Classification of a matrix to 0 and 1 matrix
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Moe
el 10 de Jul. de 2014
Comentada: Roger Stafford
el 12 de Jul. de 2014
Hello everyone,
I want matrix A to be like matrix B
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/165540/image.jpeg)
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
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Roger Stafford
el 11 de Jul. de 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 comentarios
Roger Stafford
el 12 de Jul. de 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))]);
Más respuestas (2)
Jos (10584)
el 11 de Jul. de 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 comentarios
Joseph Cheng
el 11 de Jul. de 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.
Joseph Cheng
el 10 de Jul. de 2014
Editada: Joseph Cheng
el 10 de Jul. de 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 comentarios
Joseph Cheng
el 11 de Jul. de 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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