Min and max value of many results

I have a code like below

w=0:0.001:1.393;
for i=1:length(w)
k=2;
tao=2;
T=tan((tao*w(i)-k*pi)/2)/w(i)
end

When I run this code, results like this are coming (this is part of it)

T =
       3.8973
T =
       3.9164
T =
       3.9357
T =
       3.9553
T =
        3.975    
T =
        3.995
 

How can I write a min and max value of T in one vector?

Like this for above example:

[3.8973, 3.995]

[min value, max value]

Can you help me? Thank you

[Stephen23]

http://web.archive.org/web/20210113193937/https://www.mathworks.com/matlabcentral/answers/716003-min-and-max-value-of-many-results

Answer 1

[Jan]

Use a vectorized method to obtain the results as a vector, not as a bunch of single variables:

w   = 0:0.001:1.393;
k   = 2;
tao = 2;
T   = tan((tao * w - k * pi) / 2) ./ w;

Then min and max can be applied directly:

Result = [min(T), max(T)];

Alternatively with a less efficient loop:

w   = 0:0.001:1.393;
k   = 2;  % Define constants before the loop
tao = 2;
T = zeros(1, numel(w));  % Pre-allocate
for iw = 1:numel(w)      % Prefer NUMEL instead of LENGTH
  T(iw) = tan((tao * w(iw) - k * pi) / 2) / w(iw);
end
% [EDITED] "for k" -> "for iw"

Then the above command to obtain the result is working also.

[N/A]

Thank you for your answer. But when I run this code, I took the max value but also I took -Inf value for min value. Can you help me?

-Inf        3.995

[N/A]

I should take 1.0354 for min value.

[Jan]

The -Inf was a mistake in my loop code: I've used "k" as loop counter, but it is a constant also. This is fixed now.

The vectorized code replies the same value as the loop now. For the 2nd element of w, the value of T is 1.00000033333348, which is smaller than the minimal value you assume. Why do you think it is 1.0354?

w   = 0:0.001:1.393;
iw  = 2;
k   = 2;
tao = 2;
T   = tan((tao * w(iw) - k * pi) / 2) / w(wi)

Answer 2

[Daniel Pollard]

Good news! Matlab has built in min and max commands. So it's pretty straightforward to define them and store them in a vector:

Tmin = min(T);
Tmax = max(T);
Tminmax = [Tmin, Tmax];

Just like that.

[N/A]

Thank you for your answer. But it is valid for matrices. I tried but, when I define this code, I took answer like this; (this is part of it)

Tmin =
       3.9553
Tmax =
       3.9553
Tmin =
        3.975
Tmax =
        3.975
Tmin =
        3.995
Tmax =
        3.995

[Daniel Pollard]

Ah I see. In that case I'd change your code, so that T is a vector, and my code is run outside of your loop:

w=0:0.001:1.393;
T = zeros(length(w), 1)
for i=1:length(w)
    k=2;
    tao=2;
    T(i)=tan((tao*w(i)-k*pi)/2)/w(i);
end
Tmin = min(T);
Tmax = max(T);
Tminmax = [min(T), max(T)];

That should do the trick.

As a side note, notice my semicolons; they supress the output, and are there deliberately. Without them, Matlab prints whatever you've done on that line: with them, it just stores the variables in memory. Printing values every time you go around a loop will make it take a lot longer, so it's better to store them, and then print them later (using disp) when you want/need to.

[Adam Danz]

@Dorukhan Astekin , this is the same exact conversation we had earlier today that you deleted. You can't delete this one because it has answers but the lesson to be learned is that deleting a conversation and reposting the same question isn't going to help you.

If you want free help from volunteers, you need to work with them and address their questions.

Daniel is recommending the same thing I recommended to you an hour or more ago. You've wasted your own time as well as others rather than trying to understand the suggestions and clarify the problem.

[Adam Danz]

> "Thank you for your answer. But it is valid for matrices."

This was the same response you gave me hours prior.

In the example in your question,

T =
       3.8973

T is a matrix.

T = 3.8937;
ismatrix(T)

ans = logical
   1

>you didn't understand my question

You're right that I didn't understand your question because you haven't explained the problem well and your examples do not agree with your descriptions. The addition of a for-loop in this version of your question was helpful to show the problem.

> They gave better answers than you.

I love when I see better answers. But both answers here gave the same answer I did: to use min() and max(), which didn't work for your data for reasons that were not clear until you added the for-loop explaining how you're creating T.

> Please, don't crowd under my question.

An important lesson is here to work with volunteers who are give you their time. Sometimes it's difficult to describe a problem and that often requires some back-and-forth comments so others can understand the data you're working with, your goal, and the problem. Part of helping others solve these problems is helping them to communicate them in the first place with is why I'm trying to drill this message in. When you delete a thread that contains commentary and ask a new question, it's a message that you do not value the time of other contributors whose main goal is to help you.