Python 中的循环和 choice() 函数

Question

我是一名新手程序员，我试图在我的书中为挑战编写代码，其中包含一个循环，该循环随机获取数字和/或字母以宣布彩票中奖者。我正在尝试编写代码：从元组中取出一个未被拾取的随机对象 4 次将每个对象存储在列表中打印清单from random import choice #Import choice() function from the random modulelottery_1 = (1,2,3,4,5,6,7,8,9,'a','b','c','d','e')lottery_winner = []for i in range(4): #Takes 4 random numbers and/or letters    random = choice(lottery_1)    if random not in lottery_winner:        lottery_winner.append(pulled_number)print('$1M Winner\n')print(lottery_winner)有时它只选择 2 个字符结果：$1M Winner[1, 'e']>>>为什么会这样？我可以更改什么以使其选择 4 个字符？

Answer 1

实际上，它是选择四个字符。但是，当所选字符之一已在 中时lottery_winner，不会添加它。在这种情况下，您最终得到的总结果少于四个。

lenik 的回答是最实用的解决方案。但是，如果您对如何使用该choice函数进行操作的逻辑感到好奇，请记住，您需要在帽子出现重复时再次选择，或者您需要从帽子中消除选项当你去时。

选项 #1，每当获胜者重复时再试一次：

for i in range(4):

    new_winner = False # Initially, we have not found a new winner yet.

    while not new_winner: # Repeat the following as long as new_winner is false:

        random = choice(lottery_1)

        if random not in lottery_winner:

            lottery_winner.append(pulled_number)

            new_winner = True # We're done! The while loop won't run again.

                              # (The for loop will keep going, of course.)

选项 #2，每次都从列表中删除获胜者，这样他们就不会被再次选中：

for i in range(4):

    random = choice(lottery_1)

    lottery_winner.append(pulled_number)

    lottery_1.remove(pulled_number) # Now it can't get chosen again.

请注意，remove()删除指定值的第一个实例，在值不唯一的情况下，这可能不会执行您想要的操作。

Answer 2

import random

lottery_1 = (1, 2, 3, 4, 5, 6, 7, 8, 9, 'a', 'b', 'c', 'd', 'e')

'''Instead of using the choice method which can randomly grab the same value,

i suggest that you use the sample method which ensures that you only get randomly unique values'''

# The k value represents the number of items that you want to get

lottery_winner = random.sample(lottery_1, k=4)

print('$1M Winner\n')

print(lottery_winner)

Answer 3

这对我有用：

>>> import random

>>> lottery_1 = (1,2,3,4,5,6,7,8,9,'a','b','c','d','e')

>>> random.sample(lottery_1, 4)

[1, 7, 'a', 'e']

>>>