我觉得这可能是我应该知道的,但我现在想不到它.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
问题在于,当我希望在创建手时用户输入的任何名称时,都会创建一个名为handname的列表.
有人可以帮忙吗?谢谢
您可以保留一个字典,其中键是手的名称,值是列表.
然后你可以说字典[handname]来访问特定的手.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.