Creating a Grid for cooridnates

Hello, I have an image where i have foudn the location of the spot centroid. I have the liist of all the coordimates.

What I want to do is create a set of vertical lines that is an average of the spots x positon, so in this case 4 vertical lines of which I have shown 1 below). and then do it for the horizontal lines.

In reality my images change in size and I can have upto 50x50 spots.

these are the actual coordiantes (x,y)

3.9988 76.5058

5.4914 203.4980

7.3364 330.4643

130.3843 77.1741

132.5535 204.2574

134.5383 331.2441

257.6364 77.8253

259.6390 205.0028

261.5130 331.9971

385.1154 78.8457

386.9360 205.8598

388.7871 332.6533

(Or Broken in to single column vectors)

x =

3.9988

5.4914

7.3364

130.3843

132.5535

134.5383

257.6364

259.6390

261.5130

385.1154

386.9360

388.7871

y =

76.5058

203.4980

330.4643

77.1741

204.2574

331.2441

77.8253

205.0028

331.9971

78.8457

205.8598

332.6533

Answer 1

[Matt J]

and I also do pick up a few spurious spots that are'nt part of the grid

If outliers are present, I would use the approach below, which requries the download of this FEX contribution,

https://www.mathworks.com/matlabcentral/fileexchange/78008-tools-for-processing-consecutive-repetitions-in-vectors

I don't really know the distribution of your outliers, so that may affect choices of certain parameters, as noted below in the code.

load Image_outliers

C=vertcat(regionprops(Image,'Centroid').Centroid);
x=C(:,1);
y=C(:,2);

D=pdist2([x,y],[x,y],'city','Smallest',2);
D(1,:)=[];
d=median(D(:));


Xedges=binEdges(x,d);
Yedges=binEdges(y,d);

[~,~,~,Gx,Gy]=histcounts2(x,y,Xedges,Yedges);

xl=splitapply(@median,x,Gx);
yl=splitapply(@median,y,Gy);

imshow(Image,[]);
xline(xl,'y--'); yline(yl,'y--')

function edges=binEdges(z,d)

 Z=-d/2:round(max(z)+d/2);

 T=sum(abs(Z-z(:))<=d/5,1);

 thresh=0.5*max(T); %depends on outlier rate

 [start,stop]=groupLims(groupTrue(T<=thresh),1);

 edges=(Z(start)+Z(stop))/2;

end

[Jason]

Thanks again Matt J. Im going to accept this, but could I ask something first (feel free to tell me to raise a new topic!)

You suggested using this:

D=pdist2([x,y],[x,y],'city','Smallest',2);
D(1,:)=[];
d=median(D(:));

which I have never come across before - but seems to be exactly what I need!

So if I wanted to have a seperate median for x and y distances, is this possible?

In the example below, Im trying to see if there is always a pattern to how D is calculated i.e. is it always top left and next point below?

x1 =

41.5535
43.5383
168.6390
170.5130


y1 =

51.2574
178.2441
52.0028
178.9971


D =

0         0         0         0
127.8308  127.7277  127.8308  127.7277

[Matt J]

So if I wanted to have a seperate median for x and y distances, is this possible?

If you know a fairly tight upper bound on the distance of the points from their grid lines (e.g., 10 pixels) then you can do as below,

dx=dmedian(x);
dy=dmedian(y);

function d=dmedian(u)

 D=abs(u(:)-u(:)');
 D(D<10)=inf;
 d=median(min(D));
 
end

In the example below, Im trying to see if there is always a pattern to how D is calculated i.e. is it always top left and next point below?

No, the second row of D is just the distance from each (x,y) to its nearest neighbor, whichever that happens to be.

[Jason]

Thanks Matt, it all works beautifully!

Spurious spots removed!

Can even get the angle

Answer 2

[Matt J]

Use histcounts2.

x =[3.9988
    5.4914
    7.3364
  130.3843
  132.5535
  134.5383
  257.6364
  259.6390
  261.5130
  385.1154
  386.9360
  388.7871];

y =[76.5058
  203.4980
  330.4643
   77.1741
  204.2574
  331.2441
   77.8253
  205.0028
  331.9971
   78.8457
  205.8598
  332.6533];

scatter(x,y)

[~,~,~,Gx,Gy]=histcounts2(x,y,[4,3]);

xl=splitapply(@mean,x,Gx);
yl=splitapply(@mean,y,Gy);

xline(xl,'r--'); yline(yl,'r--')

[Jason]

Hi matt, thanks, but does this assume you know the [4,3]. In reality I wont

[Matt J]

It assumes that, but there are a number of different options for specifying the binning which you will see in the doc for histcounts2. The one that applies will depend on what info you have. For example, maybe you know that the (x,y) are always separated by at least 50.

[Matt J]

Here's a morphological way you can find the number of rows and columns of points from the original binarized image:

load Image

coarsify=@(I) imdilate(cummax(I,2),ones(5,1));
R=coarsify(Image);
C=coarsify(Image')';

rows=bwconncomp(R).NumObjects

rows = 3

columns=bwconncomp(C).NumObjects

columns = 4

immontage({Image,R,C},'Bor',[15,15],'Size',[1,3],'Back','r');

[Jason]

Hi matt, wow that is beautiful and absolutely fantastic! I have some questions if thats O.K

1: Your coarsify function, what exactly does that do (in simplistic terms) , just "elongate" in 1D?

2: So I've copied your code as below:

              %%% Draw Grid if user wants ----- 
              %Calc number of discrete Rows and Cols
               coarsify=@(IM) imdilate(cummax(IM,2),ones(5,1));  %Matt J
               R=coarsify(I);    % My Image is I
               C=coarsify(I')';
               rows=bwconncomp(R).NumObjects
               columns=bwconncomp(C).NumObjects
               immontage({I,R,C},'Bor',[2,2],'Size',[1,3],'Back','r');


              [~,~,~,Gx,Gy]=histcounts2(x1,y1,[columns,rows]);
              xl=splitapply(@mean,x1,Gx);
              yl=splitapply(@mean,y1,Gy);
              xline(ax,xl,'y--'); yline(ax,yl,'y--')

But its not counting the number of rows and columns correctly - I can't see my mistake

              rows =
              
              1
              
              
              columns =
              
              1

And this is my montage and final image with lines

Also, I do know the separation but in reality there is some jitter (as my imaging system has distortion) and I also do pick up a few spurious spots that are'nt part of the grid

Thanks

Jason

[Jason]

I think I've realised why the count is wrong - bwconncomp(R) requires a binary image and not 12 bit tiff!

[Matt J]

Problem solved, then? If so, please Accept-click the answer.

[Image Analyst]

I think you could also use kmeans() to count the number of rows and column.