148
votes

How do I go about computing a factorial of an integer in Python?

9

9 Answers

202
votes

Easiest way is to use math.factorial (available in Python 2.6 and above):

import math
math.factorial(1000)

If you want/have to write it yourself, you can use an iterative approach:

def factorial(n):
    fact = 1
    for num in range(2, n + 1):
        fact *= num
    return fact

or a recursive approach:

def factorial(n):
    if n < 2:
        return 1
    else:
        return n * factorial(n-1)

Note that the factorial function is only defined for positive integers so you should also check that n >= 0 and that isinstance(n, int). If it's not, raise a ValueError or a TypeError respectively. math.factorial will take care of this for you.

115
votes

On Python 2.6 and up, try:

import math
math.factorial(n)
18
votes

Existing solution

The shortest and probably the fastest solution is:

from math import factorial
print factorial(1000)

Building your own

You can also build your own solution. Generally you have two approaches. The one that suits me best is:

from itertools import imap
def factorial(x):
    return reduce(long.__mul__, imap(long, xrange(1, x + 1)))

print factorial(1000)

(it works also for bigger numbers, when the result becomes long)

The second way of achieving the same is:

def factorial(x):
    result = 1
    for i in xrange(2, x + 1):
        result *= i
    return result

print factorial(1000)
7
votes
def factorial(n):
    if n < 2:
        return 1
    return n * factorial(n - 1)
5
votes

If you are using Python2.5 or older try

from operator import mul
def factorial(n):
    return reduce(mul, range(1,n+1))

for newer Python, there is factorial in the math module as given in other answers here

5
votes
def fact(n):
    f = 1
    for i in range(1, n + 1):
        f *= i
    return f
4
votes

For performance reasons, please do not use recursion. It would be disastrous.

def fact(n, total=1):
    while True:
        if n == 1:
            return total
        n, total = n - 1, total * n

Check running results

cProfile.run('fact(126000)')

4 function calls in 5.164 seconds

Using the stack is convenient(like recursive call), but it comes at a cost: storing detailed information can take up a lot of memory.

If the stack is high, it means that the computer stores a lot of information about function calls.

The method only takes up constant memory(like iteration).

Or Using for loop

def fact(n):
    result = 1
    for i in range(2, n + 1):
        result *= i
    return result

Check running results

cProfile.run('fact(126000)')

4 function calls in 4.708 seconds

Or Using builtin function math

def fact(n):
    return math.factorial(n)

Check running results

cProfile.run('fact(126000)')

5 function calls in 0.272 seconds
0
votes

Another way to do it is to use np.prod shown below:

def factorial(n):
    if n == 0:
        return 1
    else:
         return np.prod(np.arange(1,n+1))
0
votes

Non-recursive solution, no imports:

def factorial(x):
    return eval(' * '.join(map(str, range(1, x + 1))))