如何解决Python卡和甲板OOP
该功能可创建一副纸牌，默认情况下为52张纸牌，提供西装时仅为13张。
AND
从卡组中删除第一张card_count卡，将其作为列表返回，并在缺卡时返回相应错误的功能。
下面的预期结果。
>>> deck2 = Deck('♠')
>>> deck2.shuffle_deck()
>>> print(deck2.cards)
[A of ♠,10 of ♠,3 of ♠,7 of ♠,5 of ♠,4 of ♠,8 of ♠,J of ♠,9 of ♠,Q of ♠,6 of ♠,2 of ♠,K of ♠]
>>> deck2.deal_card(7)
[A of ♠,8 of ♠]
>>> deck2.deal_card(7)
Cannot deal 7 cards. The deck only has 6 cards left!
我的问题是，每次运行代码时，我都会返回一个空列表。我想我有第一类（PlayingCard）设置权和西装的可选参数。但是，我不太确定如何实现交易功能或为什么我的列表返回空白。我想念一些小东西吗？
import random
class PlayingCard():
def __init__(self,rank,suit):
acc_suit = ("♠","♥","♦","♣")
acc_rank = (2,3,4,5,6,7,8,9,10,"J","Q","K","A")
assert rank in acc_rank,"Not a valid rank for a playing card."
assert suit in acc_suit,"Not a valid suit for a playing card."
self.suit = suit
self.rank = rank
def __repr__(self):
return self.rank + ' of ' + self.suit
class Deck():
def __init__(self,*suit):
acc_suit = ("♠","♣")
self.cards = []
if suit == None:
for suit in range(4):
for rank in range (1,14):
card = PlayingCard(rank,suit)
self.cards.append(card)
if suit in acc_suit:
for rank in range (1,14):
card = PlayingCard(rank,suit)
self.cards.append(card)
def shuffle_deck(self):
random.shuffle(self.cards)
def deal_card(self):
return self.cards.pop(0)
def __str__(self):
res = []
for card in self.cards:
res.append(str(card))
return '\n'.join(res)
解决方法
您的列表为空的原因是，如果您提供西装，则永远都不会创建套牌。打电话
deck2 = Deck('♠')
将导致suit = ('♠',)-一个 1元组（包含1个字符串），因此它将永远不会统计if suit in acc_suit: => self.deck为空。 / p>
还有很多其他错误，使您的代码可运行-内联注释中指出：
import random
class PlayingCard():
def __init__(self,rank,suit):
# you try to create cards by range(2,14) but you assert that suit must
# be 2 to 10 or "J"-"A" for picture cards
# also: range(2,14) is [2..13] only so you miss a card
r = {11:"J",12:"Q",13:"K",14:"A"}
rank = r.get(rank,rank) # translate 10 to 14 to "JQMA"
acc_suit = ("♠","♥","♦","♣")
acc_rank = (2,3,4,5,6,7,8,9,10,"J","Q","K","A")
assert rank in acc_rank,"Not a valid rank for a playing card."
assert suit in acc_suit,"Not a valid suit for a playing card."
self.suit = suit
self.rank = rank
def __repr__(self):
# can not add a string to an integer (rank is an int for 2-10 !)
return f"{self.rank} of {self.suit}"
和您的甲板课：
class Deck():
# change this to a named param with default so you need to supply suit=' '
# omit the,* if you do not like to force suit=" " when calling this
def __init__(self,*,suit = None):
acc_suit = ("♠","♣")
self.cards = []
if not suit:
for suit in range(4):
for rank in range (1,14):
card = PlayingCard(rank,suit)
self.cards.append(card)
if suit in acc_suit:
for rank in range (2,15):
card = PlayingCard(rank,suit)
self.cards.append(card)
def shuffle_deck(self):
random.shuffle(self.cards)
# this function does not accept parameters like you posted its usage
def deal_card(self,amount = 1):
# always returns a list of cards,even if you only request one
rv,self.cards = self.cards[:amount],self.cards[amount:]
# this does NOT do the evaluation of enough cards are left,do it yourself:
# if len(rv) != amount: raise whatever
return rv
def __str__(self):
return '\n'.join(str(c) for c in self.cards)
测试：
deck2 = Deck(suit = "♠")
print(deck2.cards)
deck2.shuffle_deck()
print(deck2.cards)
print(deck2.deal_card(7))
print(deck2.deal_card(7))
输出：
# orignal
[2 of ♠,3 of ♠,4 of ♠,5 of ♠,6 of ♠,7 of ♠,8 of ♠,9 of ♠,10 of ♠,J of ♠,Q of ♠,K of ♠,A of ♠]
# shuffeled
[3 of ♠,A of ♠,2 of ♠,K of ♠]
# first 7
[3 of ♠,6 of ♠]
# next 7 (only 6 remain)
[4 of ♠,K of ♠]
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
