How to efficiently compare two matrix to get a single reference value?
显示 更早的评论
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
3 个评论
Rik
2019-1-15
Instead of changing that line of code, why don't you use a loop to process the vectors?
madhan ravi
2019-1-15
Second Rik , I can't think of any solution other than that.
balandong
2019-1-15
采纳的回答
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]))
更多回答（1 个）
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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