How to determine the minimum point of a plot?

Hi,

I have a plot as attached herewith in which the value of the point as shown by arrow mark is to be determined and compared to a reference value. It is plotted at a time step of 999 (t ranges from 1 to 1000).

global data;
cp=0; 
for i=999:max(length(data.variable.t))
    for j=60:max(length(data.variable.x))-1
            if data.variable.curvepressure(i,j) <= -10.2661     
            disp(data.variable.curvepressure(i,j))
             cp=1;
            break
        end
    end
end

The above code is not working and need your advice please.

[Athanasios Paraskevopoulos]

Good! What exactly is your question? Do you want to create this plot?

Answer 1

[Piyush Kumar]

@Rahul,

The code you have written will display the values less than or equal to -10.2661. The break you are using will break only from the inner loop. The outer loop continues execution.

From the subject "How to determine the minimum point of a plot?", it seems like you want to find global minima of the plot. However, the provided code does not find the global minimum of the plot.

Answer 2

[Abhinaya Kennedy]

A function can help you with finding the minimum value of your plot. You can define the function as follows:

Inputs:

• data: The data to be analyzed. This could be a vector, matrix, or structure depending on the data format.

Outputs:

• min_value: The minimum value found in the data.
• min_index: The index of the minimum value in the data.

function [min_value, min_index] = find_plot_minimum(data)
  % Assuming data is a vector
  if isempty(data)
    min_value = NaN;
    min_index = NaN;
    return;
  end
  
  [min_value, min_index] = min(data);
end

You can use this as an outline to find the minimum value from your plot and tweak it to suit your needs.

You can get additional info on the min function here: https://www.mathworks.com/help/matlab/ref/min.html

[Rahul]

Hi,

Thanks a lot. But, I'm little confused regarding the "min_index".

Can you please elaborate the role of min_index in the given function?

I tried the function. Its working only partially. The second output arguement (i.e, the min_index) is not showing in result.

rgds,

rahul

[Walter Roberson]

In that code, min() will be applied along the first non-scalar dimension of data . In the second output argument, min() will return the indices of the minimum value relative to that first non-scalar dimension . The result will be an array with that first non-scalar dimension "collapsed"

For example if data is 1 x 3 x 2 then the output will be 1 x 1 x 2 reflecting the indices of the minimums of (1,:,1) and (1,:,2)

It is common for min() to be applied to a vector, in which case the second output will be the index into the vector where the minimum occurs.

[Rahul]

Thanks everyone.

your comments are indeed helpful.

[Walter Roberson]

You need to assign two outputs from the function

[min_value, min_index] = find_plot_minimum(YourDataGoesHere)

[Rahul]

Hi Walter Roberson,

Thanks a lot. I followed your suggestion. Still I guess something is wrong at my side which results in only 1st arguement at output and not the second.

Kindly advice the correction plz.

code as below

function [min_value, min_index] = find_plot_minimum(data)
  % Assuming data is a vector
  global data;
  for i=999:max(length(data.variable.t))
    for j=60:max(length(data.variable.x))-1
        if isempty(data.variable.gradpressure(i,j))
            min_value = NaN;
            min_index = NaN;
            return;
        end
        [min_value, min_index] = min(data.variable.gradpressure(i,j));
    end
  end

ans =

-8.8415

[Walter Roberson]

length() is already a scalar, so max(length) is not doing anything useful for you.

Declaring data as a global while data is already passed in, is going to be an error soon.

You are overwriting min_value and min_index every iteration of your loops.

data.variable.gradpressure(i,j) will never be empty. If data.variable.gradpressure is numeric then (i,j) indexing will result in a scalar numeric output; If data.variable.gradpressure is a cell array, then (i,j) indexing will result in a cell array containing something

Anyhow, you need to run the function and record both outputs. You would invoke it with something like

[Output_min_value, Output_min_index] = find_plot_minimum(YourData)

This is invoking the code, not what should go onto the function line.

[Rahul]

Hi Walter,

Thanks once again. Its difficult to pass the data as you have shown in as above. The name of my data is "data" uploaded in workspace. This is because the no. of input data is large and hence it is uploaded in workspace. But it seems the code is unable to read it.

Can you tell me how to do it?

I'm getting error msg

Not enough input arguments.

Error in find_plot_minimum (line 5)

for i=998:length(data.variable.t)

[Walter Roberson]

If it is "uploaded in workspace" then you should be able to just reference it by name.

Not enough input arguments.

You need to pass data to the function.

Variables are passed by reference, not by value, so as long as it is known in the calling workspace, passing it in is fast and cheap.

%calling routine
YourData = .... something appropriate
[Output_min_value, Output_min_index] = find_plot_minimum(YourData)