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

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balandong
balandong 2019-1-15
评论： Rik 2019-1-17
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
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balandong
balandong 2019-1-15
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.

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采纳的回答

Rik
Rik 2019-1-15
编辑：Rik 2019-1-15
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]))
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balandong
balandong 2019-1-17
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
Rik 2019-1-17
@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.

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更多回答（1 个）

Bruno Luong
Bruno Luong 2019-1-15
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

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