這可能是當從基本數據結構在Python轉換到類一個很好的例子。
考慮:
values = ('2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K', 'A')
suits = ('H', 'C', 'D', 'S')
sRep = {'H':'Hearts', 'C':'Clubs', 'D':'Diamonds', 'S':'Spades'}
ranks = {'2':15, '3':3, '4':4,'5':5,'6':6,'7':7,'8':8,
'9':9, '0':10, '0':10, 'J':11, 'Q':12, 'K':13, 'A':14 }
class Card:
def __init__(self, value, suit):
value=str(value)
self.value, self.suit = value.upper(), suit.upper()
self.rSuit = sRep[suit.upper()]
self.rank = ranks[value.upper()]
def __repr__(self):
return "%s of %s" % (self.value, self.rSuit)
def __cmp__(self,other):
if self.rank > other.rank: return 1
if self.rank < other.rank: return -1
if self.value > other.value: return 1
if self.value < other.value: return -1
if self.rSuit > other.rSuit: return 1
if self.rSuit < other.rSuit: return -1
return 0
嘗試一些卡:
c1=Card(2,'s')
c2=Card(4,'d')
if c1>c2:
print "A", c1, "beats a", c2
elif c2>c1:
print "A", c2, "beats a", c1
else:
print "Same..."
此打印:
A 2 of Spades beats a 4 of Diamonds
因爲我們在類中定義的排序順序,複雜的排序是容易的,基於不同遊戲的排名很容易。
卡列表爲例:
a = [['3D'], ['3D', '4D', '5D'], ['4C'], ['2C'],['4C', '4D'], ['4D'], ['5D'], ['JC'], ['JC', 'JS'], ['JS']]
print sorted([Card(c[0],c[1]) for e in a for c in e])
打印:
[3 of Diamonds, 3 of Diamonds, 4 of Clubs, 4 of Clubs, 4 of Diamonds,
4 of Diamonds, 4 of Diamonds, 5 of Diamonds, 5 of Diamonds, J of Clubs,
J of Clubs, J of Spades, J of Spades, 2 of Clubs]
隨着一點點更多的工作,你可以定義手,什麼手擊敗了另一手。
你可以閱讀更多關於這個例子中的經典Python書籍如何看待一個計算機科學家:與Python HERE
學習,如果所有列表的長度是1的那會工作,但X [0]的['4C','4D']的列表是'4D',所以它贏了;不能轉換 – tcatchy
@ FRU5TR8EDD:嘗試使用列表理解。查看更新。 –
天賜,非常感謝 – tcatchy