how to accumulate results from if, end

Question

imola 2014-10-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/157544-how-to-accumulate-results-from-if-end

编辑： imola 2014-10-11

采纳的回答： Geoff Hayes

在 MATLAB Online 中打开

Dear all,

I'm checking points inside rectangle and I got result 1 for every point inside

, I want to accumulate this result but I don't no how to do it. could anyone help me. (i) is the points inside and the result is(PointsInside=1), my code is down

for m=1:11
AB=((XY(1,1)-XY(1,2))*(yy(:,m)-XY(2,2))-(XY(2,1)-XY(2,2))*(xx(:,m)-XY(1,2))>=0);
BC=((XY(1,2)-XY(1,3))*(yy(:,m)-XY(2,3))-(XY(2,2)-XY(2,3))*(xx(:,m)-XY(1,3))>=0);
CD=((XY(1,3)-XY(1,4))*(yy(:,m)-XY(2,4))-(XY(2,3)-XY(2,4))*(xx(:,m)-XY(1,4))>=0);
DA=((XY(1,4)-XY(1,1))*(yy(:,m)-XY(2,1))-(XY(2,4)-XY(2,1))*(xx(:,m)-XY(1,1))>=0);
if (AB||BC||CD||DA)==0
   PointsInside=1
end
end

I hope somebody will help me

thanks

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Answer 1

Geoff Hayes 2014-10-6

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/157544-how-to-accumulate-results-from-if-end#answer_154083

在 MATLAB Online 中打开

imola - just define an empty array outside of your for loop, and add the point that is inside the rectangle to this array whenever it satisfies the condition. Something like

 pointsInRect = [];
 for k=1:11
    AB=((XY(1,1)-XY(1,2))*(yy(:,k)-XY(2,2))-(XY(2,1)-XY(2,2))*(xx(:,k)-XY(1,2))>=0);
    BC=((XY(1,2)-XY(1,3))*(yy(:,k)-XY(2,3))-(XY(2,2)-XY(2,3))*(xx(:,k)-XY(1,3))>=0);
    CD=((XY(1,3)-XY(1,4))*(yy(:,k)-XY(2,4))-(XY(2,3)-XY(2,4))*(xx(:,k)-XY(1,4))>=0);
    DA=((XY(1,4)-XY(1,1))*(yy(:,k)-XY(2,1))-(XY(2,4)-XY(2,1))*(xx(:,k)-XY(1,1))>=0);
    if (AB||BC||CD||DA)==0
       pointsInRect = [pointsInRect ; xx(1,k) yy(1,k)];
    end
 end

As for your conditions, you are saying that if all are false then the point must be in the rectangle. Is that correct?

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imola 2014-10-6

Dear Geoff,

Thank you very much you always help me I really appreciate that, that's great when I check now 8 points where inside I got them but I need to get the number of them, I will work on it and if I couldn't I might ask you.

Regards

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