How to efficiently compare two matrix to get a single reference value?

May I know how make the following code more efficient and compact. Specifically, I want to reduce the line

ConfMat (logical (((TrueVal==xx) .*(Predicted==xx))))=xx;

Here are the complete code and its output

TrueVal=  [1 1 1 2 2 2 3 3 3 1 2]';
Predicted=[1 2 3 1 2 3 1 2 3 3 2]';

ConfMat = single(ones(length(TrueVal), 1));
ConfMat (logical (((TrueVal==1) .*(Predicted==2))))=4;
ConfMat (logical (((TrueVal==1) .*(Predicted==3))))=7;

ConfMat (logical (((TrueVal==2) .*(Predicted==1))))=2;
ConfMat (logical (((TrueVal==2) .*(Predicted==2))))=5;
ConfMat (logical (((TrueVal==2) .*(Predicted==3))))=8;

ConfMat (logical (((TrueVal==3) .*(Predicted==1))))=3;
ConfMat (logical (((TrueVal==3) .*(Predicted==2))))=6;
ConfMat (logical (((TrueVal==3) .*(Predicted==3))))=9;

% Final output
% ConfMat= [1;4;7;2;5;8;3;6;9;7;5] 

Thanks in advance

[Rik]

Instead of changing that line of code, why don't you use a loop to process the vectors?

[madhan ravi]

Second Rik , I can't think of any solution other than that.

[balandong]

Thanks both of you, I have the same idea about using for loop. I just wonder if there is an elegant ways of doing it.

Answer 1

[Rik]

I don't know if this is elegant enough for you, but it does work.

TrueVal=  [1 1 1 2 2 2 3 3 3 1 2]';
Predicted=[1 2 3 1 2 3 1 2 3 3 2]';

%legend: TrueVal Predicted value
matrix=[1 2 4
    1 3 7
    2 1 2
    2 2 5
    2 3 8
    3 1 3
    3 2 6
    3 3 9];
ConfMat = single(ones(numel(TrueVal), 1));
for n=1:size(matrix,1)
    xx_TrueVal=matrix(n,1);
    xx_Predicted=matrix(n,2);
    L=((TrueVal==xx_TrueVal) & (Predicted==xx_Predicted));
    ConfMat(L)=matrix(n,3);
end
isequal(ConfMat,single([1;4;7;2;5;8;3;6;9;7;5]))

Or maybe you think this is a more elegant method:

TrueVal=  [1 1 1 2 2 2 3 3 3 1 2]';
Predicted=[1 2 3 1 2 3 1 2 3 3 2]';

% %legend: TrueVal Predicted value
% matrix=[1 2 4
%     1 3 7
%     2 1 2
%     2 2 5
%     2 3 8
%     3 1 3
%     3 2 6
%     3 3 9];
% matrix=accumarray(matrix(:,1:2),matrix(:,3),[],[],1);
matrix = [...
    1     4     7
    2     5     8
    3     6     9];
ConfMat = single(ones(numel(TrueVal), 1));
for n_true=1:size(matrix,2)
    for n_pred=1:size(matrix,1)
        L=((TrueVal==n_true) & (Predicted==n_pred));
        ConfMat(L)=matrix(n_true,n_pred);
    end
end
clc
isequal(ConfMat,single([1;4;7;2;5;8;3;6;9;7;5]))

[madhan ravi]

+1

[balandong]

Hi Rik,

Thanks for the insight. It does look smart than my initial idea. However, the proposal by Bruno is somewhat more compact. I had to accept his answer for this.

[Rik]

@balandong no problem. Both solutions have their own situation where they are the best option. It is your code, and your question, so it is on you to choose.

Just in case someone else prefers my solution, I'll keep my answer here.

Answer 2

[Bruno Luong]

TrueVal=  [1 1 1 2 2 2 3 3 3 1 2]';
Predicted=[1 2 3 1 2 3 1 2 3 3 2]';

[ut,~,it] = unique(TrueVal);
[up,~,ip] = unique(Predicted);
ConfM = [1 4 7;
         2 5 8;
         3 6 9];
     
assert(size(ConfM,1)==length(ut),'ConfM must have same #rows than #TrueVal');
assert(size(ConfM,2)==length(up),'ConfM must have same #rows than #Predicted');
ConfMat = ConfM(sub2ind(size(ConfM),it,ip))

returns

ConfMat =

     1
     4
     7
     2
     5
     8
     3
     6
     9
     7
     5

[balandong]

Dear Bruno,

Thanks for the compact answer, appreciate it