# -*- coding: utf-8 -*-
# Nolaimport collections
from random import choiceCard = collections.namedtuple('Card', ['rank', 'suit'])  # 返回具有命名字段的元组的新子类 表示一张牌

suit_values = dict(spades=3, hearts=2, diamonds=1, clubs=0)  # ♣->♦->♥->♠ 2-A 花色和大小从小到大 0(2梅花色)-51(A黑桃色) 花色大小映射def spades_high(card):"""纸牌升序:param card::return:"""rank_value = FrenchDeck.ranks.index(card.rank)  # 获取纸牌大小indexreturn rank_value * len(suit_values) + suit_values[card.suit]  # 纸牌大小index * 4色 + 花色权重 2clubs=0 2diamonds=1class FrenchDeck(object):# ['2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K', 'A'] 13ranks = [str(n) for n in range(2, 11)] + list('JQKA')suits = 'spades diamonds clubs hearts'.split()  # ['spades', 'diamonds', 'clubs', 'hearts'] 4def __init__(self):self._cards = [Card(rank, suit) for rank in self.ranksfor suit in self.suits]def __len__(self):return len(self._cards)def __getitem__(self, position):return self._cards[position]if __name__ == '__main__':# namedtuple用法Point = collections.namedtuple('Point', ['x', 'y'])print(Point.__doc__)  # docstring   Point(x, y)p = Point(11, 22)print(p[0] + p[1])  # 使用indexx, y = p  # 像元组一样解包print(x, y)print(p.x, p.y)  # 使用字段名获取对应值d = p._asdict()  # 转为dict  OrderedDict([('x', 11), ('y', 22)]) 11 22print(d, d['x'], d['y'])print(Point(**d))  # dict打包为新子类  Point(x=11, y=22)print(p._replace(x=100))  # 类似str替换  Point(x=100, y=22)print('@.@ '*20 + "\n"*2)# 得到一个纸牌对象beer_card = Card('7', 'diamonds')print(beer_card)  # Card(rank='7', suit='diamonds')print('@.@ ' * 20 + "\n"*2)# len函数查看一叠牌有多少张deck = FrenchDeck()print(len(deck))  # 52print('@.@ ' * 20 + "\n"*2)# 使用__getitem__方法从一叠牌中抽取特定的一张牌print(deck[0])  # 第一张  Card(rank='2', suit='spandes')print(deck[-1])  # 最后一张  Card(rank='A', suit='hearts')print('@.@ ' * 20 + "\n" * 2)# 使用python内置函数random.choice随机抽取一张牌print(choice(deck))  # Card(rank='9', suit='hearts')print(choice(deck))  # Card(rank='3', suit='diamonds')print('@.@ ' * 20 + "\n" * 2)# 抽取最上面3张print(deck[:3])# 只看牌面是A的牌print(deck[-4:])print(deck[48:])print('@.@ ' * 20 + "\n" * 2)# 仅仅实现__getitem__方法，这一摞牌就编程可迭代了for card in deck[:5]:print(card)print('@.@ ' * 20 + "\n" * 2)# 反迭代reverse listfor card in reversed(deck):if card.rank == "K":breakprint(card)print('@.@ ' * 20 + "\n" * 2)# in运算符可用在FrenchDeck类上，因为它是可迭代的print(Card('Q', 'diamonds') in deck)print(Card('S', 'diamonds') in deck)print('@.@ ' * 20 + "\n" * 2)# spades_high函数对牌进行升序排序 标准库中sorted函数自增可根据指定key自增排序 reverse参数为True倒序for card in sorted(deck[:8], key=spades_high, reverse=True):print(card)# Card(rank='3', suit='spades')# Card(rank='3', suit='hearts')# Card(rank='3', suit='diamonds')# Card(rank='3', suit='clubs')# Card(rank='2', suit='spades')# Card(rank='2', suit='hearts')# Card(rank='2', suit='diamonds')# Card(rank='2', suit='clubs')print('@.@ ' * 20 + "\n" * 2)