python permutation
import itertools a = [1, 2, 3] n = 3 perm_iterator = itertools.permutations(a, n) for item in perm_iterator: print(item)
permutation python
def permutations(s): if len(s) <= 1: yield s else: for i in range(len(s)): for p in permutations(s[:i] + s[i+1:]): yield s[i] + p input = 'ABCD' for permutation in enumerate(permutations(input)): print repr(permutation[1]), print
permutation test python
# This is a possible implementation of permutation test using monte-carlo method: def exact_mc_perm_test(xs, ys, nmc): n, k = len(xs), 0 diff = np.abs(np.mean(xs) - np.mean(ys)) zs = np.concatenate([xs, ys]) for j in range(nmc): np.random.shuffle(zs) k += diff < np.abs(np.mean(zs[:n]) - np.mean(zs[n:])) return k / nmc Example: >>> xs = np.array([12.6, 11.4, 13.2, 11.2, 9.4, 12.0]) >>> ys = np.array([16.4, 14.1, 13.4, 15.4, 14.0, 11.3]) >>> exact_mc_perm_test(xs, ys, 30000) 0.019466666666666667
Source: stackoverflow.com