Statistics: combinations in Python

Question

I need to compute combinatorials (nCr) in Python but cannot find the function to do that in math, numpy or stat libraries. Something

Answer 1

If your program has an upper bound to n (say n <= N) and needs to repeatedly compute nCr (preferably for >>N times), using lru_cache can give you a huge performance boost:

from functools import lru_cache

@lru_cache(maxsize=None)
def nCr(n, r):
    return 1 if r == 0 or r == n else nCr(n - 1, r - 1) + nCr(n - 1, r)

Constructing the cache (which is done implicitly) takes up to O(N^2) time. Any subsequent calls to nCr will return in O(1).

Answer 2

A quick search on google code gives (it uses formula from @Mark Byers's answer):

def choose(n, k):
    """
    A fast way to calculate binomial coefficients by Andrew Dalke (contrib).
    """
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in xrange(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

choose() is 10 times faster (tested on all 0 <= (n,k) < 1e3 pairs) than scipy.misc.comb() if you need an exact answer.

def comb(N,k): # from scipy.comb(), but MODIFIED!
    if (k > N) or (N < 0) or (k < 0):
        return 0L
    N,k = map(long,(N,k))
    top = N
    val = 1L
    while (top > (N-k)):
        val *= top
        top -= 1
    n = 1L
    while (n < k+1L):
        val /= n
        n += 1
    return val

Answer 3

Here is an efficient algorithm for you

for i = 1.....r

   p = p * ( n - i ) / i

print(p)

For example nCr(30,7) = fact(30) / ( fact(7) * fact(23)) = ( 30 * 29 * 28 * 27 * 26 * 25 * 24 ) / (1 * 2 * 3 * 4 * 5 * 6 * 7)

So just run the loop from 1 to r can get the result.

Answer 4

Using only standard library distributed with Python:

import itertools

def nCk(n, k):
    return len(list(itertools.combinations(range(n), k)))

Answer 5

That's probably as fast as you can do it in pure python for reasonably large inputs:

def choose(n, k):
    if k == n: return 1
    if k > n: return 0
    d, q = max(k, n-k), min(k, n-k)
    num =  1
    for n in xrange(d+1, n+1): num *= n
    denom = 1
    for d in xrange(1, q+1): denom *= d
    return num / denom

Answer 6

This is @killerT2333 code using the builtin memoization decorator.

from functools import lru_cache

@lru_cache()
def factorial(n):
    """
    Calculate the factorial of an input using memoization
    :param n: int
    :rtype value: int
    """
    return 1 if n in (1, 0) else n * factorial(n-1)

@lru_cache()
def ncr(n, k):
    """
    Choose k elements from a set of n elements,
    n must be greater than or equal to k.
    :param n: int
    :param k: int
    :rtype: int
    """
    return factorial(n) / (factorial(k) * factorial(n - k))

print(ncr(6, 3))