sum of row in pattern

Data= [       1    4  0.0000;
              2    7  0.0000;
              3    9  0.0000;
              4    5  0.1760;
              4    6  0.1580;
              5    7  0.3060;
              6    9  0.3580;
              7    8  0.1490;
              8    9  0.2090]

i expect answer like this, its nothing but the similar vaule in 1st and 2nd column according to that sum of 3rd column

ANS= 1 0.0000

2 0.0000

3 0.0000

4 0.3340

5 0.4820

6 0.5160

7 0.4550

8 0.3580

9 0.5670

ex.: take number 4 => 0.0000 + 0.1760 + 0.1580 = 0.3340

take number 7 => 0.0000 + 0.3060 + 0.1490 = 0.4550

Answer 1

[David Hill]

k=unique(Data(:,1:2));
y=zeros(length(k),2);
for m=1:length(k)
    y(m,:)=[k(m),sum(Data(logical(ismember(Data(:,1),k(m))+ismember(Data(:,2),k(m))),3))];
end

Answer 2

[Image Analyst]

Try this:

Data = [ 1    4  0.0000;
	2    7  0.0000;
	3    9  0.0000;
	4    5  0.1760;
	4    6  0.1580;
	5    7  0.3060;
	6    9  0.3580;
	7    8  0.1490;
	8    9  0.2090]
col3 = Data(:, 3)
for row = 1 : max(max(Data(:, 1:2)))
	rowsToUse = any(Data(:, 1:2) == row, 2)
	theSums(row) = sum(col3(rowsToUse))
end

Answer 3

[Stephen23]

Data = [1,4,0.0000;
        2,7,0.0000;
        3,9,0.0000;
        4,5,0.1760;
        4,6,0.1580;
        5,7,0.3060;
        6,9,0.3580;
        7,8,0.1490;
        8,9,0.2090];
[U,~,X] = unique(Data(:,1:2));
Y = [Data(:,3);Data(:,3)];
Z = [U,accumarray(X,Y)]

Z = 9×2

1.0000 0 2.0000 0 3.0000 0 4.0000 0.3340 5.0000 0.4820 6.0000 0.5160 7.0000 0.4550 8.0000 0.3580 9.0000 0.5670

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Answer 4

[Adam Danz]

Data= [       1    4  0.0000;
              2    7  0.0000;
              3    9  0.0000;
              4    5  0.1760;
              4    6  0.1580;
              5    7  0.3060;
              6    9  0.3580;
              7    8  0.1490;
              8    9  0.2090];

Fastest method (so far)

D = [Data(:,[1,3]);Data(:,[2,3])];             
out = [unique(D(:,1)),accumarray(D(:,1),D(:,2))]

out = 9×2

1.0000 0 2.0000 0 3.0000 0 4.0000 0.3340 5.0000 0.4820 6.0000 0.5160 7.0000 0.4550 8.0000 0.3580 9.0000 0.5670

Slowest (1-line challenge)

out = [unique(Data(:,1:2)), arrayfun(@(i)sum(Data(any(Data(:,1:2)==i,2),3)), unique(Data(:,1:2)))]

out = 9×2

1.0000 0 2.0000 0 3.0000 0 4.0000 0.3340 5.0000 0.4820 6.0000 0.5160 7.0000 0.4550 8.0000 0.3580 9.0000 0.5670

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Comparison of 10,000 iterations between existing solutions.