溶液

Question

正在开展一个小型项目来进行二十一点游戏。我意识到有十亿如何做到这一点，但我正在努力理解编码并建立我自己的......据说，我搞砸了很多。虽然这个难题让我感到困惑......在设置我的套牌时，我尝试了这个：

suits = ['spades', 'hearts', 'clubs', 'diamonds']
ranks = ['ace', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight',     'nine', 'ten', 'Jack', 'Queen', 'King']

decks = []

for suit in suits:
    for rank in ranks:
        decks += (rank, suit)
print decks

我得到两个列表一起排序的预期结果： ['ace'，'黑桃'，'两个'，'黑桃'，'三'，'黑桃'，'四'......]

但是，当我尝试将它们组合成字典时，如下所示：

b = dict(zip(decks[1::2], decks[0::2]))
print b

我得到：{'心'：'国王'，'俱乐部'：'国王'，'黑桃'：'国王'，'钻石'：'国王'}为什么它只做国王的价值观？

当我尝试使用以下代码纠正时：

spade = ['spades']

ranks = ['ace', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'Jack', 'Queen', 'King']

spades = []

for rank in ranks:
    spades += (rank, spade)
print spades

我得到了这个输出： ['ace'，['spades']，'two'，['spades']，'three'，['spades']，...]

那是什么给出的？帮助一个菜鸟！我的目的是通过列表创建一个优雅的套牌，将值附加到卡片上，创建一个dictonary并使用这些值来计算分数......并尝试更好地理解Python编码！

谢谢！

Answer 1

谈论乘法列表应该强迫您使用itertools.product，并通过转换为list强制迭代

import itertools

suits = ['spades', 'hearts', 'clubs', 'diamonds']
ranks = ['ace', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight',     'nine', 'ten', 'Jack', 'Queen', 'King']

decks = list(itertools.product(suits,ranks))

print(decks)

结果

[('spades', 'ace'), ('spades', 'two'), ('spades', 'three'), ('spades', 'four'), ('spades', 'five'), ('spades', 'six'), ('spades', 'seven'), ('spades', 'eight'), ('spades', 'nine'), ('spades', 'ten'), ('spades', 'Jack'), ('spades', 'Queen'), ('spades', 'King'), ('hearts', 'ace'), ('hearts', 'two'), ('hearts', 'three'), ('hearts', 'four'), ('hearts', 'five'), ('hearts', 'six'), ('hearts', 'seven'), ('hearts', 'eight'), ('hearts', 'nine'), ('hearts', 'ten'), ('hearts', 'Jack'), ('hearts', 'Queen'), ('hearts', 'King'), ('clubs', 'ace'), ('clubs', 'two'), ('clubs', 'three'), ('clubs', 'four'), ('clubs', 'five'), ('clubs', 'six'), ('clubs', 'seven'), ('clubs', 'eight'), ('clubs', 'nine'), ('clubs', 'ten'), ('clubs', 'Jack'), ('clubs', 'Queen'), ('clubs', 'King'), ('diamonds', 'ace'), ('diamonds', 'two'), ('diamonds', 'three'), ('diamonds', 'four'), ('diamonds', 'five'), ('diamonds', 'six'), ('diamonds', 'seven'), ('diamonds', 'eight'), ('diamonds', 'nine'), ('diamonds', 'ten'), ('diamonds', 'Jack'), ('diamonds', 'Queen'), ('diamonds', 'King')]

或者因为您似乎需要一个平面列表，只需使用chain.from_iterable展平它：

decks = list(itertools.chain.from_iterable(itertools.product(suits,ranks)))

Answer 2

您可以使用列表组合创建一组卡片，如[[rank,suits]]。并且dict()仅从2个值[value1,value2]创建一个对象。您持续获得{'hearts': 'King', 'clubs': 'King', 'spades': 'King', 'diamonds': 'King'}的原因是因为dict具有唯一的{key:value}对，而King数组中要为这些键分配的最后一项是ranks。

suits = ['spades', 'hearts', 'clubs', 'diamonds']
ranks = ['ace', 'two', 'three', 'four', 'five', 'six', 'seven',
         'eight',     'nine', 'ten', 'Jack', 'Queen', 'King']

print [(rank, suit) for suit in suits for rank in ranks]

溶液

您可以将密钥设为ranks，将值(rank,suit)的数组设为rank

card_dict = {}
suits = ['spades', 'hearts', 'clubs', 'diamonds']
ranks = ['ace', 'two', 'three', 'four', 'five', 'six', 'seven',
         'eight', 'nine', 'ten', 'Jack', 'Queen', 'King']
deck = [(rank, suit) for suit in suits for rank in ranks]

map(
    lambda rank: card_dict.update(
        {
            rank: filter(lambda card: rank in card, deck)
        }
    ), ranks
)

print card_dict

结果

{
  'King': [('King', 'spades'), ('King', 'hearts'), ('King', 'clubs'), ('King', 'diamonds')],
  'seven': [('seven', 'spades'), ('seven', 'hearts'), ('seven', 'clubs'), ('seven', 'diamonds')],
  'Queen': [('Queen', 'spades'), ('Queen', 'hearts'), ('Queen', 'clubs'), ('Queen', 'diamonds')],
  'ace': [('ace', 'spades'), ('ace', 'hearts'), ('ace', 'clubs'), ('ace', 'diamonds')],
  'ten': [('ten', 'spades'), ('ten', 'hearts'), ('ten', 'clubs'), ('ten', 'diamonds')],
  'nine': [('nine', 'spades'), ('nine', 'hearts'), ('nine', 'clubs'), ('nine', 'diamonds')],
  'six': [('six', 'spades'), ('six', 'hearts'), ('six', 'clubs'), ('six', 'diamonds')],
  'two': [('two', 'spades'), ('two', 'hearts'), ('two', 'clubs'), ('two', 'diamonds')],
  'three': [('three', 'spades'), ('three', 'hearts'), ('three', 'clubs'), ('three', 'diamonds')],
  'four': [('four', 'spades'), ('four', 'hearts'), ('four', 'clubs'), ('four', 'diamonds')],
  'five': [('five', 'spades'), ('five', 'hearts'), ('five', 'clubs'), ('five', 'diamonds')],
  'Jack': [('Jack', 'spades'), ('Jack', 'hearts'), ('Jack', 'clubs'), ('Jack', 'diamonds')],
  'eight': [('eight', 'spades'), ('eight', 'hearts'), ('eight', 'clubs'), ('eight', 'diamonds')]
}

或者按suits代替ranks

进行映射

{
  'clubs': [('ace', 'clubs'),
    ('two', 'clubs'),
    ('three', 'clubs'),
    ('four', 'clubs'),
    ('five', 'clubs'),
    ('six', 'clubs'),
    ('seven', 'clubs'),
    ('eight', 'clubs'),
    ('nine', 'clubs'),
    ('ten', 'clubs'),
    ('Jack', 'clubs'),
    ('Queen', 'clubs'),
    ('King', 'clubs')
  ],
  'diamonds': [('ace', 'diamonds'),
    ('two', 'diamonds'),
    ('three', 'diamonds'),
    ('four', 'diamonds'),
    ('five', 'diamonds'),
    ('six', 'diamonds'),
    ('seven', 'diamonds'),
    ('eight', 'diamonds'),
    ('nine', 'diamonds'),
    ('ten', 'diamonds'),
    ('Jack', 'diamonds'),
    ('Queen', 'diamonds'),
    ('King', 'diamonds')
  ],
  'hearts': [('ace', 'hearts'),
    ('two', 'hearts'),
    ('three', 'hearts'),
    ('four', 'hearts'),
    ('five', 'hearts'),
    ('six', 'hearts'),
    ('seven', 'hearts'),
    ('eight', 'hearts'),
    ('nine', 'hearts'),
    ('ten', 'hearts'),
    ('Jack', 'hearts'),
    ('Queen', 'hearts'),
    ('King', 'hearts')
  ],
  'spades': [('ace', 'spades'),
    ('two', 'spades'),
    ('three', 'spades'),
    ('four', 'spades'),
    ('five', 'spades'),
    ('six', 'spades'),
    ('seven', 'spades'),
    ('eight', 'spades'),
    ('nine', 'spades'),
    ('ten', 'spades'),
    ('Jack', 'spades'),
    ('Queen', 'spades'),
    ('King', 'spades')
  ]
}

混洗

import random
random.shuffle(deck)
print deck

将列表相乘

2 个答案:

溶液

结果

混洗