Multiple answers into single array

Below is my current workspace in MATLAB.

theta = linspace(0,2*pi);
for a = [10:0.1:14]
    c = 30-a;
    bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));
    M = [max(bt)]
end

When I run this file it gives me the maximum value of 41 arrays. The answer shows like this:

M =

13.5983

M =

13.6983

M =

13.7983

M =

13.8983

M =

13.9982

I am wondering is there a way I can make my answer into a single array which I can then plot against a(10:14)?

I wish for it to look something like this:

M =

13.5983 13.6983 13.7983 13.8983 13.9982

OR

M =

13.5983

13.6983

13.7983

13.8983

13.9982

Any help would be great

Answer 1

[Clayton Gotberg]

This is definitely possible! Since you aren't using integers for indexing, there are a couple of ways to do it:

theta = linspace(0,2*pi);
M = []; % Create empty M
for a = [10:0.1:14]
    c = 30-a;
    bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));
    M = [M max(bt)]; % Add max(bt) to the end of M every loop
end

% Also

theta = linspace(0,2*pi);
count = 1; % Create empty M
for a = [10:0.1:14]
    c = 30-a;
    bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));
    M(count) = max(bt);
    count = count+1;
end

% Or

theta = linspace(0,2*pi);
area_of_interest = 10:0.1:14;
count = length(area_of_interest);

for k = 1:count
    a = area_of_interest(k)
    c = 30-a;
    bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));
    M(k) = max(bt); % Add max(bt) to the end of M every loop
end

For plotting against a(10:14), I want to check about what you mean.

If you want to plot the maximum against a from 10 to 14, just use plot(a,M). If you want to plot only against the five values a = [10 ,11, 12 ,13, 14], you can just change the area of interest from 10:0.1:14 to 10:14. If you want to plot the 10th through 14th elements of a and M, you can say plot(a(10:14),M(10:14)).

[Brayden Kirk]

Thanks for the reply,

When trying to generate a plot, I am hoping for every value of a (10.1, 10.2, 10.3,........, 14) I can plot the corresponding maximum value of M.

For example: when a=10.1, M=10.0987

[Clayton Gotberg]

That makes sense.

In that case, simply plot area_of_interest against the M you get by any of the methods I laid out above.

[Brayden Kirk]

Thank you so much for your help, it is greatly appreciated

[Brayden Kirk]

Clayton,

I do actually have another quick question if that is okay.

So the array for M is working and that is all good.

I was wondering whether I could also create another seperare array which will tell me the index of each of the maximums?

Therefore I will have a single array (M) with 41 maximum values and another single array which has the corresponding index of the 41 maximum values.

I hope this makes sense.

Cheers

[Clayton Gotberg]

It totally makes sense. One great feature of the max function is that you can also ask it to provide the index of the maximum value. If you tell the max function to give two outputs, the second one will be the index. For example:

[max_value, index] = max(array);

I should mention that if the maximum value occurs multiple times, index will only be for the earliest time it appears. If you have more questions about exactly how the function works, I recommend reading the documentation page.

[Brayden Kirk]

Okay awesome,

And then if I would like to put those index values into a single array by themselves, similar to what I did with the maximums, how would it be best to do this?

basically how would I get an array to look like:

index:

3 5 8 33 42

instead of

index

3

index

5

index

8

Sorry for the many questions, the help is incredibly appreciated

[Clayton Gotberg]

I'm happy to be of help!

You'd do it identically to how you made a matrix for the maximums. For now, I'll leave it at that - if you get stuck, let me know and I can explain indexing in more detail.

[Brayden Kirk]

I actually think I have worked it out myself, I now have this in my workspace:

theta = linspace(0,2*pi);

M=[];

N=[];

for a = [10:0.1:14]

c = 30-a;

bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));

M = [M max(bt)];

[max_value, index] = max(bt);

N= [N index];

end

M

N

A=(10:0.1:14);

plot(A,M)

[Clayton Gotberg]

That should totally work,but I'd change a few things to help make it cleaner and faster.

theta = linspace(0,2*pi);
A=10:0.1:14; % No need for parentheses
M=zeros(size(A)); % Preallocation saves time
N=M;
count = 1;
for a = A % Now you don't generate A twice
    c = 30-a;
    bt = -a.*sin(theta)+((a.^2*sin(theta).*cos(theta))/sqrt(c.^2-a.^2*sin(theta).^2));
    [M(count), N(count)] = max(bt); % No reason to calculate max again
end
M
N
plot(A,M)

I'd also recommend using more descriptive variable names (I'm on mobile or I'd be doing it too). Obviously you aren't going to forget what bt is tonight, but if you have to refer back to this in six months you'll be glad you don't have to basically resolve the question. It also improves readability for whoever looks at your code later.

[Brayden Kirk]

Thanks so much Clayton

[Clayton Gotberg]

I'm glad I could help! If you feel that this question has been fully answered, please accept my answer in order to keep the number of open questions minimized. If you need anything else, though, just ask! I also recommend the MATLAB online training courses (such as the MATLAB Onramp) if you feel like a more general tutorial would be helpful.