Sum elements in the first column according to repeated values in a second column

Hi, I have a two columns matrix, where the elements of the second column are repeated (here below in my example, 55, 98 and 17 are repeated several times).

I would like to sum the elements of the first column, which have (in common) the same element of the second column. Here an example... any fast way to do it?

% Input
A = [ 4 55
      3 55
      7 55
      1 55
      9 98
      1 98
      2 98
      8 98
      0 98
      1 98
      6 17
      3 17
      0 17
      4 17]

% desired output
B = [15 55
     21 98
     13 17];

where, for example, the value 15 in B(1,1) is given by 4 + 3 + 7 + 1 = 15

Answer 1

[Star Strider]

A = [ 4 55
      3 55
      7 55
      1 55
      9 98
      1 98
      2 98
      8 98
      0 98
      1 98
      6 17
      3 17
      0 17
      4 17];

[A2u,~,ix] = unique(A(:,2));
A1sums = accumarray(ix,A(:,1),[],@sum);

Desired_Result = [A1sums A2u]

Desired_Result = 3×2

13 17 15 55 21 98

The loops are still there, however they are hidden inside the accumarray call.

.

[Sim]

Cool!! Many thanks @Star Strider!

[Star Strider]

As always, my pleasure!

[Johan]

I figured a pretty obfuscated way to do that without for loop; note that it's slower than using accumarray :)

A = [ 4 55
      3 55
      7 55
      1 55
      9 98
      1 98
      2 98
      8 98
      0 98
      1 98
      6 17
      3 17
      0 17
      4 17];
 
logicalmask = A(:,2) == unique(A(:,2))';
result = [sum(A(:,1).*logicalmask)',unique(A(:,2))]

result = 3×2

13 17 15 55 21 98

[Sim]

Thanks a lot for your solution @Johan Pelloux-Prayer!! It is very interesting! :)

(I would accept all the answers/solutions :) )

Answer 2

[Sim]

I found this solution, but I would like to avoid the loop for:

[a,b] = unique(A(:,2));
[~,d] = unique(A(:,2),'last');
for i = 1 : size(a,1)
    B1(i) = sum(A(b(i):d(i),1));
end
B = [B1' a]

% result
B =

    13    17
    15    55
    21    98