Question

我这样说:

food_to_food_type = {"apple": "fruit", "green bean": "vegetable", "tomato": "controversial"}

我的职能是,在食品类型和价值是我所希望的粮食数量时,需要一条字典,例如:

num_foods_of_type = {"fruit": 1, "vegetable": 1, "controversial": 1}

在此情况下,我想要的是每类粮食的数量相等,我经常这样做。

采用我的<条码>食物_to_food_type的字典,并将其作为我的<条码>num_foods_of_type的字典中的关键,将每一方的数值固定下来,从而形成新的字典吗?

这里我要说的是:

NUM_FOODS_DESIRED = 1
num_foods_of_type = {}
for food_type in food_to_food_type.values():
    num_foods_of_type[food_type] = NUM_FOODS_DESIRED

但是,我希望以职能方式做到这一点,以便我能够把粮食换粮食换粮食换粮食换粮食——转而我的职能:

order_food_types(magic_maplike_function(food_to_food_type.values(), NUM_FOODS_DESIRED))

当然,我可以写上<条码>魔法>,像——功能<>/条码>我,但肯定必须用一种艰难的方式来做到这一点,这是正确的吗?

Answer 1

A wonderful solution was proposed here to count the number of occurence in a python list. This can be applied to your case:

values = food_to_food_type.values()
dict( zip( values, map( values.count, values ) ) )

Apparantly, I haven t read your question sufficiently careful. You meant something like this: (?)

values = food_to_food_type.values()
dict( zip( values, [NUM_FOOD_TYPES]*len(values) ) )

Answer 2

你们可以做到这一点:

dict(zip(food_to_food_type.values(), food_to_food_type.values()*[NUM_FOODS_DESIRED]))
>>> { vegetable : 1,  fruit : 1,  controversial : 1}

Answer 3

>>> help(dict.fromkeys)
Help on built-in function fromkeys:

fromkeys(...)
    dict.fromkeys(S[,v]) -> New dict with keys from S and values equal to v.
    v defaults to None.

Thus,

dict.fromkeys(food_to_food_type.values(), NUM_FOODS_DESIRED)

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