python 二项分布_用python计算二项分布概率矩阵
我相信scipy.stats.binom可以按照您想要的方式利用广播.
# Binomial PMF: Pr(X=k) = choose(n, k) * p**k * (1-p)**(n-k)
# Probability of getting exactly k successes in n trials
>>> from scipy.stats import binom
>>> n = np.arange(1, N+1, dtype=np.int64)
>>> dist = binom(p=0.25, n=n)
>>> M = dist.pmf(k=np.arange(N+1, dtype=np.int64)[:, None])
>>> M.round(2)
array([[0.75, 0.56, 0.42, 0.32, 0.24, 0.18, 0.13, 0.1 , 0.08, 0.06],
[0.25, 0.38, 0.42, 0.42, 0.4 , 0.36, 0.31, 0.27, 0.23, 0.19],
[0. , 0.06, 0.14, 0.21, 0.26, 0.3 , 0.31, 0.31, 0.3 , 0.28],
[0. , 0. , 0.02, 0.05, 0.09, 0.13, 0.17, 0.21, 0.23, 0.25],
[0. , 0. , 0. , 0. , 0.01, 0.03, 0.06, 0.09, 0.12, 0.15],
[0. , 0. , 0. , 0. , 0. , 0. , 0.01, 0.02, 0.04, 0.06],
[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0.01, 0.02],
[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ],
[0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. , 0. ]])
此处,行为k(索引为0),列为n(索引为1)：
>>> from math import factorial as fac
>>> def manual_pmf(p, n, k):
... return fac(n) / (fac(k) * fac(n - k)) * p**k * (1-p)**(n-k)
>>> manual_pmf(p=0.25, n=3, k=2)
0.140625 # (2, 2) in M because M's columns are effectively 1-indexed
您还可以将n从零开始,以获得在行和列上都为0索引的数组：
>>> n = np.arange(N+1, dtype=np.int64) # M.shape is (11, 11)