Azzera filtri
Azzera filtri

Classification of a matrix to 0 and 1 matrix

6 visualizzazioni (ultimi 30 giorni)
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

Risposta accettata

Roger Stafford
Roger Stafford il 11 Lug 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 Commenti
Moe
Moe il 11 Lug 2014
Thanks Roger. I'm wondering if we can have matrix B same as follow picture:
Roger Stafford
Roger Stafford il 12 Lug 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))]);

Accedi per commentare.

Più risposte (2)

Jos (10584)
Jos (10584) il 11 Lug 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 Commenti
Joseph Cheng
Joseph Cheng il 11 Lug 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.
Moe
Moe il 12 Lug 2014
Thanks Jos & Joseph Cheng.

Accedi per commentare.


Joseph Cheng
Joseph Cheng il 10 Lug 2014
Modificato: Joseph Cheng il 10 Lug 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 Commenti
Moe
Moe il 10 Lug 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 il 11 Lug 2014
i got lazy and all you needed to do was switch out AgeGroup and ID with A(:,2) and A(:,3)

Accedi per commentare.

Categorie

Scopri di più su Creating and Concatenating Matrices in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by