How do I go about computing a factorial of an integer in Python?
10 Answers
The 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.

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2I'm not understanding how you can use `factorial` within the `factorial` function. How can you use the same function within the function you're currently defining? I'm new to Python so I'm just trying to understand. – J82 Nov 07 '14 at 02:32
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14@J82: The concept used here is called recursion ( http://en.wikipedia.org/wiki/Recursion_(computer_science) ) - a function calling itself is perfectly fine and often useful. – schnaader Nov 07 '14 at 10:06
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5The recursive function will raise a [`RecursionError`](https://docs.python.org/library/exceptions.html#RecursionError) for any number larger than 998 (try `factorial(999)`) unless you [increase Python's recursion limit](https://stackoverflow.com/a/3323008/3064538) – Boris Verkhovskiy Dec 15 '19 at 19:15
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2Raising CPython's recursion limit is dangerous -- you can kill the interpreter. Just [don't use recursion in Python](https://stackoverflow.com/questions/67988828/why-is-python-recursion-so-expensive-and-what-can-we-do-about-it) if it can be helped (it usually can, as this example illustrates). – ggorlen Oct 14 '21 at 18:40
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factorial(999) ≈ 4.02 × 10^2564, so it's unlikely you would want to compute such a large number anyway. – snibbets Jun 22 '23 at 10:23
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1[Starting with Python 3.9](https://bugs.python.org/issue37315), passing a `float` to this function will raise a `DeprecationWarning`. If you want to do that, you need to convert `n` to an `int` explicitly: `math.factorial(int(n))`, which will discard anything after the decimal, so you might want to check that [`n.is_integer()`](https://stackoverflow.com/questions/21583758/how-to-check-if-a-float-value-is-a-whole-number) – Boris Verkhovskiy Nov 22 '19 at 11:47
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)

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`operator.mul` could be used instead of `long.__mul__` and it would work in both [Python 2](https://docs.python.org/2/library/operator.html#operator.mul) and [Python 3](https://docs.python.org/3/library/operator.html#operator.mul). – Cristian Ciupitu Nov 22 '21 at 00:55
def factorial(n):
if n < 2:
return 1
return n * factorial(n - 1)
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3`factorial(999)` (and above) will raise a `RuntimeError` unless you [increase Python's recursion limit](https://stackoverflow.com/a/3323008/3064538) – Boris Verkhovskiy Nov 22 '19 at 11:43
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 a '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 the built-in function math
def fact(n):
return math.factorial(n)
Check running results
cProfile.run('fact(126000)')
5 function calls in 0.272 seconds

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1I think this while loop looks a little bit cleaner def fact(n): ret = 1 while n > 1: n, ret = n - 1, ret * n return ret – edilio May 18 '18 at 15:13
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1Looks great, shouting (large font) that recursion is disastrous, but can you back this up? Yes, you need a lot of stack, but only for a very short time. And yesterday's "a lot" is today's "just a little", especially in computing. We write high level code in order to not waste our time, and recursion helps with that. You don't need low level code a lot for performance reasons, today – Roland Nov 07 '21 at 13:11
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It also depends on the context you're using the factorial in -- recursive functions have the benefit of being cache-able, this can be particularly helpful with factorials – Schalton Feb 19 '23 at 16:43
If you are using Python 2.5 or older, try
from operator import mul
def factorial(n):
return reduce(mul, range(1, n+1))
For newer versions of Python, there is factorial in the math module as given in other answers here.

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2This is a Python 2-only answer, `reduce` was removed from Python 3. – Boris Verkhovskiy Nov 22 '19 at 11:37
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@Boris, in Python3 you just need to add `from functools import reduce` – John La Rooy Nov 24 '19 at 22:55
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It was removed for a reason, you shouldn't use it https://www.artima.com/weblogs/viewpost.jsp?thread=98196 – Boris Verkhovskiy Nov 24 '19 at 23:43
def fact(n):
f = 1
for i in range(1, n + 1):
f *= i
return f

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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))

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Non-recursive solution, no imports:
def factorial(x):
return eval(' * '.join(map(str, range(1, x + 1))))

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3It would be interesting to compare this to some of the other methods presented here. My guess is it's off-the-charts inefficient. – Mark Ransom Jul 10 '21 at 03:16
You can also make it in one line recursively if you like it. It is just a matter of personal choice. Here we are using inline if else
in Python, which is similar to the ternary operator in Java:
Expression1 ? Expression2 : Expression3
One line
function call
approach:def factorial(n): return 1 if n == 0 else n * factorial(n-1)
One line
lambda
function approach:(although it is not recommended to assign lambda functions directly to a name, as it is considered a bad practice and may bring inconsistency to your code. It's always good to know. See PEP8.)
factorial = lambda n: 1 if n == 0 else n * factorial(n-1)

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