将一个字典中的带有空列表的键与另一个包含该键作为值的字典合并后找到最小的集合

Question

I have two dictionaries in Python 3-x.:

A = {1: [9], 7: [], 14: [], 32: [9], 40: []}

and

B = {1: [7, 9, 10], 7: [1, 14], 14: [7, 40], 32: [9, 40], 40: [14, 32]}

I want to find, for those empty lists as values in dict A, the key, contained as values in dict B. Then, merge that key from dict A to the value list, containing the key from A and output the shortest set of combined key and value merger.

For example, key 7 in dict A has an empty list as value. So we look, which value lists in B contains key 7. We see that 1: [7, 9, 10] and 14: [7, 40] contains "7" as value. So perform a merger to output the sets: {1, 7, 9, 10} and {14, 40, 7}. From these two sets, the final output should be the smallest set, ie, {14, 40, 7}. Note: If both sets are same size, I would like to choose the first one. The same goes for the keys 14 and 40 in dict A.

So far, I have tried the following:

temp3 = []
for k, v in A.items():
 if len(v) == 0:
       s3 = set()
       for i,j in B.items():
           if k in j:
               lst = []
               lst.append(j)
               min_list = min(lst)
               s3 = set(min_list)
               s3.add(k)
               s3.add(i)
               temp3.append(s3)
               C = set()
               for i in temp3:
                   C = set(i)
               print("temp3 for", k, "is", C)

The output is:

temp3 for 7 is {9, 1, 7, 10}
temp3 for 7 is {40, 14, 7}
temp3 for 14 is {1, 14, 7}
temp3 for 14 is {32, 40, 14}
temp3 for 40 is {40, 14, 7}
temp3 for 40 is {40, 9, 32}

I can perform the mergers. But I cannot figure out how to get the smallest set as output as describes above. Please help.

Answer 1

It's easier if you use an iterable of the candidate sets and use min on it:

for a in A:
    if not A[a]:
        sets = ({b, *B[b]} for b in B if a in B[b])
        print(a, min(sets, key=len))

Output:

7 {40, 14, 7}
14 {1, 14, 7}
40 {40, 14, 7}