Comments on: Python: Factors of a number

By: Rich

Rich — Sun, 28 Mar 2010 12:13:20 +0000

Actually, it can be a generator expression instead of a list comprehension, which is better for memory:

set(reduce(list.__add__, ([i, n/i] for i in range(1, sqrt(n) + 1) if n % i == 0)))

By: Rich

Rich — Sun, 28 Mar 2010 12:10:55 +0000

set(reduce(list.__add__, [[i, n/i] for i in range(1, sqrt(n) + 1) if n % i == 0]))

By: mat

mat — Thu, 12 Nov 2009 10:48:43 +0000

@mike: Thanks for pointing out the error in my code I’ve update the post now. Also the method with the prime factors is clearly much more efficient, thanks

@Ryan: I’ve not come across generators before, thanks for your code looks a hell of a lot neater than what I wrote!

By: Mike Hansen

Mike Hansen — Thu, 12 Nov 2009 05:18:19 +0000

Here are some timings. I’ve named my function factors and your function slow_factors in the timings below.

Lots of small factors:

sage: %timeit factors(2**40)
10000 loops, best of 3: 44.2 µs per loop
sage: %timeit slow_factors(2**40)
10 loops, best of 3: 255 ms per loop

Product of two primes:

sage: p = next_prime(10^5)*next_prime(10^6); p
100003300009
sage: %timeit slow_factors(p)
10 loops, best of 3: 473 ms per loop
sage: %timeit factors(p)
10 loops, best of 3: 183 ms per loop

Primes:

sage: p = next_prime(10^10); p
10000000019
sage: %timeit factors(p)
10 loops, best of 3: 96.9 ms per loop
sage: %timeit slow_factors(p)
10 loops, best of 3: 144 ms per loop

When there are lots of factors, then the method I posted gets much faster.

By: Ryan

Ryan — Wed, 11 Nov 2009 08:21:58 +0000

And a better version of above:

def genFactors2(n):
“”"Generate the factors of an integer.”"”
yield 1
for i in xrange(2, math.floor(math.sqrt(n))):
if n % i == 0:
yield i
yield n/i
yield n

def factors2(n):
“”"Return the factors of a integer.”"”
return list(genFactors(n))

By: Ryan

Ryan — Wed, 11 Nov 2009 08:21:08 +0000

Here’s a variant that uses a generator expression.

def genFactors(n):
“”"Generate the factors of an integer.”"”
yield 1
for i in xrange(2, n/2 + 1):
if n % i == 0:
yield i
yield n

def factors(n):
“”"Return the factors of a integer.”"”
return list(genFactors(n))

By: Mike Hansen

Mike Hansen — Wed, 11 Nov 2009 04:31:55 +0000

At the end, you want to do

fact.sort()
return fact

since the sort method just returns None and not the sorted list.

A more efficient method to find all the factors is to find all of the prime factors and then build up all of the factors from that. You can do this by say noticing that if p is prime then p%30 must also be prime or one. Also, p%30 can’t be either a 2, 3, or 5 since 2*3*5 = 30. Using this information, you can reduce the number of divisions you need to perform.

You can see the code for this at http://sage.math.washington.edu/home/mhansen/factor.py

Also, if you like doing Project Euler problems in Python, then you should try using Sage ( http://www.sagemath.org ) which has a lot of this functionality already built in (and much faster since it interfaces with C libraries).