Operating on One Column Based on Another

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Jeremy
Jeremy el 27 de Abr. de 2018
Comentada: Jeremy el 28 de Abr. de 2018
So I've run into a buzzsaw of a problem. I have a m x n numeric matrix (no strings, etc.) and two columns are of interest. Column 1 looks like this:
A = [1;1;1;1;2;2;2;3;3;3;3;3;4;4;4;5;5;5]
Next, column 2:
B = [1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18];
What want to do is look at where values of A are equal and operate on those values in the same position in B. In other words, A(1:4) = 1 so I want to add together (or standard deviation, variance, etc.) B(1:4). Then, I want to move on to the next set of equal values of A and operate on the corresponding positions of B, and so on. Once values have been added together, I want to compare the result against a tolerance value and if the sum falls outside of the value, I want to delete those rows.
The problem I'm running into is that all of the comparison methods I've tried automatically dismiss the last value in B. So, where A(5:1) doesn't equal A(4:1), I'll get a sum, let's say, for B(1:3) instead of B(1:4). For example, one method I tried was:
a = unique(matrix(:,col_A))
tol = 5;
for i = 1:length(matrix)
idx = find(matrix(i,col_A) == a);
if isequal(matrix(i),idx)
std_dev = std(matrix(m,B));
if std_dev < tol
matrix(matrix(m,:)) = [];
end
end
end
At first, because of how isequal works, the algorithm discards A(4,1) doesn't equal A(5,1). I understand why this is happening, but I can't think of a viable solution. I tweaked it a bit, but now when I run the algorithm as written above, nothing happens at all.
If I was unclear at any point, don't hesitate to let me know. Any help you all can provide would be most appreciated!

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Stephen23
Stephen23 el 27 de Abr. de 2018
Editada: Stephen23 el 27 de Abr. de 2018
You should look into accumarray:
>> A = [1;1;1;1;2;2;2;3;3;3;3;3;4;4;4;5;5;5];
>> B = [1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18];
>> S = accumarray(A,B,[],@sum)
S =
10
18
50
42
51
>> S(A)
ans =
10
10
10
10
18
18
18
50
50
50
50
50
42
42
42
51
51
51
If the values of A are not consecutive integers starting from one, then use unique first and use its third output argument.
  3 comentarios
Stephen23
Stephen23 el 28 de Abr. de 2018
Editada: Stephen23 el 28 de Abr. de 2018
[~,~,newA] = unique(A,'stable')
Jeremy
Jeremy el 28 de Abr. de 2018
You are my new hero. That is perfect and I sincerely appreciate your time. Take care and have a good weekend!

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