生日悖论优化代码速度

Question

我正在编写一个模拟器，该模拟器可以检查每个盒子需要容纳多少个球，或者每个盒子至少需要有2个球（生日悖论）。我用python编写了脚本，它是如此之慢。我的朋友用C ++编写了一个程序，脚本速度更快。对于箱数= 1000，在我的python脚本中，它需要几分钟才能达到“永恒”。

有我的代码：

import numpy as np
import matplotlib.pyplot as plt


def check_every_box_is_occupied(boxes):
    for box in boxes:
        if box == 0:
            return False
    return True


def check_birthday_paradox(boxes):
    for box in boxes:
        if box >= 2:
            return True
    return False


def main():
    number_of_tests = 100
    birthday_paradox_graph = [[], []]
    every_box_is_occupied_graph = [[], []]
    boxes_max_num = 1000
    for number_of_boxes in range(10, boxes_max_num + 1, 1):
        print(number_of_boxes)
        average_frequency_birthday_paradox = 0
        average_frequency_every_box_is_occupied = 0
        for index in range(number_of_tests):
            for number_of_balls in range(1, number_of_boxes ** 2, 1):
                boxes = np.array([0] * number_of_boxes)
                for i in range(number_of_balls):
                    boxes[np.random.randint(number_of_boxes)] += 1
                if check_birthday_paradox(boxes):
                    average_frequency_birthday_paradox += number_of_balls
                    break
            for number_of_balls in range(number_of_boxes, number_of_boxes ** 2, 1):
                boxes = np.array([0] * number_of_boxes)
                for i in range(number_of_balls):
                    boxes[np.random.randint(number_of_boxes)] += 1
                if check_every_box_is_occupied(boxes):
                    average_frequency_every_box_is_occupied += number_of_balls
                    break

        plt.rcParams.update({'font.size': 15})
        birthday_paradox_graph[0].append(number_of_boxes)
        birthday_paradox_graph[1].append(average_frequency_birthday_paradox / number_of_tests)
        every_box_is_occupied_graph[0].append(number_of_boxes)
        every_box_is_occupied_graph[1].append(average_frequency_every_box_is_occupied / number_of_tests)
    plt.figure(1)
    plt.plot(birthday_paradox_graph[0], birthday_paradox_graph[1], 'ko')
    plt.figure(2)
    plt.plot(every_box_is_occupied_graph[0], every_box_is_occupied_graph[1], 'ko')
    plt.show()


if __name__ == '__main__':
    main()

您能帮我改进代码以使其更快吗？可以在原始python中使用吗？