制作python程序来处理卡片的麻烦。

Question

我想要打一场纸牌游戏。我坚持的是处理卡片。我所做的就是用每张卡片制作一张单词并给它一个值，因为有些东西比其他卡片更有价值。我想到的是将字典分为4个部分，或者每个字典制作4个副本，然后从每个字典中删除39张卡片（每个人留下13张卡片）。这甚至可能还是我以错误的方式解决这个问题？

from random import randint
deck = {}
def makeDeck(deck):
  suit = ['Club', 'Spade', 'Heart', 'Diamond']
  whichSuit = 0
  whichNum = 2
  count = 1
  while count != 52:
    if whichNum == 11:
      whichNum = 'Jack'
    if whichNum == 12:
      whichNum = 'Queen'
    if whichNum == 13:
      whichNum = 'King'
    if whichNum == 14:
      whichNum = 'Ace'
    deck[str(whichNum)+' '+suit[whichSuit]] = count
    count += 1
    if whichNum == 'Jack':
      whichNum = 11
    if whichNum == 'Queen':
      whichNum = 12
    if whichNum == 'King':
      whichNum = 13
    if whichNum == 'Ace':
      whichNum = 14
    whichNum += 1
    if count == 13 or count == 26 or count == 39:
     whichSuit += 1
     whichNum = 2
def dealCards(deck):
  me = deck
  comp1 = deck
  comp2 = deck
  comp2 = deck

（对不起，如果代码错了，这是我的第一篇文章，谢谢）

Answer 1

听起来像是一个使用课程的好机会！我会这样做：

from random import shuffle

class Cards:
    def __init__(self):
        values = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K']
        suites = ['H', 'S', 'C', 'D']
        self.deck = [j + i for j in values for i in suites]

    def shuffle(self):
        shuffle(self.deck)

    def deal(self, n_players):
        self.hands = [self.deck[i::n_players] for i in range(0, n_players)]

c = Cards()
print c.deck
c.shuffle()
print c.deck
c.deal(4)
print c.hands

Answer 2

我对Python中的字典函数不太熟悉，但我要做的是使用卡片对象和 set 列表与shuffle。

from random import shuffle    
class Card:
    def __init__(self,suit,num):
        self.suit = suit
        self.num = num

deck = list()
suits = ['Diamond', 'Heart', 'Spade', 'Club']

nums = ['2','3','4','5','6','7','8','9','10','J','Q','K','A']

for suit in suits: #This is the code that actually makes a deck
    for num in nums:
        deck.append(Card(suit,num))

shuffle(deck)
for number in range(13):
    for player in range(4):
        #deal cards here using deck.pop()
        print(deck.pop().num) #just to prove it works randomly =P        

我希望能回答你的问题（因为这是你的第一个问题，这是我的第一个答案）

编辑：不推荐使用Oops集。改为使用内置集。

Edit2：并且set.pop（）并不是真正随机的，它从进一步阅读中出现，只是随意的。男孩脸红了。

Answer 3

您可以选择使用python的内置函数random.shuffle。不要打扰词典;只需创建一个卡片列表并将其整理：

>>> import random
>>> ranks = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'K', 'Q']
>>> suits = ['C', 'D', 'H', 'S']
>>> cards = [[rank, suit] for rank in ranks for suit in suits]
>>> random.shuffle(cards)
>>> cards
[['J', 'S'], ['2', 'S'], ['3', 'S'], ['9', 'S'], ['9', 'D'], ['5', 'S'], 
 ['8', 'H'], ['A', 'C'], ['4', 'D'], ['Q', 'H'], ['2', 'C'], ['Q', 'D'], 
 ['7', 'H'], ['4', 'C'], ['7', 'S'], ['6', 'C'], ['K', 'H'], ['6', 'S'], 
 ['9', 'C'], ['9', 'H'], ['A', 'H'], ['J', 'C'], ['2', 'D'], ['J', 'H'], 
 ['3', 'H'], ['4', 'H'], ['8', 'C'], ['Q', 'S'], ['10', 'S'], ['A', 'S'], 
 ['K', 'S'], ['5', 'D'], ['10', 'D'], ['8', 'D'], ['7', 'C'], ['5', 'C'], 
 ['Q', 'C'], ['3', 'D'], ['8', 'S'], ['6', 'H'], ['A', 'D'], ['2', 'H'], 
 ['6', 'D'], ['K', 'D'], ['10', 'C'], ['5', 'H'], ['4', 'S'], ['K', 'C'], 
 ['7', 'D'], ['10', 'H'], ['3', 'C'], ['J', 'D']]

如果您需要自己动手，请考虑Fisher-Yates shuffle。这很简单。

冒着说明可怕的明显的风险，一旦你有一个洗牌清单，你可以简单地通过切片来处理它：

>>> hand1 = cards[0:13]
>>> hand2 = cards[13:26]
# ...and so on...

或者以您需要的任何更复杂的方式。 （但是，请注意，没有必要循环通过手或类似的东西;因为它已经是随机的，简单的切片就足够了。）

Answer 4

您可以使用more_itertools库在{n}个玩家中import itertools as it import more_itertools as mit # Build a Deck suits = "♥♠♣♦" ranks = list(range(2, 11)) + list("JQKA") cards = list(it.product(suits, ranks)) print("Number of cards:", len(cards)) # Out: Number of cards: 52 # Shuffle and Distribute players = 5 random.shuffle(cards) hands = [list(hand) for hand in list(mit.distribute(players, cards))] hands[0] # player 1 张卡。

扑克牌

[('♥', 'A'),
('♥', 6),
('♦', 9),
('♠', 'A'),
('♥', 7),
('♠', 8),
('♣', 10),
('♦', 'K'),
('♥', 4),
('♠', 4),
('♠', 'Q')]

输出

>>> n, iterable = 3, [1, 2, 3, 4, 5, 6, 7]
>>> children = distribute(n, iterable)
>>> [list(c) for c in children]
[[1, 4, 7], [2, 5], [3, 6]]

看起来玩家1有2对。

这个工具在做什么？

more_itertools.distribute平均分配来自n个子组中的可迭代项目。

来自the docs的修改示例：

more_itertools

sessionid是第三方软件包，其中包含itertools recipes和其他许多useful tools。

Answer 5

我也有类似的问题，在这里检查。我必须创建一个程序，该程序创建并洗牌，然后发牌。我已经检查了所有示例，并且Benjamin here提供的示例结构合理，为我提供了继续的基础。

我在这里分享练习的完整解决方案：

代码：

from random import shuffle

class Cards:
    def __init__(self):
        ranks = ['A', '2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K']
        suites = ['♥', '♠', '♣', '♦']
        self.deck = [j + i for j in ranks for i in suites]
        self.deal = []

    def shuffle(self):
        shuffle(self.deck)

    def deal_out(self,num_cards, num_players):
        deal = [[0 for x in range(num_cards)] for y in range(num_players)]
        for i in range(num_cards):
            for k in range(num_players):
                deal[k][i]=self.deck.pop()
        self.deal = deal


#declare number of players and cards to be dealt
num_cards = int(input("Please enter a number of cards to be dealt: "))
num_players = int(input("Please enter a number of players: "))

#create new game with new deck
first_game = Cards()
print('\nDeck:\n')
print(first_game.deck)

#shuffle cards
first_game.shuffle()
print('\nShuffled Deck:\n')
print(first_game.deck)

#deal cards
first_game.deal_out(num_cards, num_players)

#Print the dealt cards
print('\nDealt cards:\n')
print(first_game.deal)

#Print cards left after the deal
print('\nCards left in the deck after the deal:\n')
print(first_game.deck)

输出：

Please enter a number of cards to be dealt: 4
Please enter a number of players: 5

Deck:

['A♥', 'A♠', 'A♣', 'A♦', '2♥', '2♠', '2♣', '2♦', '3♥', '3♠', '3♣', '3♦', '4♥', '4♠', '4♣', '4♦', '5♥', '5♠', '5♣', '5♦', '6♥', '6♠', '6♣', '6♦', '7♥', '7♠', '7♣', '7♦', '8♥', '8♠', '8♣', '8♦', '9♥', '9♠', '9♣', '9♦', '10♥', '10♠', '10♣', '10♦', 'J♥', 'J♠', 'J♣', 'J♦', 'Q♥', 'Q♠', 'Q♣', 'Q♦', 'K♥', 'K♠', 'K♣', 'K♦']

Shuffled Deck:

['6♥', '5♠', '6♣', '8♦', '10♠', '2♣', '4♦', '6♦', 'J♦', '3♥', '8♥', 'K♥', '10♦', 'A♦', '10♣', '2♥', '3♣', '4♣', '10♥', '8♠', '9♦', '3♦', 'A♣', '9♥', 'Q♦', '5♥', 'Q♠', '9♣', 'K♦', '7♥', 'J♥', '7♠', '2♦', '7♣', 'A♠', 'J♣', '4♠', '5♣', '9♠', '7♦', '2♠', 'Q♥', 'J♠', 'A♥', '4♥', '5♦', 'K♠', '8♣', '3♠', 'Q♣', '6♠', 'K♣']

Dealt cards:

[['K♣', 'K♠', 'Q♥', '4♠'], ['6♠', '5♦', '2♠', 'J♣'], ['Q♣', '4♥', '7♦', 'A♠'], ['3♠', 'A♥', '9♠', '7♣'], ['8♣', 'J♠', '5♣', '2♦']]

Cards left in the deck after the deal:

['6♥', '5♠', '6♣', '8♦', '10♠', '2♣', '4♦', '6♦', 'J♦', '3♥', '8♥', 'K♥', '10♦', 'A♦', '10♣', '2♥', '3♣', '4♣', '10♥', '8♠', '9♦', '3♦', 'A♣', '9♥', 'Q♦', '5♥', 'Q♠', '9♣', 'K♦', '7♥', 'J♥', '7♠']