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Classification of a matrix to 0 and 1 matrix

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Moe
Moe 2014-7-10
评论： Roger Stafford 2014-7-12
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

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采纳的回答

Roger Stafford
Roger Stafford 2014-7-11
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 个评论
Moe
Moe 2014-7-11
Thanks Roger. I'm wondering if we can have matrix B same as follow picture:
Roger Stafford
Roger Stafford 2014-7-12
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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更多回答（2 个）

Jos (10584)
Jos (10584) 2014-7-11
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)
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Joseph Cheng
Joseph Cheng 2014-7-11
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 2014-7-10
编辑：Joseph Cheng 2014-7-10
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 个评论
Moe
Moe 2014-7-10
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 2014-7-11
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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