Python 我的程序的递归？

这段代码的作用是打印出一种组合中可以播放的所有n个组合。其中

S=Spades

，

H=Hearts

，

D=Diamonds

，和

C=Clubs

。因此，在这种情况下，手会产生：

[['3H'], ['3H', '3C'], ['3H', '3D'], ['3H', '3C', '3D'], ['3C'], ['3C', '3D'], ['3D'], ['4S'], ['6D'], ['7D'], ['9S']]

对于所有可玩的n个组合

我想知道是否有一种方法可以让我在递归循环中完成这段代码？因为如果一个卡片组中有4套以上的套装，那么继续重新键入迭代会变得很乏味

hand = ['3H', '3C', '3D', '4S', '6D', '7D', '9S']

def generate_plays(sorted_hand_value):
    playable_card = []
    for i in range(len(hand)):  
        playable_card.append([sorted_hand_value[i]]) # appends 1 of a kind to the     playable_cards list

    if i+3 <= (len(hand)-1): #need this restriction of that the interation won't index something that is out of the range of the list

        if sorted_hand_value[i][0] == sorted_hand_value[i+1][0]: #checks if first and second card have the same value
            playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+1]))) #appends 2 of a kind to the playable_card list

            if sorted_hand_value[i][0] == sorted_hand_value[i+2][0]:
                playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+2]))) #checks if first and third card have the same value
                playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+1],sorted_hand_value[i+2]))) #appends 3 of a kind to the playable_card list 

                if sorted_hand_value[i][0] == sorted_hand_value[i+3][0]:
                    playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+3])))#checks if first and fourth card have the same value
                    playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+1],sorted_hand_value[i+2],sorted_hand_value[i+3]))) #appends 4 of a kind to the playable_card list

    elif i+2 <= (len(hand)-1):

        if sorted_hand_value[i][0] == sorted_hand_value[i+1][0]:
            playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+1])))

            if sorted_hand_value[i][0] == sorted_hand_value[i+2][0]:
                playable_card.append(list((sorted_hand[i],sorted_hand[i+2])))
                playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+1],sorted_hand_value[i+2])))

    elif i+1 <= (len(hand)-1):

        if sorted_hand_value[i][0] == sorted_hand_value[i+1][0]:
            playable_card.append(list((sorted_hand_value[i],sorted_hand_value[i+1])))
return playable_card

print generate_plays(hand) # 

hand=['3H'，'3C'，'3D'，'4S'，'6D'，'7D'，'9S']
def生成播放（已排序的手牌值）：
可玩的_卡=[]
对于范围内的i（len（hand））：
可玩的卡片。追加（[sorted_hand_value[i]]）#将一种类型的卡片追加到可玩的卡片列表中
如果我+3
然后

给予

您可以使用groupby来选择具有相同值的卡片组（首先对它们进行排序，以确保它们都在一起），然后使用powerset配方来生成所有种类的n（不包括空集）

输出为：

[['3C'], ['3D'], ['3H'], ['3C', '3D'], ['3C', '3H'], ['3D', '3H'],
 ['3C', '3D', '3H'], ['4S'], ['6D'], ['7D'], ['9S']]

你说的“n”是什么意思？这通常与面值（即三个十）有关，而不是与西装有关。你说的“可以玩”是什么意思？游戏规则是什么？
print(n_of_a_kind(hand))

[('4S',), ('7D',), ('6D',), ('9S',), ('3H',), ('3C',), ('3D',), ('3H', '3C'), ('3H', '3D'), ('3C', '3D'), ('3H', '3C', '3D')]

from itertools import groupby, combinations

def kinds(cards):
    for _, cs in groupby(sorted(cards), lambda x: x[0]):
        s = list(cs)
        for r in xrange(1, len(s)+1):
            for p in combinations(s, r):
                yield list(p)

print list(kinds(['3H', '3C', '3D', '4S', '6D', '7D', '9S']))

[['3C'], ['3D'], ['3H'], ['3C', '3D'], ['3C', '3H'], ['3D', '3H'],
 ['3C', '3D', '3H'], ['4S'], ['6D'], ['7D'], ['9S']]