From equality of a list's rows to summation over different elements of a matrix

Question

My own code (for the following aim) which I think is not so good written is:

 Do[desired[k] = 0;
    Do[
       Do[
      If[list1[[i]] == list2[[j]], desired[k] += matrix[[i, j]]]

   , {i, 1, Length@list2}]
 , {j, 1, Length@list1}]
  , {k, 1, 5}]

Please see two below lists (8*5) (which are equal to each other):

list1 = {{1, 0, 0, 3, 0}, {1, 0, 0, 3, 0}, {3, 0, 0, 2, 0}, {3, 0, 0, 
1, 0}, {2, 0, 0, 3, 0}, {3, 0, 0, 3, 0}, {3, 0, 0, 3, 0}, {3, 0, 
0, 2, 0}};
list2 = {{1, 0, 0, 3, 0}, {1, 0, 0, 3, 0}, {3, 0, 0, 2, 0}, {3, 0, 0, 
1, 0}, {2, 0, 0, 3, 0}, {3, 0, 0, 3, 0}, {3, 0, 0, 3, 0}, {3, 0, 
0, 2, 0}};

Some elements of these matrices are equal, For instance: {1, 0, 0, 3, 0}, {1, 0, 0, 3, 0}. For the future, I take i index from the first list and j from the second list.

I have another matrix as

matrix={{7, 6, 9, 9, 8, 8, 10, 10}, {6, 8, 10, 6, 6, 6, 6, 8}, {8, 9, 6, 8, 
  6, 9, 5, 8}, {5, 9, 7, 6, 7, 7, 5, 8}, {10, 5, 10, 8, 8, 5, 10, 
 6}, {10, 9, 5, 5, 10, 8, 8, 7}, {7, 6, 7, 10, 7, 9, 7, 8}, {6, 7, 6,
  5, 7, 8, 7, 10}}

It is an 8*8 matrix.

I am going to try sum over some elements of the matrix that their indices are obeyed by the equality of the rows of the two upper lists: (sum over matrix[[i,j]] in which i is from the first list and j from the second one).I mean

    (*{1,0,0,3,0}*)
     desired[1] = matrix[[1, 1]] + matrix[[1, 2]] + matrix[[2, 1]] +  matrix[[2,2]];
   (*{3,0,0,2,0} *)
    desired[2] = matrix[[3, 3]] + matrix[[3, 8]] + matrix[[8, 3]]+ matrix[[8, 8]];
   (*{3,0,0,1,0}*)
    desired[3] = matrix[[4, 4]];
   (*{2,0,0,3,0} *)
    desired[4] = matrix[[5, 5]];
   (*{3,0,0,3,0}*)
     desired[5] =matrix[[6, 6]] + matrix[[6, 7]] + matrix[[7, 6]] + matrix[[7, 7]];

for the last case ((*{3,0,0,2,0}*)) we do not need a new calculation because it was considered.

Is an efficient code? Also I put the {k, 1, 5} by hand. I saw that the k cannot exceed form 5.

Answer 1

Edit2: This answer only looks at the case where list2 is identical to list1, because that was the only example given in the question. It seems that Coolwater's answer handles multiple lists the way OP would like.

---

If I understand correctly, you first want to group the row indices of like elements of the first list. You can use GroupBy :

Thread[list1 -> Range@Length@list1];
GroupBy[%, First -> Last]

<|{1, 0, 0, 3, 0} -> {1, 2}, {3, 0, 0, 2, 0} -> {3, 8}, {3, 0, 0, 1, 0} -> {4}, {2, 0, 0, 3, 0} -> {5}, {3, 0, 0, 3, 0} -> {6, 7}|>

Then you can generate all combinations of matrix indices from each group:

Tuples[#, 2] & /@ %

<|{1, 0, 0, 3, 0} -> {{1, 1}, {1, 2}, {2, 1}, {2, 2}}, {3, 0, 0, 2, 0} -> {{3, 3}, {3, 8}, {8, 3}, {8, 8}}, {3, 0, 0, 1, 0} -> {{4, 4}}, {2, 0, 0, 3, 0} -> {{5, 5}}, {3, 0, 0, 3, 0} -> {{6, 6}, {6, 7}, {7, 6}, {7, 7}}|>

Finally you can perform the sum over the specified matrix elements:

Sum[Extract[matrix, inds], {inds, #}] & /@ %

<|{1, 0, 0, 3, 0} -> 27, {3, 0, 0, 2, 0} -> 30, {3, 0, 0, 1, 0} -> 6, {2, 0, 0, 3, 0} -> 8, {3, 0, 0, 3, 0} -> 32|>

Values@%

{27, 30, 6, 8, 32}

This agrees with your expected result:

 Array[ desired, 5 ]

{27, 30, 6, 8, 32}

---

All together, for easier copying:

Thread[list1 -> Range@Length@list1];
GroupBy[%, First -> Last]
Tuples[#, 2] & /@ %
Sum[Extract[matrix, inds], {inds, #}] & /@ %
Values@%

---

Edit: Coolwater's answer below taught me about PositionIndex, which much more compactly gives exactly the same output as the first two lines of my code. So my code can be replaced with:

PositionIndex[list1]
Tuples[#, 2] & /@ %
Sum[Extract[matrix, inds], {inds, #}] & /@ %
Values@%

Answer 2

Total[Extract[matrix, {##}] & @@@ MapThread[Tuples@*List,
        Lookup[{##}, Union[Keys[#], Keys[#2]], {}]], {2}] &[
                   PositionIndex[list1], PositionIndex[list2]]

{27, 8, 6, 30, 32}

If you only need the sum of them replace the last List by {1, 2}