Python，我如何获得两个列表列表之间的百分比精度

Question

Supoose I have 2 matrices (list of lists)

list1 = [[1, 2, 3, 4], [2, 6, 7, 7], [3, 6, 2, 9], [4, 7, 4, 3]]

list2 = [[1, 2, 3, 4], [9, 4, 3, 5], [3, 5, 2, 7], [1, 9, 8, 3]]

How would I get the percentage accuracy between the 2. So if both matrices have exactly the same values (in exactly the same lists) then the accuracy is 100%

I have already tried using

len(set(test_list1) & set(test_list2)) / float(len(set(test_list1) | set(test_list2))) * 100

but it only works for singular lists not list of lists

Answer 1

This works fine

list1 = [[1, 2, 3, 4], [2, 6, 7, 7], [3, 6, 2, 9], [4, 7, 4, 3]]

list2 = [[1, 2, 3, 4], [2, 6, 7, 7], [3, 6, 2, 9], [4, 7, 4, 3]]

first_list = [item for sublist in list1 for item in sublist] 
second_list = [item for sublist in list2 for item in sublist] 

common = set(first_list).intersection(second_list)
total = set(first_list).union(second_list)

print((len(common)/len(total))*100)

Answer 2

Hop it helps:

result = []
for test_list1, test_list2 in zip(list1, list2):
    result.append(len(set(test_list1) & set(test_list2)) / float(len(set(test_list1) | set(test_list2))) * 100)
print(sum(result) / len(result))

Answer 3

You can try

print(sum([sum([1 if x == list2[inx1][inx2] else 0 for inx2, x in enumerate(i)]) for inx1, i in enumerate(list1)]) / sum([len(i) for i in list1]) * 100)

That output

43.75

This is counting the matches between the inner lists of list1 and list2 and for every match, it will count it and on miss-match it won't. Then it's dividing the results with the count of all the elements in the list1 and multiply it by 100.