我正在尝试为类编写一个方法,该方法将创建类的现有实例的新实例。问题是当我尝试使用new_handname时,我无法在控制台中访问新实例。
这是用于在python中创建二十一点游戏。该代码的想法是,当分手时,将创建一个新实例来创建新手
import random
class Card(object):
def __init__(self, value, suit,nvalue):
self.value = value
self.suit = suit
self.nvalue = nvalue
suit = ['Hearts','Spades','Clubs','Diamonds']
value = ['2','3','4','5','6','7','8','9','10','J','Q','K','A']
nvalue = [2,3,4,5,6,7,8,9,10,10,10,10,11]
class Hand(object):
def __init__(self,current_hand):
self.current_hand = current_hand
def hand_total(self):
current_sum = 0
for i in range(0,len(self.current_hand)):
current_sum += self.current_hand[i].nvalue
return current_sum
def hand_type(self):
if self.current_hand[0].value == self.current_hand[1].value:
return('pair')
elif self.current_hand[0].value == 'A' or self.current_hand[1].value == 'A':
return('soft')
else:
return('hard')
def append(self,current_hand,some_card):
self.current_hand = self.current_hand + some_card
def hit(self):
self.current_hand.append(deck[0])
deck.pop(0)
def double(self,new_handname):
new_handname = Hand(self)
def deal_start_hand():
player_hand.append(deck[0])
deck.pop(0)
dealer_hand.append(deck[0])
deck.pop(0)
player_hand.append(deck[0]) #### player gets two cards ### assuming europe no hole card rules
deck.pop(0)
def gen_deck():
for v,n in zip(value,nvalue):
for s in suit:
deck.append(Card(v,s,n))
### variable initiation ###
deck = []
player_hand = []
dealer_hand = []
##program start ##
gen_deck()
random.shuffle(deck)
deal_start_hand()
p1 = Hand(player_hand)
p1.double('p2')
p2 ### I expect p2 to return an instance but does not
>>> p1
<__main__.Hand object at 0x00000006A80F0898>
>>> p2
Traceback (most recent call last):
File "<pyshell#182>", line 1, in <module>
p2
NameError: name 'p2' is not defined
注意:current_hand是纸牌对象的列表。
我希望p2返回该类的实例,但未定义变量p2
答案 0 :(得分:1)
您的split
例程如下所示,其中返回了该类的新实例:
class Hand(object):
def __init__(self, current_hand):
self.current_hand = current_hand
def split(self):
return Hand(self.current_hand)
只需创建一个实例,然后将其拆分:
# You need to define "some_default" somewhere
myhand = Hand(some_default)
second_hand = myhand.split()
但是,您的split
例程需要考虑已经玩过哪些牌,以及仍然在牌堆中的哪些牌,而您的代码都不会考虑。我可能会建议绘制游戏的“状态”(将其视为状态机),将其绘制在纸上,然后考虑如何编写每个状态和过渡。像这样的纸牌游戏比乍看起来看起来要复杂得多。