How to build a new matrix based on some available information

Assume matrix E as follows:

E = [
1  260  120
1  340  60
1  455  300
1  555  60
2  620  120
2  705  60
2  720  360
2  1025  60
2  1045  60
3  1130  60
3  1345  60
3  1375  60
];

Based on the unique ID (column one) and sum up of second and third columns of matrix E, I want to create a new matrix D.

D = [
1  260  120  260
1  340  60  380
1  455  300  440
1  555  60  740
2  620  120  620
2  705  60  740
2  720  360  800
2  1025  60  1160
2  1045  60  1220
3  1130  60  1130
3  1345  60  1190
3  1375  60  1250
];

% D(1,4) = D(1,2)
% D(2,4) = D(1,4) + D(1,3)
% D(3,4) = D(2,4) + D(2,3)
% D(4,4) = D(3,4) + D(3,3)
% D(5,4) = D(5,2)                 %because ID is changed then we should reset number D(5,2) = D(5,4)
% D(6,4) = D(5,2) + D(6,3)
.
.
.

[Akira Agata]

Let me clarify.

You mentioned that D(2,4) = D(1,4) + D(2,3). But looking at the matrix D you provided, it seems that D(2,4) = D(1,4) + D(1,3), and D(3,4) = D(2,4) + D(2,3), and so forth.

[Alex Rob]

@Akira Agata Thank you. I corrected them in the question.

Answer 1

[the cyclist]

This is straightforwardly accomplished with a simple for loop:

E = [
1  260  120
1  340  60
1  455  300
1  555  60
2  620  120
2  705  60
2  720  360
2  1025  60
2  1045  60
3  1130  60
3  1345  60
3  1375  60
];

nrows = size(E,1);

E3 = zeros(nrows,1);

E3(1) = E(1,2);

for nr = 2:nrows
    if E(nr,1)==E(nr-1,1)
        E3(nr) = E3(nr-1) + E(nr-1,3);
    else
        E3(nr) = E(nr,2);
    end

end

D = [E,E3]

Answer 2

[Akira Agata]

Let me post an alternative solution:

E = [
1  260  120
1  340  60
1  455  300
1  555  60
2  620  120
2  705  60
2  720  360
2  1025  60
2  1045  60
3  1130  60
3  1345  60
3  1375  60
];

 C = zeros(size(E,1), 1);
 [G, ID] = findgroups(E(:, 1));

 for kk = 1:numel(ID)
    idx = G == ID(kk);
    temp = [E(find(idx, 1), 2); E(find(idx, nnz(idx)-1), 3)];
    C(idx) = cumsum(temp);
 end

 D = [E, C];

Answer 3

[Andrei Bobrov]

E = [
1  260  120
1  340  60
1  455  300
1  555  60
2  620  120
2  705  60
2  720  360
2  1025  60
2  1045  60
3  1130  60
3  1345  60
3  1375  60
];

D = E;
ii = find([true;diff(E(:,1)) ~= 0]);
D(:,4) = [0;D(1:end-1,3)];
D(ii,4) = D(ii,2);
add = accumarray(D(:,1),D(:,4));
D(ii(2:end),4) = D(ii(2:end),4) - add(1:end-1);
D(:,4) = cumsum(D(:,4));