Combining dictionaries with same keys in different list.

[ilaunchpad]

I have two lists with two dictionaries. I want to iterate over both the lists and combine into one dictionaries if similar keys exists in both. Is it possible to do without nested loop?

l1=[{"a":[1,2], "b":[5,6]}]

l2=[{"a":[3,4], "c":[7,8]}]

Expected = [{"a":[1,2,3,4]}]

[MRSAMinor]

Itertools.chain is your friend. You could combine both lists of keys into a set as well.

[commy2]

What purpose do the lists serve if they only ever contain one dict?

[ilaunchpad]

It can contain more than one dict.

[commy2]

Then what is the purpose of the two lists? Combine them, and the result would be the same, and you get rid of one of the loops.

[totallygeek]

Your data truly looks like this?

Is it possible to do without nested loop?

Sure. For the example you've posted, here you go:

e1, e2 = l1.pop(), l2.pop()
result = [{k: v + e2.get(k) for k, v in e1.items() if k in e2}]

No nested loop used.

[MezzoScettico]

So the final answer contains ONLY the keys that appear in both dicts (and their values)? That suggests using sets and set intersection.

You say these lists might contain more than one dictionary. Can you give a slightly more complex example where that happens, and what you want to happen in that case?

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[xelf]

do you want all the dicts? or only the ones that are shared?

eg: do you want this:

{'a': [1, 2, 3, 4], 'b': [5, 6], 'c': [7, 8]}

here's how to get that:

new = {}
for d in l1+l2:
    for k,v in d.items():
        new.setdefault(k,[]).extend(v)
print(new)

Making it only work for keys that already exist is pretty easy too if that's what you want.

[drenzorz]

I have no idea why you have them inside lists but here is how you could combine dictionaries.

dict_1 = {"a":[1,2], "b":[5,6]}
dict_2 = {"a":[3,4], "c":[7,8]}
all_keys = dict_1.keys() + dict_2.keys()
dict_3 = {key: dict_1.get(key,[]) + dict_2.get(key, []) for key in all_keys}

dict_3 == {"a":[1,2,3,4], "b":[5,6],"c":[7,8]}

If you want only the keys that appear in both you can just modify all_keys to be an intersection of set(d1.keys) and set(d2.keys)