How do I go about computing a factorial of an integer in Python?
Function for factorial in Python
Each Answer to this Q is separated by one/two green lines.
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(n1)
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.
On Python 2.6 and up, try:
import math
math.factorial(n)
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)
def factorial(n):
if n < 2:
return 1
return n * factorial(n  1)
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 builtin function math
def fact(n):
return math.factorial(n)
Check running results
cProfile.run('fact(126000)')
5 function calls in 0.272 seconds
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.
def fact(n):
f = 1
for i in range(1, n + 1):
f *= i
return f
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))
Nonrecursive solution, no imports:
def factorial(x):
return eval(' * '.join(map(str, range(1, x + 1))))
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(n1)

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(n1)