If statement not executing

here, In my code, the statement to be executed after checking if abs(r1)==abs(v(i)) is not being executed. Why is that so? It is showing that Bm2 is an unrecognised variable, but clearly Bm2 will have a value if the if statement is executed.

r1 = round(r,1);
 
for i=1:length(v)
y2(i) = abs(p1* abs(v(i)*100)^4 + p2* abs(v(i)*100)^3 + p3*abs(v(i)*100)^2 + p4*abs(v(i)*100) + p5)/fd1;
B(i) = y2(i) * fd1;
end
 
 
for i=1:length(v)
if abs(r1) == abs((v(i)))
Bm2 = B(i) *10 ^-6;
else
end
end
 
for i=1:numel(u)
if u(i)==0
Bx(i)= Bm2;
else
y3(i) = abs((p11* abs(u(i))^3 + p12*abs(u(i))^2 + p13*abs(u(i)) + p14)/fd2) ;  
Bx(i) = y3(i) * Bm2;
end
end
 
figure(2);
plot(u,Bx);

Answer 1

[Fangjun Jiang]

Most likely a "floating point data equal comparison" issue. see (1/3)==(1-2/3). They are not equal if you run it in MATLAB.

[GEON G BASTIAN]

Thank you for your answer. How can this issue be solved?

because code is logically correct i guess. Just that the if statement is not being executed.

[Fangjun Jiang]

The if statement is being executed but your code is not robust. Change to

if (abs(r1)-abs(v(i)))<=eps

[Steven Lord]

It could be a floating-point issue, or it could be that r is a non-scalar and not all elements of r are equal to any of the v(i)'s.

If it's a floating-point issue, comparing with a tolerance is the usual approach. Using ismembertol instead of the loop and the equality testing is another option.

x = 0:0.1:1;
y = 0.3;
y == x(4) % false
abs(y-x(4)) < eps % true

ismembertol(y, x, eps) % true

If a vector issue, calling any is a potential solution.

if x == 0.5
    disp('Yes')
else
    disp('No')
end

if any(x == 0.5)
    disp('Yes')
else
    disp('No')
end

Answer 2

[David Hill]

You are overriding Bm2 unless there is only one case where abs(v)==abs(r1). You should be able to execute without loops. If you are having floating point issues, just add a tolerance.

r1 = round(r,1);
y2 = abs([p1,p2,p3,p4,p5]*abs((repmat(v,5,1)*100).^((4:-1:0)')))/fd1;
B = y2 * fd1;
 
Bm2 = B(abs(v)==abs(r1)) *10 ^-6;%is there only one case where v==r1? otherwise Bm2 will not be scalar

Bx(u==0)= Bm2;%Bm2 assumed to be scalar. Is Bx preallocated (Bx=zeros(1,length(u)))?
y3 = abs([p11,p12,p13,p14]*abs(repmat(u(u~=0),4,1).^((3:-1:0)')))/fd2);  
Bx(u~=0) = y3 * Bm2;%Bm2 assume to be scalar

 
figure(2);
plot(u,Bx);