1

我正在制作一个modified blackjack game. 只玩牌 2 到 9。这副牌中没有 10、面牌或 A。

首先，我创建了变量 deck，它采用一个空列表。
然后我创建了 4for loops来将卡片列表附加到套牌中。

我想知道是否有更短的方法来编写以下内容：

import random

deck = []
for list_of_cards in range(2, 10):
    spades = 'spades'
    list_of_spades = str(list_of_cards) + ' of' + ' ' + spades
    deck.append(list_of_spades)
for list_of_cards in range(2, 10):
    clubs = 'clubs'
    list_of_clubs = str(list_of_cards) + ' of' + ' ' + clubs
    deck.append(list_of_clubs)
for list_of_cards in range(2, 10):
    diamonds = 'diamonds'
    list_of_diamonds = str(list_of_cards) + ' of' + ' ' + diamonds
    deck.append(list_of_diamonds)
for list_of_cards in range(2, 10):
    hearts = 'hearts'
    list_of_hearts = str(list_of_cards) + ' of' + ' ' + hearts
    deck.append(list_of_hearts)

结果就像我想要的那样，即：

['2 of spades', '3 of spades', '4 of spades', '5 of spades', '6 of spades', '7 of spades', '8 of spades', '9 of spades', '2 of clubs', '3 of clubs', '4 of clubs', '5 of clubs', '6 of clubs', '7 of clubs', '8 of clubs', '9 of clubs', '2 of diamonds', '3 of diamonds', '4 of diamonds', '5 of diamonds', '6 of diamonds', '7 of diamonds', '8 of diamonds', '9 of diamonds', '2 of hearts', '3 of hearts', '4 of hearts', '5 of hearts', '6 of hearts', '7 of hearts', '8 of hearts', '9 of hearts']

由于我是 Python 新手，我假设有一种方法可以写得更短。谢谢您的帮助！

4

5 回答 5

5

为什么需要 4 个循环？为什么你需要创建那些无用的字符串？（和...+ ' of' + ' ' + spades一样' of spades'）（我也修复了一些其他的东西）

deck = []
for n in range(2, 10):
    deck.append('%i of spades' % n)
    deck.append('%i of hearts' % n)
    deck.append('%i of diamonds' % n)
    deck.append('%i of clubs' % n)

或者，使用嵌套循环：

deck = []
for n in range(2, 10):
    for s in ['spades', 'hearts', 'diamonds', 'clubs']:
        deck.append('%i of %s' % (n, s))
于 2013-12-05T13:58:54.197 回答
4

例如，您不需要为“of”和“”使用单独的字符串。您不需要单独的字符串来保存单词“diamonds”。只写“钻石”。

也就是说，这是一个使用字符串格式化、列表理解和拆分而不是手动制作一组西装的解决方案。

deck = ['%d of %s' % (face, suit) for suit in 'spades hearts clubs diamonds'.split() for face in range(2,10)]
于 2013-12-05T14:02:50.527 回答
2

我会使用列表理解：

>>> suits = ["hearts","diamonds","clubs","spades"]
>>> deck = ["{} of {}".format(num, suit) for suit in suits for num in range(2,10)]
>>> len(deck)
32
>>> deck
['2 of hearts', '3 of hearts', '4 of hearts', '5 of hearts', '6 of hearts', '7 of hearts', '8 of hearts', '9 of hearts', '2 of diamonds', '3 of diamonds', '4 of diamonds', '5 of diamonds', '6 of diamonds', '7 of diamonds', '8 of diamonds', '9 of diamonds', '2 of clubs', '3 of clubs', '4 of clubs', '5 of clubs', '6 of clubs', '7 of clubs', '8 of clubs', '9 of clubs', '2 of spades', '3 of spades', '4 of spades', '5 of spades', '6 of spades', '7 of spades', '8 of spades', '9 of spades']

这里的 list comp 等价于

deck = []
for suit in suits:
    for num in range(2, 10):
        deck.append("{} of {}".format(num, suit))
于 2013-12-05T14:02:52.023 回答
1 
deck = []
colors = ["spades", "clubs", "diamonds", "hearts"]
for color in colors:
    for list_of_cards in range(2, 10):
        deck.append("{0} of {1}".format(list_of_cards, color))
于 2013-12-05T14:02:36.140 回答
1

两个 for 循环就足够了：

deck = []
for suit in ('spades', 'hearts', 'diamonds', 'clubs'):
    for n in range(2, 10):
        deck.append('%i of %s' % (n, suit))

您也可以使用列表推导来编写它：

deck = ['%i of %s' % (n, suit) for suit in ('spades', 'hearts', 'diamonds', 'clubs') for n in range(2, 10)]
于 2013-12-05T14:03:21.847 回答