用Python制作一个交易程序集。

Question

主要是因为我不相信经典游戏的结果，做了一个交易(见https://math.stackexchange.com/questions/608957/monty-hall-problem-extended)我做了这个小程序.事实上，如果有3扇门，而问答大师打开一扇门，然后你从你最初选择的门上切换，你的机会从33%上升到67%！此程序是一个通用版本，您可以选择门的数目和测验大师将打开多少扇门，它计算出最初选择的价格(基本上是比门的数目多1)的机会，以及在打开门后更改所选门的机会。

我被使用Python集的优雅解决方案所吸引，但我不知道这是否是最有效的方法。与其他方法相比，随着更多的门和门的打开，它看起来变得更高效了。

谢谢你的意见..。

#!/usr/bin/env python
'''  application of Make a deal statistics
     application is using sets {}
     for reference:
 https://math.stackexchange.com/questions/608957/monty-hall-problem-extended
'''
import random


def Make_a_Deal(doors, doors_to_open):
    '''  Generalised function of Make_a_Deal. Logic should be self explanatory
         Returns win_1 for the option when no change is made in the choice of
         door and win_2 when the option to change is taken.'''

    win_1, win_2 = False, False

    doors_set = set(range(1, doors+1))
    price_set = set(random.sample(doors_set, 1))
    choice1_set = set(random.sample(doors_set, 1))
    open_set = set(random.sample(doors_set.difference(price_set).
                   difference(choice1_set), doors_to_open))
    choice2_set = set(random.sample(doors_set.difference(open_set).
                      difference(choice1_set), 1))
    win_1 = choice1_set.issubset(price_set)
    win_2 = choice2_set.issubset(price_set)

    return win_1, win_2


def main():
    '''  input:
         - throws: number of times to Make_a_Deal (must be > 0)
         - doors: number of doors to choose from (must be > 2)
         - doors_to_open: number of doors to be opened before giving the option
           to change the initial choice (must be > 0 and <= doors-2)'''

    try:
        throws = int(input('how many throws: '))
        doors = int(input('how many doors: '))
        doors_to_open = int(input('how many doors to open: '))
        if (throws < 1) or (doors < 3) or \
                (doors_to_open > doors-2) or (doors_to_open < 1):
            print('invalid input')
            return

    except Exception as e:
        print('invalid input: ', e)
        return

    number_of_wins_1, number_of_wins_2, counter = 0, 0, 0

    while counter < throws:
        win_1, win_2 = Make_a_Deal(doors, doors_to_open)

        if win_1:
            number_of_wins_1 += 1
        if win_2:
            number_of_wins_2 += 1

        counter += 1
        print('completion is {:.2f}%'.
              format(100*counter/throws), end='\r')

    print('number of wins option 1 is {:.2f}%: '.
          format(100*number_of_wins_1/counter))
    print('number of wins option 2 is {:.2f}%: '.
          format(100*number_of_wins_2/counter))


if __name__ == '__main__':
    main()

Answer 1

我第二次使用@Graipher的答案函数作为sets采样的辅助函数，但是我们可以通过使用以下定义来改进它的解决方案，该定义自动将random.sample返回的值转换为set：

def random_choice(population, k=1):
    return {*random.sample(population, k)}

对于代码的读者来说，您的文档字符串没有多大意义。相反，为用户编写docstring，记录函数/模块正在做什么，而不是如何。

您不需要在其名称中明确变量的类型，选择反映存储内容的名称，而不是如何存储。

我会将输入收集与main分开。它的验证是好的，尽管有点过了:如果抛出不是正的，那么就不会有任何抛出，而统计数据将是0，这是一致的。让您的main与数据的来源无关，以简化可重用性和测试。如果您确实想报告进度，则应该接受回调，以使行为是可选的，并且易于互变。

如果将输入收集与main分离，就可以很容易地将input切换到argparse或其他任何GUI。

不要抓住通用的Exceptions在这里，你只关心ValueErrors，它将避免打印Invalid input的例外，例如KeyboardInterrupt。

您还应该将错误消息打印到sys.stderr，并且可以选择使用sys.exit()使用非零返回代码退出程序。

喜欢使用for循环比whiles显式计数器。

拟议改进：

#!/usr/bin/env python3

"""Application of Make a deal statistics

Application is using sets {}

For reference:
    https://math.stackexchange.com/questions/608957/monty-hall-problem-extended
"""

import sys
import random


def random_choice(population, k=1):
    return {*random.sample(population, k)}


def make_a_deal(doors, doors_to_open):
    """Generalised function of Make a Deal.

    Returns a pair of boolean indicating:
     - wether the original choice was a win for the first element
     - wether the changed choice was a win for the second element
    """

    doors = set(range(doors))
    prize = random_choice(doors)
    original_choice = random_choice(doors)

    opened = random_choice(doors - prize - original_choice, doors_to_open)
    changed_choice = random_choice(doors - opened - original_choice)

    return original_choice <= prize, changed_choice <= prize


def main(throws, doors, doors_to_open, progression_callback=None):
    """Gather statistics about repeated simulations of Make a Deal game.

    Parameters:
     - throws: number of simulations to run
     - doors: number of doors to choose from (must be > 2)
     - doors_to_open: number of doors to be opened before giving the
    option to change the initial choice (must be > 0 and <= doors - 2)
    """
    if doors < 3:
        raise ValueError('doors must be at least 3')

    if not (0 < doors_to_open <= doors - 2):
        raise ValueError('doors to open mismatch the number of doors')

    original_wins = changed_wins = 0
    for throw in range(throws):
        original_win, changed_win = make_a_deal(doors, doors_to_open)

        if original_win:
            original_wins += 1

        if changed_win:
            changed_wins += 1

        if progression_callback is not None:
            progression_callback(throw / throws)

    return original_wins / throws, changed_wins / throws


def show_completion(ratio):
    print('Completion is {:.2%}'.format(ratio), end='\r')


if __name__ == '__main__':
    try:
        throws = int(input('How many throws: '))
        doors = int(input('How many doors: '))
        doors_to_open = int(input('How many doors to open: '))
        wins1, wins2 = main(throws, doors, doors_to_open, show_completion)
    except ValueError as e:
        print('Invalid input:', file=sys.stderr, end=' ')
        sys.exit(e)

    print('Number of wins without changing is {:.2%}'.format(wins1))
    print('Number of wins after changing is {:.2%}'.format(wins2))

Answer 2

Python有一个正式的风格指南，PEP8。它建议对变量和函数名使用lower_case。docstrings还有一个样式指南，PEP257，它建议使用三重双引号("""So, like this""")。

您的模块docstring可能应该包含一个句子摘要，比如“扩展Monty问题的n扇门，其中k是由游戏大师打开的。”

遗憾的是，random.choice不适用于sets，因为它们是不可索引的。但是，您经常使用它的替代random.sample(x, 1)，因此您可能希望将其放入一个函数中：

def random_choice(x):
    return random.sample(x, 1)[0]

这也意味着您需要使用{random_choice(x)}，而不是set(random_choice(x))，因为现在这是一个单独的元素，不再是一个包含一个元素的列表。

doors_set.difference(price_set).difference(choice1_set)和doors_set - price_set - choice1_set是一样的，它更短，至少也一样可读性。

类似地，choice1_set.issubset(price_set)等同于choice1_set <= price_set (尽管这里更有表现力的版本可能更容易理解)。

您不需要在Python中声明变量，特别是如果您以任何方式覆盖它的值(win_1和win_2)。

我还会将option_1和option_2重命名为更容易理解的no_change和change。

由于您再也不使用doors，所以我将从名称中删除所有尾随的_set。

如果您从0或1开始枚举您的门，则没有什么不同，因此为了简单起见，我将使用set(range(doors))。

这样，您的make_a_deal函数将变成：

def make_a_deal(doors, doors_to_open):
    '''  Generalised function of Make_a_Deal. Logic should be self explanatory
         Returns win_1 for the option when no change is made in the choice of
         door and win_2 when the option to change is taken.'''

    doors = set(range(doors))
    price = {random_choice(doors)}
    choice1 = {random_choice(doors)}
    open_doors = set(random.sample(doors - price - choice1, doors_to_open))
    choice2 = {random_choice(doors - open_doors - choice1)}
    no_change = choice1 <= price
    change = choice2 <= price

    return no_change, change