Classification of a matrix to 0 and 1 matrix

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Hello everyone,
I want matrix A to be like matrix B
ID, age and sex groups are repeated in matrix A. Matrix B is classified age based on the sex group with counting the value from matrix A. if any value in any group was repeated more than 1, then total will appear in matrix B. For example, in matrix A: ID=5, Age group=2, Sex group=2--->Then in matrix B: the value (4,5) is equal by 2

Accepted Answer

Roger Stafford
Roger Stafford on 11 Jul 2014
Assuming A and B are numerical arrays arranged as shown in your diagram,
B = accumarray([2*A(:,2)+A(:,3)-2,A(:,1)],1,[2*max(A(:,2)),max(A(:,1))]);
  2 Comments
Moe
Moe on 11 Jul 2014
Thanks Roger. I'm wondering if we can have matrix B same as follow picture:
Roger Stafford
Roger Stafford on 12 Jul 2014
For the second version, just interchange the first and second columns in the formula:
B = accumarray([2*A(:,1)+A(:,3)-2,A(:,2)],1,[2*max(A(:,1)),max(A(:,2))]);

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More Answers (2)

Jos (10584)
Jos (10584) on 11 Jul 2014
A = [1,7,1 ; 2,2,1 ; 2,4,2 ; 3,13,2 ; 3,11,2 ; 4,6,2 ; 5,2,2 ; 5,2,2 ; 5,9,1 ; 6,7,1 ; 6,10,2 ; 7,8,1 ; 7,6,2 ; 7,6,2 ; 7,6,1 ; 7,1,1 ; 7,12,2];
% If A is as above:
nID = 7 ;
nAge = 13 ;
nSex = 2 ;
B = reshape(accumarray(A(:,[3 2 1]),1,[nSex nAge nID]) ,[],nID)
  4 Comments
Joseph Cheng
Joseph Cheng on 11 Jul 2014
hmmm? there seems to be a missing comment before mine where the person posting asked for it the other way around. so I commented how to rewrite your function. Jos you had it correct the first time corresponding to the question asked.

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Joseph Cheng
Joseph Cheng on 10 Jul 2014
Edited: Joseph Cheng on 10 Jul 2014
I would first create a matrix Btemp of size max(AgeGroup) by max(ID) by max(SexGroup) full of zeros. then do a loop for each row of A to add Btemp(AgeGroup,ID,SexGroup) with 1; after you loop for each row of A then make your B matrix by stagering both sexgroup
Air coding so pardon any syntax mistakes:
Btemp = zeros(max(A(:,2)),max(A(:,1)),max(A(:,3))); %create all zeros
%add 1 for each instance listed in matrix A.
for row=1:size(A,1)
Btemp(A(row,2),A(row,1),A(row,3)) = Btemp(A(row,2),A(row,1),A(row,3))+1;
end
B=zeros(max(AgeGroup)*2,max(ID));
%every other row (even and odd) are the sexgroups. sexgroup1 is 1,3,5.... sexgroup2 is 2,4,6...
B(1:2:end,:) = Btemp(:,:,1);
B(2:2:end,:) = Btemp(:,:,2);
i think that should do it. or at least gives you a good starting point to correct my 5 min code.
  2 Comments
Moe
Moe on 10 Jul 2014
Joseph! I couldn't run/edit your code, could you edit it?
A= [1,7,1;2,2,1;2,4,2;3,13,2;3,11,2;4,6,2;5,2,2;5,2,2;5,9,1;6,7,1;6,10,2;7,8,1;7,6,2;7,6,2;7,6,1;7,1,1;7,12,2];
Joseph Cheng
Joseph Cheng on 11 Jul 2014
i got lazy and all you needed to do was switch out AgeGroup and ID with A(:,2) and A(:,3)

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