Checking each element in a vector
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GEON G BASTIAN
el 13 de En. de 2020
Comentada: Adam Danz
el 14 de En. de 2020
Hello guys, I want to check each element of a vector v and if any element in the vector is equal to or in between -0.9 and 0.9, then i want to calculate d using one formula and if the element is not between -0.9 and 0.9, I want to use another formula. Also after each iteration, I want to display the variable d.
So for example if my vector v = -1:0.1:1 then the result should be
0.35
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.3354
0.35
but for me, the 0.35 value is not being displayed but for every iteration, the same value 0.3354 is being displayed .
v = -1 :0.1: 1;
for i= 1:length(v)
if any(v >= -0.9 & v<=0.9)
d = sqrt((h^2)+(s^2));
if any(v<-0.9 & v>0.9)
g= sqrt(s^2 + (v-0.9)^2);
d = sqrt(g^2 + h^2);
end
end
disp(d);
end
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Respuesta aceptada
Adam Danz
el 13 de En. de 2020
Editada: Adam Danz
el 14 de En. de 2020
Note, we don't know what h, s, and g are but I'm guessing they are numeric vectors of the same size as v in which case, either of these two solutions should work (obviously not tested). If they are not vectors or not the same size as v, it would probably be easy to adapt either of these solutions to their size and class.
Loop method
If you're using a loop, you should only access a single element of v at a time by indexing v(i).
v = -1 :0.1: 1;
for i= 1:numel(v) % replaced length() with numel()
if v(i) >= -0.9 || v(i)<=0.9
d = sqrt((h^2)+(s^2));
elseif v(i)<-0.9 || v(i)>0.9 % this line can be replace with just 'else'
g= sqrt(s^2 + (v(i)-0.9)^2);
d = sqrt(g^2 + h^2);
end
end
disp(d);
Vectorization method
However, you don't need a loop. The method below is more efficient and easier to read. Vectorization is kind of the point of using Matlab over competitors.
v = -1 :0.1: 1;
idx = v <= -0.9 | v >= 0.9;
d = nan(size(v));
d(idx) = (sqrt(s(idx).^2 + (v(idx)-0.9).^2) + h(idx).^2).^2;
d(~idx) = sqrt((h(~idx).^2)+(s(~idx).^2));
1 comentario
Adam Danz
el 14 de En. de 2020
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