From multiple dictionaries, generate all combinations of keys while summing their values

Let's say I have n dictionaries of different sizes, dictA, ..., dictN (n may be as great as 18). They all have strings as their keys and positive numbers as their values. What I need is a dictionary dictComb that has as its keys all combinations of keys from dictA to dictN (one key from each dictionary) with the value of each combination equal to the sum of the constituent keys' values.

Example (n=2):

dictA = {"Beefsteak": 20, "Campari": 100}
dictB = {"McIntosh": 5, "Fuji": 11, "Honeycrisp": 32}

dictComb = {
    "Beefsteak McIntosh": 25, "Beefsteak Fuji": 31, "Beefsteak Honeycrisp": 52,
    "Campari McIntosh": 105, "Campari Fuji": 111, "Campari Honeycrisp": 132
}

I tried something like this:

for (a, b) in zip(dictA, dictB):
     dictComb.update({" ".join([a, b]): sum([dictA[a], dictB[b]])})

Unfortunately, that gives:

dictComb = {"Beefsteak McIntosh": 25, "Campari Fuji": 111}

Solution

One more product usage (with dicts = dictA, dictB):

from itertools import product

dictComb = dict(zip(
    map(' '.join, product(*dicts)),
    map(sum, product(*map(dict.values, dicts)))
))

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Benchmark

With 18 dicts of two items each:

0.54 seconds  no_comment
0.70 seconds  no_comment_2
0.57 seconds  no_comment_3
1.37 seconds  Olivier_1
1.63 seconds  Olivier_2

Benchmark code:

def no_comment(dicts):
    return dict(zip(
        map(' '.join, product(*dicts)),
        map(sum, product(*map(dict.values, dicts)))
    ))

def no_comment_2(dicts):
    result = {}
    for d in dicts:
        result = d.copy() if not result else {
            f'{K} {k}': V + v
            for K, V in result.items()
            for k, v in d.items()
        }
    return result

def no_comment_3(dicts):
    result = {}
    for d in dicts:
        result = d.copy() if not result else {
            f'{K} {k}': V + v
            for k, v in d.items()
            for K, V in result.items()
        }
    return result

def Olivier_1(dicts):
    dictComb = {} 
    for keys in itertools.product(*dicts):
        keyComb = " ".join(keys)
        sum = 0
        for i, key in enumerate(keys):
            sum += dicts[i][key]
        dictComb[keyComb] = sum
    return dictComb

def Olivier_2(dicts):
    dictComb = {} 
    for keys in itertools.product(*dicts):
        keyComb = " ".join(keys)
        dictComb[keyComb] = sum(dicts[i][key] for i, key in enumerate(keys))
    return dictComb

funcs = no_comment, no_comment_2, no_comment_3, Olivier_1, Olivier_2

from itertools import product
import itertools
from timeit import repeat

# Correctness
dictA = {"Beefsteak": 20, "Campari": 100}
dictB = {"McIntosh": 5, "Fuji": 11, "Honeycrisp": 32}
dicts = dictA, dictB
want = {
    "Beefsteak McIntosh": 25, "Beefsteak Fuji": 31, "Beefsteak Honeycrisp": 52,
    "Campari McIntosh": 105, "Campari Fuji": 111, "Campari Honeycrisp": 132
}
for f in funcs:
    print(f(dicts) == want, f.__name__)
print()

# Speed
dicts = [dictA] * 18
for _ in range(2):
    for f in funcs:
        t = min(repeat(lambda: f(dicts), number=1))
        print(f'{t:.2f} seconds ', f.__name__)
    print()

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(Bah, my no_comment_2 and no_comment_3 are wrong if a middle dict is empty. Oh well, since they're not better anyway, I'm too lazy to fix that. Keeping them just to show the speed of that approach.)