如何将矩阵的行值与参数的索引相关联

时间:2016-06-05 14:23:34

标签: python numpy matrix simulation

我想为参数创建一个表(对于1到3的产品,我们可以说X_n) 参数可以采用给定概率值的某些值。

考虑n = 1我可以使用以下代码模拟X_1可以采用的值

from scipy.stats import rv_discrete 
values = [-2, 0, 2]
probabilities = [0.40, 0.2, 0.4]
distrib = rv_discrete(values=(values, probabilities))
distrib.rvs(size=10)

此代码返回以下结果:

  

数组([2,-2,0,0,2,0,0,0,0,-2])

现在我想针对真实案例,n = 3

我尝试按如下方式创建表但失败了;

num_products=3
for n in range(num_products):
Xvalues_[n] = np.array([(-2,0,2), (0,2,3),(-2,0,2)])
Xprobabilities[n] = np.array([(0.4,0.2,0.4), (0.2,0.4,0.4),(0.4,0.4,0.2)])

1 个答案:

答案 0 :(得分:0)

我不确定你想要做什么,所以这只是打印出每个

的结果

import numpy as np
from scipy.stats import rv_discrete 

Xvalues = np.array([(-2,0,2), (0,2,3),(-2,0,2)])
Xprobabilities = np.array([(0.4,0.2,0.4), (0.2,0.4,0.4),(0.4,0.4,0.2)])
for values,probabilities in zip(Xvalues, Xprobabilities):
    distrib = rv_discrete(values=(values,probabilities))
    print(distrib.rvs(size=10))