Question

好的，所以我试图用Python制作一个扑克游戏..我想将已用过的牌从列表中移动到另一个列表，我将存储用过的牌。我使用此代码：

deck = ['AS', 'KS', 'QS', 'JS', '10S', '9S', '8S', '7S', '6S', '5S', '4S', '3S', '2S',\
         'AD', 'KD', 'QD', 'JD', '10D', '9D', '8D', '7D', '6D', '5D', '4D', '3D', '2D',\
         'AC', 'KC', 'QC', 'JC', '10C', '9C', '8C', '7C', '6C', '5C', '4C', '3C', '2C',\
         'AH', 'KH', 'QH', 'JH', '10H', '9H', '8H', '7H', '6H', '5H', '4H', '3H', '2H']

used = []
p1 = []
p2 = []
a = 0

while (a < 2):
    drawn_card = random.choice(deck)
    deck.append(drawn_card)
    deck = [f for f in deck if f not in used]
    p1.append(drawn_card)
    a+=1

# Don't read anything below this line

card1 = p1[0]
card2 = p1[1]
f_card1 = "Name/Suit"
f_card2 = "Name/Suit"

# Formatting Player1's first drawn card

if card1[0] == "A":
    f_card1 = "Ace"
elif card1[0] == "K":
    f_card1 = "King"
elif card1[0] == "Q":
    f_card1 = "Queen"
elif card1[0] == "J":
    f_card1 = "Jack"
else:
    f_card1 = card1[0]


if card1[1] == "S":
    f_card1 = f_card1 + " of Spades"
elif card1[1] == "D":
    f_card1 = f_card1 + " of Diamonds"
elif card1[1] == "C":
    f_card1 = f_card1 + " of Clubs"
else:
    f_card1 = f_card1 + " of Hearts"

# Formatting Player1's second drawn card

if card2[0] == "A":
    f_card2 = "Ace"
elif card2[0] == "K":
    f_card2 = "King"
elif card2[0] == "Q":
    f_card2 = "Queen"
elif card2[0] == "J":
    f_card2 = "Jack"
else:
    f_card2 = card2[0]


if card2[1] == "S":
    f_card2 = f_card2 + " of Spades"
elif card2[1] == "D":
    f_card2 = f_card2 + " of Diamonds"
elif card2[1] == "C":
    f_card2 = f_card2 + " of Clubs"
else:
    f_card2 = f_card2 + " of Hearts"

print(f_card1)
print(f_card2)

while函数是我的问题所在。我打印时有时会得到相同的卡片。我怎么能解决这个问题？我的错误在哪里？提前谢谢

Answer 1

您只需将提取的卡片附加到used列表，而不是回到deck：

used.append(drawn_card)   #instead of deck.append(..)

Answer 2

while (a < 2):
    drawn_card = random.choice(deck)
    used.append(drawn_card)
    deck = [f for f in deck if f not in used]
    p1.append(drawn_card)
    a+=1

<强> used.append（drawn_card）

Answer 3

简单的解决方法是使用其他人建议的used.append(drawn_card)。更好的方法是随机洗牌，然后从中切割所需数量的卡片：

CARDS_PER_HAND = 2

random.shuffle(deck)
hand = deck[:CARDS_PER_HAND]

但是，你可能在同一套牌上交易多手牌。你可以通过跟踪进入洗牌的牌组的偏移来处理，然后从那里切片，但这很麻烦。或者你可以制作deck的副本，随意播放，并为每只手弹出所需数量的卡片：

shuffled_deck = deck[:]    # make a copy of deck
random.shuffle(shuffled_deck)
hand1 = [shuffled_deck.pop() for i in range(CARDS_PER_HAND)]
hand2 = [shuffled_deck.pop() for i in range(CARDS_PER_HAND)]
... etc.

最简单的方法可能是使用itertools.islice()从迭代器中获取第一个 n 项：

from itertools import islice

random.shuffle(deck)
shuffled_deck = iter(deck)    # returns an iterator
hand1 = list(islice(shuffled_deck, CARDS_PER_HAND))
hand2 = list(islice(shuffled_deck, CARDS_PER_HAND))
... etc.

对于纸牌游戏Python，将一个列表中的项目移动到另一个列表

3 个答案: