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андрей марченков
Asked:2025-04-14 23:27:53 +0000 UTC2025-04-14 23:27:53 +0000 UTC

附加限制 - 仅 1 行！我不明白需要什么 - 一行代码还是一行函数输出，或者其他什么！

  • 772

任务：https://www.codewars.com/kata/535742c7e727388cdc000297/train/python

您需要生成一副 52 张牌的牌 - 四种花色，每花色 13 张牌。
牌的等级是：A、二、三、四、五、六、七、八、九、十、J、Q、K

结论应该是这样的

地图的每一行都将采用以下格式

"ace of hearts"  
"ace of spades"  
"ace of diamonds"  
"ace of clubs"

他们不必按顺序进行。

附加限制 - 仅一行

一行代码不通过

import random
dict_card_format = {1: "hearts", 2: "spades", 3: "diamonds", 4: "clubs"}
dict_rank = {1: "ace", 2: "two", 3: "three", 4: "four", 5: "five", 6: "six", 7: "seven",
             8: "eight", 9: "nine", 10: "ten", 11: "jack", 12: "queen", 13: "king"
             }

nums = list(range(1, 5))
random.shuffle(nums)
nums1 = list(range(1, 14))                                                                         
random.shuffle(nums1)

def deck_of_cards():
     return [f"{''.join(dict_rank[j])} of {''.join(dict_card_format[i])}" for j in nums1 for i in nums]

由 CodesWars 服务器发布

Test Results:
  Code lines
    Tests
       1 line only!: 13 should equal 1  - этот Тест не проходит
Completed in 0.09ms
Completed in 0.14ms
Sample tests                            - этот Тест проходит
    Tests

如果单行代码不起作用，那么如何在一行中输出函数输出？

我尝试用不同的方式写它

def deck_of_cards(): 
    n_card_format = 4
    n_rank = 13        
    nums = list(range(1, 5))                                                    
    random.shuffle(nums)
    card_format = nums[:4]                                                                            
    nums1 = list(range(1, 14))                                                                          
    random.shuffle(nums1)
    rank_list = nums1[:13] 

    # s = ""
    lst = []
    for i in card_format:
        for j in rank_list:
            if i in dict_card_format and j in dict_rank:
                lst.append(f"{''.join(dict_rank[j])} of {''.join(dict_card_format[i])}")
                # lst.append(''.join(dict_rank[j]) + " " + "of" + " " + ''.join(dict_card_format[i]))
                # s += (''.join(dict_rank[j]) + " " + "of" + " " + ''.join(dict_card_format[i]) + " ")
                # s += "\n"f"{''.join(dict_rank[j])} of {''.join(dict_card_format[i])} "
    return lst

帮我找到正确的解决方案

python

2 个回答

  1. Best Answer

    这是通过所有测试的单行解决方案：

    def deck_of_cards(): return [f'{rank} of {suit}s' for rank in ('ace', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten', 'jack', 'queen', 'king') for suit in ('heart', 'spade', 'diamond', 'club')]

    PS “它们不需要按顺序排列。”意思是“他们不需要按照特定的顺序来遵循。”也就是说，您可以按照自己方便的顺序发布 52 个短语的列表。

    • 4

  2. 您需要学习数到 1 并用一行写出整个代码。

    • 0

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