如果字典中的2个值属于同一个键，它们是否属于另一个字典中的相同键？

Question

I am trying to compare to two dictionary to check the accuracy on a big dataset

I want to see if two points belong to the same key in dictionary 1, they belong to the same key in dictionary 2

I have way to much point to do a double for loop with "if points in both dictionaries" i'm looking for a faster way to compare both dictionary

dict_1 has only 1 key for each point_id where dict_2 can have multiple key for 1 point_id

both dictionary look like:

{key1 : [list of point id], key2 : [list of point id], etc}

dict_1 = {key1 : [1,2,3,4,5,6], key2 : [7,8,9,10,11,12]}  
dict_2 = {key3 : [1,2,4,6,8,11,12], key4 :[2,5,7,9,10,11,12]}

def accuracy_from_dict_to_dict(dict_1,dict_2):
    total, truth = 0,0
    for key_dict_1 in dict_1:
        point_of_key = dict_1.get(key_dict_1)
        i=0
        while i < len(point_of_key): #for each point of the key_dict_1 list
          j = i+1
          while j < len(point of key):
              for key_dict_2 in dict_2:
                  point_i = point_of_key[i]
                  point_j = point_of_key[j]
                  if point_i in key_dict_2 and point_j in key_dict_2:
                      truth += 1
                  total += 1
                  j += 1  
          i+=1

The problem is not the code itself but more the computation time. unless the data set is small enough, it's to long to be run

Answer 1

Looks like you are just checking 2-item combinations from the both dicts. You can do it better using itertools module from the standart library:

from itertools import combinations, chain

dict_1 = {'key1' : [1,2,3,4,5,6], 'key2' : [7,8,9,10,11,12]}  
dict_2 = {'key3' : [1,2,4,6,8,11,12], 'key4' :[2,5,7,9,10,11,12]}

c_dict1 = set(chain.from_iterable(combinations(v, 2) for v in dict_1.values()))
c_dict2 = set(chain.from_iterable(combinations(v, 2) for v in dict_2.values()))
total = len(c_dict1) + len(c_dict2)
similarity = len(c_dict1 & c_dict2) / total
print(total, similarity)

will print you:

69 0.2753623188405797