How to interpolate a cloud of points?

Hello all,

I have an set of 2D mean values from which I would like to compute the largest values. My 2D matrix is store so:

[rows, columns, count_there] = find(store == max(store));

By plotting rows vs columns I get the following picture:

I was wondering if there is any way to interpolate these data to get something like:

and to compute the length of it.

Answer 1

[Matt J]

I've no way of testing this without your row/column data, but:

[rows, columns, count_there] = find(store == max(store));

[idx,edges]=discretize(columns,20);
edgectr=conv2(edges,[1,1]/2,'valid')';

G=findgroups(idx);
rowctr=splitapply(@median,rows,G);
colctr=splitapply(@(x) x(1),edgectr(idx),G);

scatter(rows,columns); hold on
plot(rowctr, colctr,'k-o','LineWidth',3); hold off

Length=sum(vecnorm(diff([rowctr(:),edgectr(:)]),2,2))

[Francesco Pignatelli]

Thanks for your reply Matt J. I attach the .mat for variable store.

[Matt J]

load store
[rows, columns, count_there] = find(store == max(store));

[idx,edges]=discretize(columns,20);
edgectr=conv2(edges,[1,1]/2,'valid')';

G=findgroups(idx);
rowctr=splitapply(@median,rows,G);
colctr=splitapply(@(x) x(1),edgectr(idx),G);

scatter(rows,columns); hold on
plot(rowctr, colctr,'k-o','LineWidth',3); hold off

Length=sum(vecnorm(diff([rowctr(:),edgectr(:)]),2,2))

Length = 1.3059e+03

[Francesco Pignatelli]

It workes perfectely! Thank you so much :)

Just want to ask you the last question: how to sum up the value along the black line at different length, as shown in the following picture?

[Matt J]

It workes perfectely! Thank you so much :)

You're welcome, but please Accept click the answer to indicate so

how to sum up the value along the black line at different length

I think you would probably need to first use interparc from the File Exchange to get more row/column samples

https://www.mathworks.com/matlabcentral/fileexchange/34874-interparc?s_tid=srchtitle

Then you can use discretize() again, similar to the above

[idx,edges]=discretize(rows,20);

G=findgroups(idx);
rowctr=splitapply(@sum,columns,G);
colctr=splitapply(@(x) x(1),edgectr(idx),G);

[Francesco Pignatelli]

Hello @Matt J,

again, thank you very much. I have implemented the following part of the code:

[rowNew columnNew] = interparc(100,rows,columns,'spline');

[idx,edges]=discretize(rowNew,20);
G=findgroups(idx);
rowctr=splitapply(@sum,columnNew,G);
colctr=splitapply(@(x) x(1),edgectr(idx),G);

But it doesn't seems clear to me if it does the right thing.

[Matt J]

Is it doing the wrong thing? How do you know?

[Francesco Pignatelli]

Because when I perform

[rowNew columnNew] = interparc(0.5,rows,columns,'spline');

it gives me an error. Furthermore I do not really understand how I can get the length of the black line at a specific spot like column 700.

[Matt J]

What is "the length of the black line at a specific spot"? Do you mean you want the length over certain intervals? What defines the interval boundaries?

[Francesco Pignatelli]

Sorry if I was not clear. Till the following stage everything is ok:

now I want to sum the values of the black line at specific coulums I want to define e.g., 500, 700, 900:

basically I wanna do:

column=500 -> sum1 = 1+2

column=700 -> sum2 = 3+4

column=900 -> sum 3 = 5+6

[Matt J]

This File Exchange file seems to do a decent job,

https://www.mathworks.com/matlabcentral/fileexchange/11837-fast-and-robust-curve-intersections?s_tid=srchtitle

load store

[rows, columns, count_there] = find(store == max(store));

[idx,edges]=discretize(columns,20);
edgectr=conv2(edges,[1,1]/2,'valid')';

G=findgroups(idx);
rowctr=splitapply(@median,rows,G);
colctr=splitapply(@(x) x(1),edgectr(idx),G);

scatter(rows,columns); hold on
plot(rowctr, colctr,'k-o','LineWidth',3);

verticals=[600,700,800];
for i=1:numel(verticals)
 [x0,y0]=intersections(rowctr,colctr,verticals(i)*[1,1],[0,1000]);
 plot(x0,y0,'x--r');
 valuesum(i)=sum(y0);
end; hold off

valuesum

valuesum = 1×3

1.0e+03 * 1.0162 1.0008 1.0144

[Francesco Pignatelli]

Just tested and it is amazing! Thanks @Matt J, you helped me a lot! :)