如何在两个字典中使用多个键列表值

Question

Eg I have the following

d1 = {'a': [100,300,34], 'b': [200,98,45], 'c':[450,567,300]}

d2 = {'d': [300,500,234], 'e': [300,500,234], 'f':[300,500,234]}

Want to get

d3 = {'ad': [400,800,268], 'be': [500,598,279], 'cf':[750,1067,534]}

I have the following in mind to start: for i, j in d1.items():
for x, y in d2.items():

Answer 1

Here, I use a nested comprehension to achieve what you want. First, I take corresponding items in d1 and d2 using zip() . Then, I concatenate their keys, making that the key of the new dict, and use a list comprehension to sum the elements of their respective lists, which gets set as the value in the new dict.

d3 = {
        i1[0] + i2[0]: [v1 + v2 for v1,v2 in zip(i1[1], i2[1])]
        for i1,i2 in zip(d1.items(), d2.items())
    }
# {'ad': [400, 800, 268], 'be': [500, 598, 279], 'cf': [750, 1067, 534]}

Note that, before Python 3.7, the 'order' of a dict was arbitrary. Now, when you call .items() , you'll get keys in the order they were most recently added or defined, but before that you could get them in any order, and so the above would not work.

Answer 2

You can do something like this:

d3 = {k1 + k2: [p + q for p, q in zip(d1[k1], d2[k2])] for k1, k2 in zip(d1.keys(), d2.keys())}

or

Without using dictionary comprehension:

d3 = {}
for k1, k2 in zip(d1.keys(), d2.keys()):
    d3[k1 + k2] = [p + q for p, q in zip(d1[k1], d2[k2])]

Output:

{'ad': [400, 800, 268], 'be': [500, 598, 279], 'cf': [750, 1067, 534]}

Make sure to use an OrderedDict if you want to maintain the order of the keys.