I need to classify my result in catagories

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Mahmoud Sami
Mahmoud Sami 2019-4-6
评论： dpb 2019-4-6
I have this in my code
N=ones(1,numel(ma));
% ma is a matrix can be given from other equations.
N(ma==45 | ma==135)=0.5;
N(0< ma<45 | 45<ma<90)=1.5;
N(0< ma<45 | 45<ma<90)=1.5;
N(90<ma<135 | 135<ma<180)=1.5;
N(ma==0 | ma==180)=0;
N(ma==90)=1;
NA=N';
NN=sum(NA(:));
Pn=ones(1,numel(NN));
Pn(NN>5)= 100000;
Pn(NN==5)=100000;
Pn(NN<5)=1;
I had the pervious fcn, but is doesn’t work.
The results as
ma=[-35.2644,0,0,-45.0000,-45.0000,0,35.2644,54.7356,30.0000,45.0000,0,-45.0000,30.0000,-54.7356,-54.7356,-54.7356,-35.2644]
N=[1.5000,0,0,1.5000,1.5000,0,1.5000,1.5000,1.5000,1.5000,0,1.5000,1.5000,1.5000,1.5000,1.5000,1.5000]
In other words I want to say:-
If ma=45 or 135 (with negative or positive signs) make N =0.5
If ma=0 or 180 , make N =0
If ma=90 or 275 (with negative or positive signs) make N =1
Otherwise make N= 1.5
Then Pn like that
If NN>or =5 make Pn=100000
If NN<5 Make Pn=1

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采纳的回答

Star Strider
Star Strider 2019-4-6
Try this:
ma=[-35.2644,0,0,-45.0000,-45.0000,0,35.2644,54.7356,30.0000,45.0000,0,-45.0000,30.0000,-54.7356,-54.7356,-54.7356,-35.2644,NaN,90 275,180];
sdv = abs(sind(ma));
cls = 1.5*ones(size(sdv)); % Classification Vector
cls((sdv >= 0.69) & (sdv <= 0.72)) = 0.5;
cls(sdv > 0.9) = 1;
cls(sdv < 0.1) = 0;
To see the result:
Out = [ma; cls] % Result & Matching Inputs
Experiment to get the result you want.

更多回答（1 个）

dpb
dpb 2019-4-6
N=1.5*ones(size(ma));
N(abs(ma)==45 | abs(ma)==135)=-0.5;
...
The rest should be self-evident from the above; lett as "exercise for student" :)
Read sections on 'logical indexing' for how and why the above works but it is a most powerful Matlab coding idiom/syntax.
  2 个评论
Mahmoud Sami
Mahmoud Sami 2019-4-6
That's not working as i was thought.
The problem is that the result can't be as written in code.
I think there is something wrong
dpb
dpb 2019-4-6
Works just fine here...
>> N=1.5*ones(size(ma))
N =
1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000
>> N(abs(ma)==45 | abs(ma)==135)=-0.5
N =
1.5000 1.5000 1.5000 -0.5000 -0.5000 1.5000 1.5000 1.5000 1.5000 -0.5000 1.5000 -0.5000 1.5000 1.5000 1.5000 1.5000 1.5000
>> [ma;N]
ans =
-35.2644 0 0 -45.0000 -45.0000 0 35.2644 54.7356 30.0000 45.0000 0 -45.0000 30.0000 -54.7356 -54.7356 -54.7356 -35.2644
1.5000 1.5000 1.5000 -0.5000 -0.5000 1.5000 1.5000 1.5000 1.5000 -0.5000 1.5000 -0.5000 1.5000 1.5000 1.5000 1.5000 1.5000
>>
from which can see the substitution of -0.5 for the 45-deg positions...same will work for the other values as well...
Post your code and error messages...

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