How to sum up of sum unique arrays in a matrix

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Moe
Moe 2014-11-14
回答： Roger Stafford 2014-11-14
Suppose I have a matrix a:
a = [12,7,1,1,1,1;28,5,2,1,1,1;28,4,2,2,1,1;32,10,2,1,1,1;32,10,2,2,1,1;37,2,4,1,1,1;48,11,4,1,1,1;72,10,2,1,1,1;72,10,2,2,1,1;73,1,4,1,1,1;73,6,2,1,1,1;73,7,2,2,1,1];
first array in each row is a unique id of that row. So, some of rows have same unique id. I need a matrix that give me sum up all arrays with unique id (for example: (2,2) with (3,2); (2,4) with (3,4)). such as:
b = [12,7,1,1,1,1;28,9,4,3,2,2;32,20,4,3,2,2;37,2,4,1,1,1;48,11,4,1,1,1;72,20,4,3,2,2;73,14,8,4,3,3];
Such as this example:

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2014-11-14
编辑：Azzi Abdelmalek 2014-11-14
[ii,jj,kk]=unique(a(:,1))
out=[ii cell2mat(accumarray(kk,1:numel(kk), [],@(x) {sum(a(x,2:end),1)}))]

更多回答（2 个）

Roger Stafford
Roger Stafford 2014-11-14
Assuming like IDs are already grouped together as in your example,
t = [true;diff(a(:,1))~=0;true];
b = [zeros(1,size(a,2)-1);cumsum(a(:,2:end),1)];
b = [a(t(2:length(t)),1),diff(b(t,:))];
Matt J
Matt J 2014-11-14
编辑：Matt J 2014-11-14
fun=@(v) { [ a(v(1),1), sum(a(v,2:end),1) ] };
result = cell2mat( accumarray(a(:,1) , (1:size(a,1)).', [] , fun) )

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