Python 3 中缀运算符

Question

我正在研究python中缀运算符，我发现了这个： Python：定义我自己的运算符？

很久以前就有人问过这个问题，但我找不到最近的问题。这似乎比我通常更舒服的黑客。使用这个技巧，我改变了我的代码：

if __name__ == '__main__':
of = Infix(lambda x, y: Card(Value[x], Suit[y]))
card1 = 'EIGHT' | of | 'DIAMONDS'
card2 = 'THREE' | of | 'SPADES'
card3 = 'QUEEN' | of | 'SPADES'
assert card1.suit != card2.suit
assert card3.suit == card2.suit
assert ('NINE' | of | 'HEARTS').beats('EIGHT' | of | 'HEARTS')
assert ('NINE' | of | 'HEARTS').beats('EIGHT' | of | 'HEARTS', trump=Suit.HEARTS)
assert ('NINE' | of | 'HEARTS').beats('EIGHT' | of | 'HEARTS', trump=Suit.DIAMONDS)
assert not Card(Value.NINE, Suit.HEARTS).beats(Card(Value.TEN, Suit.HEARTS))
assert not Card(Value.NINE, Suit.HEARTS).beats(Card(Value.TEN, Suit.HEARTS), trump=Suit.HEARTS)
assert not Card(Value.NINE, Suit.HEARTS).beats(Card(Value.TEN, Suit.HEARTS), trump=Suit.DIAMONDS)
assert not Card(Value.NINE, Suit.DIAMONDS).beats(Card(Value.EIGHT, Suit.HEARTS))
assert not Card(Value.NINE, Suit.DIAMONDS).beats(Card(Value.EIGHT, Suit.HEARTS), Suit.HEARTS)
assert not Card(Value.NINE, Suit.DIAMONDS).beats(Card(Value.EIGHT, Suit.HEARTS), Suit.CLUBS)

你可以看到代码在有和没有的情况下是什么样子的，我认为像这样更具可读性。尽管如此，这是一个黑客，我也不喜欢通过字符串访问枚举。我仍在考虑：

EIGHT = Value.EIGHT
SPADES = Suit.SPADES
card = Card(EIGHT, SPADES)

我甚至可以将这两种方法结合起来，并将中缀更改为：

of = Infix(lambda x,y: Card(x, y))
card = EIGHT | of | SPADES

有没有更好的方法来做 Python 3 中的中缀？值得做吗？

编辑：

assert not (NINE | of | HEARTS).beats(TEN | of | HEARTS)
assert not (NINE | of | HEARTS).beats(TEN | of | HEARTS, trump=HEARTS)
assert not (NINE | of | HEARTS).beats(TEN | of | HEARTS, trump=DIAMONDS)
assert not Card(NINE, DIAMONDS).beats(Card(EIGHT, HEARTS))
assert not Card(NINE, DIAMONDS).beats(Card(EIGHT, HEARTS), trump=HEARTS)
assert not Card(NINE, DIAMONDS).beats(Card(EIGHT, HEARTS), trump=CLUBS)

一条评论可能暗示第一个块并不比第二个块好。我可以看到那里的逻辑。这包括我认为好的整个EIGHT = Value.EIGHT事情。(你能看出我来自编译语言领域吗？

Answer 1

只需让类通过工厂方法进行解析：

class Card:
# other stuff
@staticmethod
def fromstring(s):
value, of, suit = s.split()
return Card(Value[value], Suit[suit])