python - 从一副纸牌中生成52张随机纸牌，而不会产生重复

到目前为止，这是我提出的：

import time
from random import randint

Suits = [
    ["Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Joker", "Queen", "King"], #hearts
    ["Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Joker", "Queen", "King"], #clubs
    ["Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Joker", "Queen", "King"], #spades
    ["Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10", "Joker", "Queen", "King"]  #diamonds
    ]


for x in range(0,52):
    #selection of random card and suit
    Suit = randint(0,3)
    Card = randint(0,12)

    # prints what card was received from the deck
    if Suit == 0:
        print("You got a", Suits[0][Card], "of Hearts")
    elif Suit == 1:
        print("You got a", Suits[1][Card], "of Clubs")
    elif Suit == 2:
        print("You got a", Suits[2][Card], "of Spades")
    else:
        print("You got a", Suits[3][Card], "of Diamonds")

这使我可以从一副纸牌中生成一张随机纸牌53次，但最终却得到了重复。我将如何避免这种情况？

最佳答案

如果您确实不需要二维数组，则可以更简单地执行此操作。如果您只有一个平面列表，则可以使用Python的random库轻松完成此操作：

import random

cards = [(s, v) for s in ['H', 'S', 'C', 'D']
         for v in [str(i) for i in range(2, 11)] + list("JKQA")]

random.shuffle(cards)

列表推导用于将cards设置为具有西服和等级组合的元组。然后使用random.shuffle来随机化就位卡列表，以便您可以迭代/拉出列表末尾的卡。

关于python - 从一副纸牌中生成52张随机纸牌，而不会产生重复，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/48677430/