python - 生成 5 张扑克牌

Question

我一直在尝试为 IRC 制作一个扑克游戏机器人，但我似乎无法处理这些牌。

我知道这个算法效率很低，但这是我使用我目前的 Python 技能所能想到的最好的算法。欢迎任何改进！

玩家是一个字典，其中键是玩家的昵称，值是他们拥有的金额。

当代码运行时，如果只有 1 名玩家，它应该给出 5 张牌。但如果有 2 名玩家，则每人会生成 4 到 6 张牌。我还没有测试更多的玩家。

一些预先初始化的变量：

numberOfPlayers = 0 #The numerical value of the amount of players in the game
players = {} #The nickname of each player and the amount of money they have
bets = {} #How much each player has bet
decks = 1 #The number of decks in play
suits = ["C","D","H","S"] #All the possible suits (Clubs, Diamonds, Hearts, Spades)
ranks = ["2","3","4","5","4","6","7","8","9","J","Q","K","A"] #All the possible ranks (Excluding jokers)
cardsGiven = {} #The cards that have been dealt from the deck, and the amount of times they have been given. If one deck is in play, the max is 1, if two decks are in play, the max is 2 and so on...
hands = {} #Each players cards

编码：

def deal(channel, msgnick):
    try:
        s.send("PRIVMSG " + channel + " :Dealing...\n")
        for k, v in players.iteritems():
               for c in range(0, 5):
                suit = random.randrange(1, 4)
                rank = random.randrange(0,12)
                suit = suits[suit]
                rank = ranks[rank]
                card = rank + suit
                print(card)
                if card in cardsGiven:
                    if cardsGiven[card] < decks:
                        if c == 5:
                            hands[k] = hands[k] + card
                            cardsGiven[card] += 1
                        else:
                            hands[k] = hands[k] + card + ", "
                            cardsGiven[card] += 1
                    else:
                        c -= 1
                else:
                    if c == 5:
                        hands[k] = hands[k] + card
                        cardsGiven[card] = 1
                    else:
                        hands[k] = hands[k] + card + ", "
                        cardsGiven[card] = 1
            s.send("NOTICE " + k + " :Your hand: " + hands[k] + "\n")
    except Exception:
        excHandler(s, channel)

如果需要任何示例或进一步解释，请询问:)

score 4 · Accepted Answer

`for`循环

for c in range(0, 5):

...

    c -= 1

不幸的是，这不是for循环在 Python 中的工作方式——递减 c 不会导致循环的另一个复飞。for循环遍历在循环开始之前固定的一组项目（例如range(0,5)，在循环期间未修改的 5 个项目的固定范围）。

如果你想做你正在做的事情，你需要使用一个while循环（它通过一个可以在循环期间修改的变量和条件工作）：

c = 0
while c < 5:
    c += 1

    ...

    c -= 1

`range()`编号

if c == 5:

这种情况目前不会被击中，因为会range(N)生成一个从0to N-1- 例如range(5)generate的数字序列0,1,2,3,4。

score 2 · Accepted Answer

我会使用itertools.product将卡片的可能价值放入列表中，然后将其洗牌并为您想要的每个玩家一次取 5 张卡片。

from itertools import product
from random import shuffle

suits = ["C","D","H","S"] 
ranks = ["2","3","4","5","4","6","7","8","9","J","Q","K","A"]

cards = list(r + s for r, s in product(ranks, suits))
shuffle(cards)

print cards[:5], cards[5:10] # change this into a suitable for loop to slice
['4D', 'KC', '5H', '9H', '7D'] ['2D', '4S', '8D', '8S', '4C']

你可以使用下面的配方itertools来获得接下来的 5 张牌，具体取决于玩家人数。

def grouper(n, iterable, fillvalue=None):
    from itertools import izip_longest
    "Collect data into fixed-length chunks or blocks"
    # grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

hand = grouper(5, cards) 
for i in xrange(5): # deal for 5 players...
    print next(hand) # each time we call this, we get another 5 cards

('4D', 'KC', '5H', '9H', '7D')
('2D', '4S', '8D', '8S', '4C')
('AD', '2H', '4S', 'KS', '9S')
('6H', 'AH', '4H', '5S', 'KD')
('6S', 'QD', '3C', 'QC', '8H')

score 0 · Accepted Answer

使用切片来完成此操作的一种简单方法可能是：

num_players = 5
num_cards_per_player = 5
remaining_cards = list(range(100)) # Imagine a list backing a Deck object
num_cards_to_deal = num_players * num_cards_per_player
cards_to_deal, remaining_cards = (remaining_cards[0:num_cards_to_deal],
        remaining_cards[num_cards_to_deal:])
cards_for_each_player = [cards_to_deal[player_num::num_players] for
        player_num in range(num_players)]
cards_for_each_player
[[0, 5, 10, 15, 20], [1, 6, 11, 16, 21], [2, 7, 12, 17, 22], [3, 8, 13, 18, 23], [4, 9, 14, 19, 24]]

对于特定类型的卡片组，您可以实现一个卡片组对象并使用上述代码稍作修改以用于方法中。卡组对象将需要处理剩余卡牌不足的情况:)，这可能是您正在玩的游戏所特有的。

score 0 · Accepted Answer

#! /usr/bin/python3.2

import random

players = ['Alice', 'Bob', 'Mallroy']
suits = ["C","D","H","S"]
ranks = ["2","3","4","5","4","6","7","8","9","J","Q","K","A"]
decks = 2
cards = ['{}{}'.format (s, r)
         for s in suits
         for r in ranks
         for d in range (decks) ]

random.shuffle (cards)
for player in players:
    print ('{}\'s hand is {}'.format (player, cards [:5] ) )
    cards = cards [5:]

score 0 · Accepted Answer

只需使用 Pythondeuces库，它是专门为这种用途而构建的：

from deuces import Card
from deuces import Deck

# create deck and shuffle it
deck = Deck()
deck.shuffle()

# play a texas hold 'em game with two players
board = deck.draw(5)
player1_hand = deck.draw(2)
player2_hand = deck.draw(2)

# now see which hand won
from deuces import Evaluator
evaluator = Evaluator()
p1handscore = evaluator.evaluate(board, player1_hand)
p2handscore = evaluator.evaluate(board, player2_hand)

score 0 · Accepted Answer

from random import randint

# This code below creates a deck
cards = []
for suit in ["diamond","club","heart","spade"]:
    for card in ['A','2','3','4','5','6','7','8','9','10','J','Q','K']:
        cards.append(card+' '+suit)
# To deal the cards we will take random numbers
for i in range(5):
    a = randint(0,len(cards)-1)
    print cards(a)
    del cards[a] # We will delete the card which drew

首先我们创建一个牌组，然后我们取一个随机数，randint然后我们用随机数抽一张牌

我有一个程序来绘制翻牌和手牌

def create_card():
    cards = []
    for suit in   [r'$\diamondsuit$',r'$\clubsuit$',r'$\heartsuit$',r'$\spadesuit$']:
        for card in  ['A','2','3','4','5','6','7','8','9','10','J','Q','K']:
            cards.append(card+suit)
    return cards

def create_board_and_hand():
    cards = create_card()
    board = []
    for i in [0,1,2,3,4]:
    a  = randint(0,len(cards)-1)
    board.append(cards[a])
    del cards[a]
    hand = []
    for i in [0,1]:
        a  = randint(0,len(cards)-1)
        hand.append(cards[a])
        del cards[a]
    return board,hand

board,hand = create_board_and_hand()

python - 生成 5 张扑克牌

6 回答 6

for循环

range()编号

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Reference

`for`循环

`range()`编号