How to find the range of a contour along each dimension of a plane with least possible error?

I have a matrix of data which I plotted as a contour using the contour function. The y-axis of the plot is taken as log of the variable, i.e. I used:

contour(x,log2(y),Data,[-99,1],'k');

The resulting plot has an irregular shape. By "irregular" I mean that it's not a nice smooth elliptical or rectangular shape, it's quite jagged at places and not evenly spread everywhere. I am posting one such plot below.

Here, I want to find out the left most point and the right most point of this contour to find the range along x-axis. Similarly, I want to find the bottom most point and the top most point of this contour to find the range along y-axis. Obviously for such shape, the x coordinates of the y_bottom and y_top will be different; y coordinates of the x_left and x_right will be different.

So, how do I find out the four points? One way is to do it by observation and select the points myself to note down the values, but there is a chance of error there. How to find it exactly with code?

Also I have another minor doubt: since I have used log2 of the y data, is there any way to minimize approximation error that I should be careful about?

Answer 1

[KSSV]

C = contour(x,log2(y),Data,[-99,1],'k');
x = C(1,:) ;
y = C(2,:) ;
iwant = [min(x) min(y) ;
        max(x) min(y) ;
        max(x) max(y) ;
        min(x) max(y)] ;

[Sourin]

This is working. However, it gave rise to another question: what if I get two or more different such contours of different size in some case? As I proceeded further in my data, that actually happened. Is there any way to find the same for each contour separately?

Also, should I post this as a separate question?

[KSSV]

Run a loop for each level...and give this code snippet inside the loop.

[Sourin]

Thank you, I will try this myself and if I fail to obtain results, I will post with my codes.

[Walter Roberson]

When you call contour and assign the output like this, the result is a contour matrix https://www.mathworks.com/help/matlab/ref/contour.html#f19-795863_sep_mw_d9e727e2-79e4-4cf6-bfaf-431a164d82b0

There are File Exchange contributions that decode these matrices into struct, after which you could loop over the struct entries

[Sourin]

No, I was not talking about different levels within a same contour. Such contours represent peak of a given height in my data, I am trying to detect the peaks. I just need the outermost level and its maximum spread. In some cases I got side by side contours (attaching a pic below for reference) and I am now trying to detect each of their ranges.

I am working on this on my own using the trick mentioned above to write a code that will separately give me the x_left, x__right; y_top, y_bottom points for each contour. However, I ran into a snag while using the technique suggested in the accepted answer so I am writing about it below:

I was first checking for the cases of a single contour on the plane (like the one I posted in main question). Quite often the first value of x-axis info, i.e. C(1,:) is giving wrong values, and sometimes the first value of y-axis info, i.e. C(2,:) is coming out as a weird number. I'm posting a pic of an output:

Here, it somehow gave C(1,:) 's first value as 1 when my x-axis data started from 14.7 and spacing between two data points was 0.01 units. Similarly, all my y-axis data were negative values but the code returned a weird large value of 1379 in the C(2,:). Only these two locations are getting affected somehow. Everything in the middle are fine.

What is going wrong here? How can I fix this?

[Walter Roberson]

If you use https://www.mathworks.com/matlabcentral/fileexchange/38863-extract-contour-data-from-contour-matrix-c then you would get out a struct array that has the per-contour information.

[Sourin]

@Walter Roberson Would you write this as answer in a separate comment? This is working without any error. It works on a single contour or multiple ones as well. If you post as a separate answer, I will be able to accept it as answer to my question. So other users can see it easily later.