Question

我打算创建称为order的此方法，该方法需要以下功能。

对列表或排序的卡片组进行排序时，它从最低到最高为2C（三叶草中的2个）。

import random

class Card(object):
    def __init__(self, num, suit):
        self.num = num
        self.suit = suit

er.num）返回t1 == t2

    def num_rank(num):
        if num[0] == "A":
            return 14
        if num[0] == "J":
            return 11
        if num[0] == "Q":
            return 12
        if num[0] == "K":
            return 13
        return int(num)

class Deck(object):
    def __init__
        self.m for s in self.suit]


    def isOrdered(self):
        if self. str('2C'):
            return True

Answer 1

您可以将self.deck和sorted(self.deck)中的列表进行比较。如果它们相等，则对牌组进行排序：

from functools import total_ordering

@total_ordering
class Card(object):
    def __init__(self, num, suit):
        self.num = num
        self.suit = suit

    def __str__(self):
        return '%s%s' % (self.num,
                         self.suit)

    def __repr__(self): return str(self)

    def __lt__(self, other):
        t1 = self.suit, self.num_rank
        t2 = other.suit, other.num_rank
        return t1 < t2

    def __eq__(self, other):
        t1 = self.suit, self.num_rank
        t2 = other.suit, other.num_rank
        return t1 == t2

    @property
    def num_rank(self):
        if self.num[0] == "A":
            return 14
        if self.num[0] == "J":
            return 11
        if self.num[0] == "Q":
            return 12
        if self.num[0] == "K":
            return 13
        return int(self.num)

class Deck(object):
    def __init__(self):
        self.num = ['2','3','4','5','6','7','8','9','10','J','Q','K','A']
        self.suit = ['C', 'D', 'H', 'S']
        self.deck = [Card(r, s) for r in self.num for s in self.suit]

    def isOrdered(self):
        print('My deck :', self.deck)
        print('My sorted deck :', sorted(self.deck))

        return self.deck == sorted(self.deck)

d = Deck()
print('Deck.isOrdered() ==', d.isOrdered())

打印：

My deck : [2C, 2D, 2H, 2S, 3C, 3D, 3H, 3S, 4C, 4D, 4H, 4S, 5C, 5D, 5H, 5S, 6C, 6D, 6H, 6S, 7C, 7D, 7H, 7S, 8C, 8D, 8H, 8S, 9C, 9D, 9H, 9S, 10C, 10D, 10H, 10S, JC, JD, JH, JS, QC, QD, QH, QS, KC, KD, KH, KS, AC, AD, AH, AS]
My sorted deck : [2C, 3C, 4C, 5C, 6C, 7C, 8C, 9C, 10C, JC, QC, KC, AC, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, 10D, JD, QD, KD, AD, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AH, 2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS]
Deck.isOrdered() == False

注意：

我使用了functools.total_ordering（doc），因此只需__eq__和__lt__即可实现
通过num_rank装饰器制作@property属性
排序现在通过(suit, num_rank)进行-这就是__eq__和__lt__的定义方式。也许应该考虑参数化isOrdered()-isOrdered(by suit or by num etc...)

Answer 2

因此您可能需要做的是在卡座中循环浏览，看看是否有任何卡出现故障。换句话说，

in_order = True 
for c in range(len(self.deck)-1): 
    if self.deck[c] > self.deck[c+1]:
        in_order = False
        break
return in_order

Answer 3

如果必须实现这一点，我会将所有卡都放置为数字，这样可以简化比较，并在必要时使用__str__函数将数字转换为名称。在卡座对象中，不需要num或rank。

# only the relevant code portion
from itertools import product

FIGURES = {11: 'J', 12: 'Q', 13: 'K', 14: 'A'}

class Deck(object):
    def __init__(self):
        self.deck = [Card(r, s) for r, s in product(range(2, 15), ['C', 'D', 'H', 'S'])]

    def isOrdered(self):
        print('My deck :', self.deck)
        print('My sorted deck :', sorted(self.deck))

        return self.deck == sorted(self.deck)

class Card(object):
    def __init__(self, num, suit):
        self.num = num
        self.suit = suit

    def __str__(self):
        if num < 11:
            return '%s%s' % (self.num, self.suit)
        else:
            return '%s%s' % (FIGURES[self.num], self.suit)

    def __repr__(self): return str(self)

    # with number only comparison are easier
    def __lt__(self, other):
        return  (self.suit, self.num) < (other.suit, other.num)

    def __gt__(self, other):
        return  (self.suit, self.num) > (other.suit, other.num)

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

PS：如果您使用的是python 3，则对象的继承不是强制性的。

检查以特定字符串开头的列表

3 个答案: