Vectorising equations in loop

I have made the following code (C1) using arrays. I undertsand this is bad practise so i tried to write it again this time vectorising the equations but i keep getting "Array indices must be positive integers or logical values."

%C1: Loop with array
n=0.1;
v=[];
for i=1:n:4.9
    v=[v 0.1553567*((i)^6)-2.0416*((i)^5)+9.1837*((i)^4)-14.829*((i)^3)-1.3703*((i)^2)+32.821*(i)-1.3155];
end
for i=4.9:n:15.3
    v=[v 0.003980879*(i^5)-0.2247*(i^4)+4.8682*(i^3)-50.442*(i^2)+254.67*(i)-430.66];
end
for i=15.4:n:25
    v=[v -0.073*(i^2)+6.1802*(i)+40.423];
end
disp(v)
t=0.9:n:25;
plot(t,v)


%C2: Loop with vectors
n=0.1;
v=[];
for i=1:n:4.9
    v(i)= 0.1553567*((i)^6)-2.0416*((i)^5)+9.1837*((i)^4)-14.829*((i)^3)-1.3703*((i)^2)+32.821*(i)-1.3155;
end
for i=4.9:n:15.3
    v(i)=0.003980879*(i^5)-0.2247*(i^4)+4.8682*(i^3)-50.442*(i^2)+254.67*(i)-430.66;
end
for i=15.4:n:25
    v(i)= -0.073*(i^2)+6.1802*(i)+40.423;
end
disp(v)
t=0.9:n:25;
plot(t,v)

Answer 1

[Ameer Hamza]

The main problem in your code is not using for-loop; rather, it is the dynamic growth of the v array. You need to use pre-allocation. Following shows one of the way

n=0.1;
I = 1:0.1:25;
v=zeros(size(I)); % pre-allocating the memory
for i = 1:numel(I)
    if I(i) <= 4.9
        v(i) = 0.1553567*((I(i))^6)-2.0416*((I(i))^5)+9.1837*((I(i))^4)-14.829*((I(i))^3)-1.3703*((I(i))^2)+32.821*(I(i))-1.3155;
    elseif I(i) <= 15.3
        v(i) = 0.003980879*(I(i)^5)-0.2247*(I(i)^4)+4.8682*(I(i)^3)-50.442*(I(i)^2)+254.67*(I(i))-430.66;
    else
        v(i) = -0.073*(I(i)^2)+6.1802*(I(i))+40.423;
    end
end
disp(v)
t=1:n:25;
plot(t,v)

Or if you want loop-free vectorized version

n=0.1;
I = 1:0.1:25;
v = zeros(size(I)); % pre-allocating the memory
idx1 = I<=4.9;
v(idx1) = 0.1553567*((I(idx1)).^6)-2.0416*((I(idx1)).^5)+9.1837*((I(idx1)).^4)-14.829*((I(idx1)).^3)-1.3703*((I(idx1)).^2)+32.821*(I(idx1))-1.3155;
idx2 = (I>4.9)&(I<=15.3);
v(idx2) = 0.003980879*(I(idx2).^5)-0.2247*(I(idx2).^4)+4.8682*(I(idx2).^3)-50.442*(I(idx2).^2)+254.67*(I(idx2))-430.66;
idx3 = I>15.3;
v(idx3) = -0.073*(I(idx3).^2)+6.1802*(I(idx3))+40.423;
disp(v)
t=1:n:25;
plot(t,v)