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我正在为我的第二个作业编写一些代码，即 Python 中的 Go Fish 游戏。对于我的一个函数定义，我需要从 target_player 卡片列表中取出 target_rank，并将其附加到当前玩家卡片列表中。我尝试使用 for 循环，但收到更改列表长度的错误。到目前为止，我拥有的代码是

def Hit(player_number, target_player, target_rank, pHands):
    while str(target_rank) in pHands[target_player]:
       pHands[player_number].append(pHands[target_player].pop(target_rank))
    print pHands[player_number]
    print pHands[target_player]
pHands = [['a', '2', '3', '4', '4', '5', '6', '7', '7', 't'], ['2', 'q', '6', '9', '5', 'k', 'k', 'a', '3', '8'], ['j', '9', 't', 't', '2', 't', '7', 'j', '5', '9'], ['8', '8', 'a', 'q', 'k', '4', '6', '9', 'q', '2']]
Hit (0,1,'a',pHands)

其中 pHands 是玩家卡列表的列表。我知道我不能使用 pop 因为目标排名是一个字符串而不是位置值，但任何帮助将不胜感激。感谢大家！

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2 回答 2

3

pop我认为不将与链接起来可能更清楚append。而且，pop需要一个索引号，而不是卡值本身，所以在这里用起来有点别扭。相反，您可以使用remove：

def Hit(player_number, target_player, target_rank, pHands):
    while target_rank in pHands[target_player]:
        pHands[target_player].remove(target_rank)
        pHands[player_number].append(target_rank)
    print pHands[player_number]
    print pHands[target_player]

但是，pop和remove都是O(n)操作。对中的每张卡都这样做一次，pHands[target_player]就可以得到while-loop O(n**2). 如果我们可以在 中执行任务会更好O(n)，并且可以使用append，这是一个O(1)操作。因此，如果我们只需要为append中的每张卡做一次pHands[target_player]，那么for-loop（下）是O(n)：

def Hit(player_number, target_player, target_rank, pHands):
    new_target_hand = []
    for card in pHands[target_player]:
        if target_rank == card:
            pHands[player_number].append(card)
        else:
            new_target_hand.append(card)
    pHands[target_player] = new_target_hand
    print pHands[player_number]
    print pHands[target_player]
于 2013-07-26T18:26:44.170 回答
1

编辑：我的作品，但这个更干净。

def Hit(player_number, target_player, target_rank, pHands):
    try:
        while True:
           pHands[player_number].append(pHands[target_player].pop(pHands.index(target_rank)))
        print pHands[player_number]
        print pHands[target_player]
    except ValueError:
        pass # value no longer in hand
pHands = [['a', '2', '3', '4', '4', '5', '6', '7', '7', 't'], ['2', 'q', '6', '9', '5', 'k', 'k', 'a', '3', '8'], ['j', '9', 't', 't', '2', 't', '7', 'j', '5', '9'], ['8', '8', 'a', 'q', 'k', '4', '6', '9', 'q', '2']]
Hit (0,1,'a',pHands)
于 2013-07-26T18:25:30.257 回答