5*n will always end in either 0, which is not in the range, or 5 which would be a duplicate digit), but I'll leave such refinements to you.

As it stands, this code runs almost 8x faster than the original.

StackExchange Code Review Q#47931, answer score: 8