Python,如何排序对象列表？

Question

我有一个看起来像这样的对象列表.

hand = [ Card(10, 'H'), Card(2,'h'), Card(12,'h'), Card(13, 'h'), Card(14, 'h') ]

卡(10,'H)在这里不是元组,而是一个对象.我知道如果列表中的每个项目都是元组形式,如此排序,如何对此列表进行排序,

hand = sorted(hand, key = lambda x: x[0])

但我不知道如何对对象列表进行排序.我想按第一个输入值排序我的列表,这是Card()中的数字

我怎样才能做到这一点？

编辑:这是Card()的定义.

class Card(object):

    RANKS = (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14)

    SUITS = ('C', 'D', 'H', 'S')

    def __init__(self, rank=12, suit='S'):

        if (rank in Card.RANKS):
            self.rank = rank
        else:
            self.rank = 12

        if (suit in Card.SUITS):
            self.suit = suit.upper()
        else:
            self.suit = 'S'

    def __str__(self):
        if (self.rank == 14):
            rank = 'A'
        elif (self.rank == 13):
            rank = 'K'
        elif (self.rank == 12):
            rank = 'Q'
        elif (self.rank == 11):
            rank = 'J'
        else:
            rank = str(self.rank)
        return rank + self.suit

    def __eq__(self, other):
        return (self.rank == other.rank)

    def __ne__(self, other):
        return (self.rank != other.rank)

    def __lt__(self, other):
        return (self.rank < other.rank)

    def __le__(self, other):
        return (self.rank <= other.rank)

    def __gt__(self, other):
        return (self.rank > other.rank)

    def __ge__(self, other):
        return (self.rank >= other.rank)

Answer 1

这个想法仍然是一样的.只是你将在类对象中寻找一个特定的属性.

对于您的卡类,您可以执行以下操作:

hand = [ Card(10, 'H'), Card(2,'h'), Card(12,'h'), Card(13, 'h'), Card(14, 'h') ]

然后你可以做到

sorted_cards = sorted(hand, key=lambda x: x.rank)

输出看起来像这样:

>>> [card.number for card in sorted_cards]
[2, 10, 12, 13, 14]

Answer 2

这是面向对象的方法.至少,您应该为此指定__eq__和__lt__操作.然后就用吧sorted(hand).

class Card(object):

    def __init__(self, rank, suit):
        self.rank = rank
        self.suit = suit

    def __eq__(self, other):
        return self.rank == other.rank and self.suit == other.suit

    def __lt__(self, other):
        return self.rank < other.rank

hand = [Card(10, 'H'), Card(2, 'h'), Card(12, 'h'), Card(13, 'h'), Card(14, 'h')]
hand_order = [c.rank for c in hand]  # [10, 2, 12, 13, 14]

hand_sorted = sorted(hand)
hand_sorted_order = [c.rank for c in hand_sorted]  # [2, 10, 12, 13, 14]

最好的做法是使对象排序逻辑(如果适用)成为类的属性,而不是在每个实例中包含所需的顺序.这样你sorted(list_of_objects)每次都可以打电话.