按字母表对表示卡片的元组列表进行排序，然后排名

Question

我想先按字母顺序（梅花、钻石、红心、黑桃）对一手牌进行排序，然后再按等级（从 A 到 K）。我有以下（非常冗长且效率低下）的代码，但输出似乎错误。我觉得这是我使用的冒泡排序的问题。我在这里缺少什么，以及这样做的更好方法是什么？对于我的最终输出，我还将把 2、11、12 和 13 转换回 Ace、Jack、Queen 和 King，但我真的不想重做一开始就将它们转换为数值的操作。 I/O 应如下所示：

样本输入：[('h', 'q'), ('c', 'j'), ('d', '2'), ('s', '3'), ('h ', '5'), ('d', 'j'), ('d', '10')]

样本输出：[('c', 'j'), ('d', '2'), ('d', '10'), ('d', 'j'), ('h ', '5'), ('h', 'q'), ('s', '3')]

这是我目前的代码和结果输出：

def sort_hand(hand):
    """
    Sorts the a hand of cards by the value of each suit/type of card. 
    """
    
    # A=1,2,3,4,5,6,7,8,9,10,J=11,Q=12,K=13
    # UPDATING CARDS SO THAT THEY CAN BE ORDERED BY RANK
    new_hand = []
 
    # GIVING NUMERIC VALUES TO JACK, QUEEN, KING, AND ACE
    for element in hand: 
        if element[1] == 'j':
            val = element[1].replace('j', '11')
            new_element = (element[0], val)
            new_hand.append(new_element)
        elif element[1] == 'q':
            val = element[1].replace('q', '12')
            new_element = (element[0], val)
            new_hand.append(new_element)
        elif element[1] == 'k':
            val = element[1].replace('k', '13')
            new_element = (element[0], val)
            new_hand.append(new_element)
        elif element[1] == 'a':
            val = element[1].replace('a', '1')
            new_element = (element[0], val)
            new_hand.append(new_element)
        else:
            pass
            new_hand.append(element)
    
    # BUBBLE SORT USED TO ORDER BY RANK 
    for ix in range(1, len(new_hand)):
        value_to_sort = new_hand[ix][1]

        while new_hand[ix-1][1] > value_to_sort and ix > 0:
            new_hand[ix], new_hand[ix-1] = new_hand[ix-1], new_hand[ix]
            ix -= 1
            
    # MAKE SUBLISTS FOR EACH SUIT      
    c_list = []
    d_list = []
    h_list = []
    s_list = []
            
    for element in new_hand:
        if element[0] == 'c':
            c_list.append(element)
        elif element[0] == 'd':
            d_list.append(element)
        elif element[0] == 'h':
            h_list.append(element)
        else: 
            s_list.append(element)
    
    # COMBINE ORDERED SUIT SUBLISTS TO MAKE FINAL SORTED HAND BY RANK
    final_hand = c_list + d_list + h_list + s_list
            
    return final_hand

>>> hand = [('h', 'q'), ('c', 'j'), ('d', '2'), ('s', '3'), ('h', '5'), ('d', 'j'), ('d', '10')]  
>>> sort_hand(hand)
[('c', '11'),
 ('d', '10'),
 ('d', '11'),
 ('d', '2'),
 ('h', '12'),
 ('h', '5'),
 ('s', '3')]

我觉得我很接近这段代码，但不知道从这里去哪里。希望得到一个解释清楚的答案，谢谢。

Answer 1

您可以使用内置的sort 或sorted 函数来实现这一点

您需要使用上述功能的附加功能是比较器逻辑，可以使用key 参数提供。

def sort_hand(inp):
  ##### I have created the order of the suites arbitrarily
  suite_map = {
     'h':1
     ,'c':2
     ,'s':3
     ,'d':4
  }

  rank_map = {
        'a':1
        'k':2
        'q':3
        'j':4
        '10':5
        '9':6
        '8':7
        '7':8
        '6':9
        '5':10
        '4':11
        '3':12
        '2':13
        '1':14
    }
    
    return suite_map[inp[0]],rank_map[inp[1]]

O/P --->

>>> hand
[('h', 'q'), ('c', 'j'), ('d', '2'), ('s', '3'), ('h', '5'), ('d', 'j'), ('d', '10')]
>>> sorted(hand,key=sort_hand)
[('h', 'q'), ('h', '5'), ('c', 'j'), ('s', '3'), ('d', 'j'), ('d', '10'), ('d', '2')]

您可以根据您的顺序控制基于suite_map和rank_map的字典顺序并进行相应的排序

Answer 2

使用 lambda 进行双键排序。

rank = ['2', '3', '4', '5', '6', '7', '8', '9', '10', 'j', 'q', 'k', 'a']
suit = ['c', 'd', 'h', 's']

inp = [('h', 'q'), ('c', 'j'), ('d', '2'), ('s', '3'), ('h', '5'), ('d', 'j'), ('d', '10')]

outp = sorted(inp,key = lambda x: 
    (suit.index(x[0]), rank.index(x[1])))

print(outp)

输出：

[('c', 'j'), ('d', '2'), ('d', '10'), ('d', 'j'), ('h', '5'), ('h', 'q'), ('s', '3')]