for loop only shows the value of last iterations

Hi everyone, I'm having a problem while using 'for' loop in my coding. Suppose I will get a set of result in 1x12 matrix with a combination of values and empty variable but the results only shows the value of the last iteration. Is it possible if I can obtain and store every values on P1 with respect to its R value for the equation shown in the last line?

Thank you very much for your help.

Po = 101325;
V = 12000;

E = V .* (3.5.*10.^6);

Z = [25 50 100 200 300 400 600 1000 1500 2500 5000 10000];
R = Z / ((E/Po).^(1/3));

for i = min(R):max(R);
    if i <= 0.6;
       P1 = 0.01;
    elseif R <= 7;
       P1 = (-0.0014) .* i + 0.01;
    else
       P1 = [];
    end
end

Ps1 = P1 .* Po;

[Matt J]

No, a matrix cannot contain an []. For example,

P1=1:5,

P1 = 1×5

1 2 3 4 5

P1(5)=[]

P1 = 1×4

1 2 3 4

[Han Xian Chia]

Alright! Thanks Matt!

Answer 1

[Yongjian Feng]

Hello Han Xia,

You want P1 to be an array, right? How about this?

Po = 101325;
V = 12000;
E = V .* (3.5.*10.^6);
Z = [25 50 100 200 300 400 600 1000 1500 2500 5000 10000];
R = Z / ((E/Po).^(1/3));
P1=[];
for i = min(R):max(R);
    if i <= 0.6;
       P1(end+1) = 0.01;
    elseif R <= 7;
       P1(end+1) = (-0.0014) .* i + 0.01;
    else
       P1(end+1) = [];
    end
end
Ps1 = P1 .* Po;

[Han Xian Chia]

Thank you sir! I have tried it out and modified a little bit. I think this is the solution to one of my problems.

Regarding the 'for' loop, is there any way I can loop the 12 output R value only? As it is in an irregular pattern, I tried this but still couldn't get it loop within these 12 values.

R = [0.3353 0.6706 1.3412 2.6824 4.0236 5.3467 8.0471 13.4119 20.1178 33.5297 67.0593 134.1186]
for i = 1:length(R)
...
end

[Yongjian Feng]

You meant something like this:

function foo()
    Po = 101325;
    V = 12000;
    E = V .* (3.5.*10.^6);
    Z = [25 50 100 200 300 400 600 1000 1500 2500 5000 10000];
    R = Z / ((E/Po).^(1/3));
    P1=[];
    for i = 1:numel(R)
        disp(i)
        if R(i) <= 0.6
           P1(end+1) = 0.01;
        elseif R(i) <= 7
           P1(end+1) = (-0.0014) .* i + 0.01;
        else
           P1(end+1) = 0;
        end
    end
    Ps1 = P1 .* Po;

end

Answer 2

[Matt J]

Here is what you could do.

I=min(R):max(R);

G=discretize(I,[-inf,0.6,7,inf],  'IncludedEdge','right');

P1=nan(size(I));
P1(G==1)=0.01;
P1(G==2)=(-0.0014) .*I(G==2)+0.01;

[Han Xian Chia]

I tried this but it doesn't work. Thanks anyway!