python之如何优化 numpy 调用的数据生成

我想知道如何使以下代码更短和/或更高效。我可以(或者我应该)通过使用功能方法摆脱 for 循环，还是我应该使用 numpy 中的方法？

代码计算整数数组的期望值。

vals = np.arange(self.n+1) 
 
# array of probability of each value in vals 
parr = np.ones(len(vals)) 
for i in range(len(vals)): 
    parr[i] *= self.prob(vals[i]) 
 
return np.dot(vals,parr) 

按照评论中的要求，方法 prob() 的实现:

def prob(self, x): 
 
    """Computes probability of removing x items 
 
    :param x: number of items to remove 
    :returns: probability of removing x items 
    """ 
 
    # p is the probability of removing an item 
    # sl.choose computes n choose x 
    return sl.choose(self.n, x) * (self.p**x) * \ 
           (1-self.p)**(self.n-x) 

请您参考如下方法:

我认为它会更快:

vals = np.arange(self.n+1) 
 
# array of probability of each value in vals 
parr = self.prob(vals)      
 
return np.dot(vals,parr) 

和函数:

def prob(list_of_x): 
 
    """Computes probability of removing x items 
 
    :param list_of_x: numbers of items to remove 
    :returns: probability of removing x items 
    """ 
 
    # p is the probability of removing an item 
    # sl.choose computes n choose x 
    return np.asarray([sl.choose(self.n, e) for e in list_of_x]) * (self.p ** list_of_x) * \ 
           (1-self.p)**(self.n - list_of_x) 

因为 numpy 更快:

import timeit 
 
import numpy as np 
 
list_a = [1, 2, 3] * 1000 
list_b = [4, 5, 6] * 1000 
 
np_list_a = np.asarray(list_a) 
np_list_b = np.asarray(list_b) 
 
print(timeit.timeit('[a * b for a, b in zip(list_a, list_b)]', 'from __main__ import list_a, list_b', number=1000)) 
print(timeit.timeit('np_list_a * np_list_b', 'from __main__ import np_list_a, np_list_b', number=1000)) 

结果:

0.19378583212707723 
0.004333830584755033