如何在Python中生成一副牌

Question

如何在Python中以列表格式最有效地生成52张卡片，以便列表如下所示：

['1 of Spades', '1 of Hearts', '1 of Clubs', '1 of Diamonds', '2 of Spades', '2 of Hearts'等。

Answer 1

我更喜欢以下代码，如Python Readability Counts

>>> faces = range(2,11) + ["King","Queen","Jack","Ace"]
>>> colour = ["Spades", "Hearts", "Clubs", "Diamonds"]
>>> from itertools import product
>>> ["{} of {}".format(*card) for card in product(faces, colour)]

Answer 2

要以列表格式轻松（并有效）生成一副牌，您可以输入：

deck = [str(x)+y for x in range(1,14) for y in ["S","H","C","D"]]

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当你print(deck)时，你会得到这样的输出：

['1S', '1H', '1C', '1D', '2S', '2H', '2C', '2D', '3S', '3H', '3C', '3D'.....

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例如，要将输出从"3C"更改为"3 of Clubs"，请更改

["S","H","C","D"]至[" of Spades"," of Hearts"," of Clubs"," of Diamonds"]

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这会使您的列表看起来像：['1 of Spades', '1 of Hearts', '1 of Clubs', '1 of Diamonds', '2 of Spades', '2 of Hearts'.....等等。

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注意：最初的例子可能是最短的例子......

Answer 3

>>> suits = ['spades', 'hearts', 'diamonds', 'clubs']
>>> faces = ['ace'] + range(2, 11) + ['jack', 'queen', 'king']
>>> deck = ['%s of %s'%(f, s) for f in faces for s in suits]