# Sum matrix with repeating indices

2 views (last 30 days)
Emma Kuttler on 20 Mar 2022
Edited: Stephen23 on 21 Mar 2022
Hi! I have one matrix, A, that has one column of identifier values (with repeats and some missing numbers) and a column of values. I'd like to produce a column of all of the unique values and then another column of their sum. For example, if A is as follows, i want to produce B.
A = [1 2
1 -3
2 0
3 3
5 6
2 -4
4 0.5
7 9
7 2]
% I want to produce B
B = [1 -1
2 -4
3 3
4 0.5
5 6
7 11]

Stephen23 on 20 Mar 2022
Edited: Stephen23 on 21 Mar 2022
A simple MATLAB approach:
A = [1,2;1,-3;2,0;3,3;5,6;2,-4;4,0.5;7,9;7,2]
A = 9×2
1.0000 2.0000 1.0000 -3.0000 2.0000 0 3.0000 3.0000 5.0000 6.0000 2.0000 -4.0000 4.0000 0.5000 7.0000 9.0000 7.0000 2.0000
[G,U] = findgroups(A(:,1));
M = [U,accumarray(G,A(:,2))]
M = 6×2
1.0000 -1.0000 2.0000 -4.0000 3.0000 3.0000 4.0000 0.5000 5.0000 6.0000 7.0000 11.0000
##### 2 CommentsShowHide 1 older comment
Steven Lord on 21 Mar 2022
Instead of using findgroups and accumarray I would probably use groupsummary.
A = [1,2;1,-3;2,0;3,3;5,6;2,-4;4,0.5;7,9;7,2];
[values, groups] = groupsummary(A(:, 2), A(:, 1), @sum);
results = [groups, values]
results = 6×2
1.0000 -1.0000 2.0000 -4.0000 3.0000 3.0000 4.0000 0.5000 5.0000 6.0000 7.0000 11.0000

Voss on 20 Mar 2022
Edited: Voss on 20 Mar 2022
One perfectly valid MATLAB approach:
A = [1 2
1 -3
2 0
3 3
5 6
2 -4
4 0.5
7 9
7 2];
[uA,~,jj] = unique(A(:,1));
n_uA = numel(uA);
B = [uA zeros(n_uA,1)];
for ii = 1:n_uA
B(ii,2) = sum(A(jj == ii,2));
end
disp(B)
1.0000 -1.0000 2.0000 -4.0000 3.0000 3.0000 4.0000 0.5000 5.0000 6.0000 7.0000 11.0000
% % I want to produce B
% B = [1 -1
% 2 -4
% 3 3
% 4 0.5
% 5 6
% 7 11];

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