我觉得这可能是我应该知道的,但我现在想不起来。我正在尝试获取一个函数来构建一个列表,其中列表的名称是函数中给出的参数;
例如
def make_hand(deck, handname):
handname = []
for c in range(5):
handname.append(deck.pop())
return handname
# deck being a list containing all the cards in a deck of cards earlier
问题是这会创建一个名为 handname 的列表,当我希望它被称为用户在创建手时输入的 handname 时。
有人可以帮忙吗?谢谢
请您参考如下方法:
您可以保留一个字典,其中键是手的名称,值是列表。
然后您只需说字典[handname] 即可访问特定的手。沿着:
hands = {} # Create a new dictionary to hold the hands.
hands["flush"] = make_hand(deck) # Generate some hands using your function.
hands["straight"] = make_hand(deck) # Generate another hand with a different name.
print hands["flush"] # Access the hand later.
