如何在 Python 中与字符串进行交叉乘法？答案

【问题标题】：How can I do cross multiplication with strings in Python?如何在 Python 中与字符串进行交叉乘法？
【发布时间】：2019-07-23 15:27:04
【问题描述】：

我有两个字符串列表：

Deck13Sample = [
    'Two', 'Three', 'Four', 'Five', 'Six', 'Seven',
    'Eight', 'Nine', 'Ten', 'Jack', 'Queen', 'King', 'Ace'
]
CardTypes = [' of Hearts', ' of Spades', ' of Diamonds', ' of Clubs']

我想将列表相乘，得到一个完整的牌组，看起来像

Deck52Sample = ['Two of Hearts', 'Three of Hearts', 'Four of Hearts', ...]

由于 Python 不能内置交叉乘法字符串，我完全不知道我现在应该做什么。

【问题讨论】：

您至少可以创建嵌套的 for 循环 - for a in Deck13Sample: for b in CardTypes: print( a + b )。所以你可以用它来创建Deck52Sample
几乎相同：stackoverflow.com/questions/2541401/…
@MikeScotty。答案没有解决字符串连接问题，所以不是真正的欺骗。
这个问题与numpy无关。我建议删除这个标签

标签： python list numpy cross-product

【解决方案1】：

将其中一个转换为2D char 数组，然后使用np.char.add 将它们加入broadcasted-way -

In [18]: np.char.add(Deck13Sample,np.array(CardTypes)[:,None])
Out[18]: 
array([['Two of Hearts', 'Three of Hearts', 'Four of Hearts',
        'Five of Hearts', 'Six of Hearts', 'Seven of Hearts',
        'Eight of Hearts', 'Nine of Hearts', 'Ten of Hearts',
        'Jack of Hearts', 'Queen of Hearts', 'King of Hearts',
        'Ace of Hearts'],
       ['Two of Spades', 'Three of Spades', 'Four of Spades',
        'Five of Spades', 'Six of Spades', 'Seven of Spades',
        'Eight of Spades', 'Nine of Spades', 'Ten of Spades',
        'Jack of Spades', 'Queen of Spades', 'King of Spades',
        'Ace of Spades'],
       ['Two of Diamonds', 'Three of Diamonds', 'Four of Diamonds',
        'Five of Diamonds', 'Six of Diamonds', 'Seven of Diamonds',
        'Eight of Diamonds', 'Nine of Diamonds', 'Ten of Diamonds',
        'Jack of Diamonds', 'Queen of Diamonds', 'King of Diamonds',
        'Ace of Diamonds'],
       ['Two of Clubs', 'Three of Clubs', 'Four of Clubs',
        'Five of Clubs', 'Six of Clubs', 'Seven of Clubs',
        'Eight of Clubs', 'Nine of Clubs', 'Ten of Clubs',
        'Jack of Clubs', 'Queen of Clubs', 'King of Clubs',
        'Ace of Clubs']], dtype='<U17')

【讨论】：

【解决方案2】：

使用itertools.product

>>> from itertools import product as pd

>>> Deck13Sample = ['Two','Three','Four','Five','Six','Seven','Eight','Nine','Ten','Jack','Queen','King','Ace']
>>> CardTypes = [' of Hearts',' of Spades',' of Diamonds',' of Clubs']

>>> [ i+j for i, j in pd(Deck13Sample, CardTypes)]
['Two of Hearts', 'Two of Spades', 'Two of Diamonds', 'Two of Clubs', 'Three of Hearts', 'Three of Spades', 'Three of Diamonds', 'Three of Clubs', 'Four of Hearts', 'Four of Spades', 'Four of Diamonds', 'Four of Clubs', 'Five of Hearts', 'Five of Spades', 'Five of Diamonds', 'Five of Clubs', 'Six of Hearts', 'Six of Spades', 'Six of Diamonds', 'Six of Clubs', 'Seven of Hearts', 'Seven of Spades', 'Seven of Diamonds', 'Seven of Clubs', 'Eight of Hearts', 'Eight of Spades', 'Eight of Diamonds', 'Eight of Clubs', 'Nine of Hearts', 'Nine of Spades', 'Nine of Diamonds', 'Nine of Clubs', 'Ten of Hearts', 'Ten of Spades', 'Ten of Diamonds', 'Ten of Clubs', 'Jack of Hearts', 'Jack of Spades', 'Jack of Diamonds', 'Jack of Clubs', 'Queen of Hearts', 'Queen of Spades', 'Queen of Diamonds', 'Queen of Clubs', 'King of Hearts', 'King of Spades', 'King of Diamonds', 'King of Clubs', 'Ace of Hearts', 'Ace of Spades', 'Ace of Diamonds', 'Ace of Clubs']

【讨论】：

【解决方案3】：

我想到了 vanilla Python 中的三个选项。对于所有这些选项，您真的不需要在西装名称前面加上' of '。我将处理 CardTypes = ['Hearts', 'Spades', 'Diamonds', 'Clubs'] 的暗示。

使用嵌套的for 循环。毕竟，这就是叉积：

deck = []
for rank in Deck13Sample:
    for suit in CardTypes:
        deck.append(f'{rank} of {suit}')

同样的事情可以更简洁地表达为列表推导式。这是我推荐的选项：
```
deck = [f'{rank} of {suit}' for rank in Deck13Sample for suit in CardTypes]
```
请注意，循环的顺序与 #1 中的相同。
最后，如果你想使用一个花哨的库导入（但是 Python 自带的），你可以使用itertools.product，它基本上是一组无限嵌套的for 循环，因此对于这个问题来说有点矫枉过正：
```
deck = [f'{rank} of {suit}' for rank, suit in itertools.product(Deck13Sample, CardTypes)]
```

作为参考，一副牌上的数字称为“等级”，符号称为“花色”。

【讨论】：

【解决方案4】：

您可以使用嵌套的 for 循环：

Deck52Sample = []
for number in Deck13Sample:
    for suit in CardTypes:
        Deck52Sample.append(number + suit)

【讨论】：

【解决方案5】：

一种方法是使用itertools.product、列表推导和使用+ 运算符的字符串连接：

import itertools

deck = [rank + suit for rank, suit in itertools.product(Deck13Sample, CardTypes)]

结果：

In [28]: deck
Out[28]: 
['Two of Hearts',
 'Two of Spades',
 'Two of Diamonds',
...
'Ace of Spades',
'Ace of Diamonds',
'Ace of Clubs']

【讨论】：

【解决方案6】：

使用对象 dtype 字符串，广播解决方案可以使用加法（字符串连接）：

In [45]: Deck13Sample = ['Two','Three','Four','Five','Six','Seven','Eight','Nine','Ten','Jack','Queen','King'
    ...: ,'Ace'] 
    ...: CardTypes = [' of Hearts',' of Spades',' of Diamonds',' of Clubs']                                  
In [46]: A=np.array(Deck13Sample,object)                                                                     
In [47]: B=np.array(CardTypes,object)                                                                        
In [48]: A[None,:]+B[:,None]                                                                                 
Out[48]: 
array([['Two of Hearts', 'Three of Hearts', 'Four of Hearts',
        'Five of Hearts', 'Six of Hearts', 'Seven of Hearts',
        'Eight of Hearts', 'Nine of Hearts', 'Ten of Hearts',
        'Jack of Hearts', 'Queen of Hearts', 'King of Hearts',
        'Ace of Hearts'],
       ['Two of Spades', 'Three of Spades', 'Four of Spades',
        'Five of Spades', 'Six of Spades', 'Seven of Spades',
        'Eight of Spades', 'Nine of Spades', 'Ten of Spades',
        'Jack of Spades', 'Queen of Spades', 'King of Spades',
        'Ace of Spades'],
       ['Two of Diamonds', 'Three of Diamonds', 'Four of Diamonds',
        'Five of Diamonds', 'Six of Diamonds', 'Seven of Diamonds',
        'Eight of Diamonds', 'Nine of Diamonds', 'Ten of Diamonds',
        'Jack of Diamonds', 'Queen of Diamonds', 'King of Diamonds',
        'Ace of Diamonds'],
       ['Two of Clubs', 'Three of Clubs', 'Four of Clubs',
        'Five of Clubs', 'Six of Clubs', 'Seven of Clubs',
        'Eight of Clubs', 'Nine of Clubs', 'Ten of Clubs',
        'Jack of Clubs', 'Queen of Clubs', 'King of Clubs',
        'Ace of Clubs']], dtype=object)

但我认为列表理解版本没有任何问题。我们只生成 52 个项目。考虑到转换为数组，列表版本可能同样快，甚至更快。

In [54]: timeit [[i+j for j in CardTypes] for i in Deck13Sample]                                             
10.3 µs ± 262 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [55]: timeit np.array(Deck13Sample,object)[None,:]+np.array(CardTypes,object)[:,None]                     
21.5 µs ± 37.2 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [56]: timeit np.char.add(Deck13Sample,np.array(CardTypes)[:,None])                                        
65.7 µs ± 1.72 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

【讨论】：