一副牌，交替发牌

Question

import random
import itertools    
suit = 'CSHD'
rank = '23456789TJQKA'
deck_of_cards = [''.join(card) for card in itertools.product(rank, suit)]

def draw_card(deck,hand, count):
    for i in range(count):
        hand.append(deck.pop())

def hand_selection(deck):
    shuffled_deck = random.sample(deck, len(deck))
    dealer_hand = []
    player_hand = []
    draw_card(shuffled_deck, dealer_hand, 1)
    draw_card(shuffled_deck, player_hand, 1)
    draw_card(shuffled_deck, dealer_hand, 1)
    draw_card(shuffled_deck, player_hand, 1)
    return dealer_hand,player_hand

这里尽量遵循发牌原则。最初我想随机生成一张牌，将其从shuffled_deck 中删除，但我们不会通过随机翻牌来发牌，是吗？出于我的目的，代码做了它应该做的事情，但它看起来不太好。 （是的，shuffled_deck、player_hand 和 dealer_hand 实际上都是全局变量，但暂时无关紧要。）

以这种方式发牌的更优雅（阅读：pythonic）的解决方案是什么？例如，我们是否在 3 名玩家之间玩类似中国扑克的游戏。像这样一个一个地处理似乎太乏味了。

Answer 1

我会将卡片组设为一个类，以将其相关功能保持在一起：

class Deck(object):

    def __init__(self, suits='CSHD', ranks='23456789TJQKA'):
        self.cards = [''.join(card) for card in itertools.product(ranks, suits)]

    def shuffle(self):
        random.shuffle(self.cards)

    def deal_cards(self, hand, count=1):
        for _ in range(count):
            hand.append(self.cards.pop())

    def return_cards(self, hand):
        self.cards.extend(hand)

为了简化交易：

deck = Deck()
deck.shuffle()
dealer_hand = []
player_hand = []
for _ in range(2): # number of rounds to deal
    for hand in (dealer_hand, player_hand): # hands to deal to
         deck.deal(hand)

这更容易适应更多玩家和/或更多回合。