Remove duplicate locations and replace by a single one

Question

Jason 2017-1-6

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https://ww2.mathworks.cn/matlabcentral/answers/319279-remove-duplicate-locations-and-replace-by-a-single-one

评论： Jason 2017-1-11

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hi.

I am trying to locate the 6 dots in my frequency domain image (its the FFT of a hexagonal array of spots)

Using a technique called "template matching" via matlabs function c=normxcorr2(template,Raw), I'm able to locate the 6 best matches. However, as below some of the matches occur that are 2 for the same object (just displaced by say 1 pixel in the image)

the location x, y and actual intensity of the 6 matches are:
0000  204.0000    0.7882
0000  205.0000    0.7882
0000  315.0000    0.7607
0000   93.0000    0.7607
0000  203.0000    0.7555
0000  204.0000    0.7555

How can I replace similar locations that are within say 10 pixels in either x or y of each other with the one that has the highest intensity. If the intensities are the same just pick the first one?

Thanks for any help Jason

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Adam 2017-1-6

It would rather depend on the specific case as to what is best to do with cases up to 10 pixels away and with different intensities. In your example it is easy as they are 1 pixel apart and the same intensities.

The surface of the correlations could vary hugely over 10 pixels in both x and y though with massive dips between the two or it could be very smooth in which case averaging the x and y to take a centred location would make sense. Clearly in the first example with a very varied correlation this would not make sense though. But that comes back to the specific interpretation. Maybe if your correlation surface is so varied as to have big dips between two points within 10 pixels of each other then these should be considered as independent of each other anyway. Maybe this is a scenario that can never happen in your case?

In the example you give a simple variation of the unique function where you give a distance tolerance and pick the first or average would work fine.

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Answer 1

Walter Roberson 2017-1-6

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https://ww2.mathworks.cn/matlabcentral/answers/319279-remove-duplicate-locations-and-replace-by-a-single-one#answer_249578

https://www.mathworks.com/help/matlab/ref/uniquetol.html

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Jason 2017-1-9

编辑：Jason 2017-1-9

在 MATLAB Online 中打开

Walter. I took a look and see the way where I can ignore my 3rd column using inf as the tolerance. when I then follow the example and swap from mean to max by:

for k = 1:length(ia)
    %aveA(k,:) = mean(A(ia{k},:),1);
    aveA(k,:) = max(A(ia{k},:));
end

I get some erroneous values for my x,y coordinates (my image is only 200x200) and I output e.g. the x coordinate following the loop above

aveA=x,y, Intensity

which gives me several x (& y values) greater than 200

   199    4982
   157    4318
    26    3308
   129    5596
    17    3968
   136    5945
    86    7340
    98    4260
   106    1454
   147    3321
   174    3145
   188    2778
    38    5834
    45    3824
    72    1283
   113    4025
   725     725
   846     846
  1042    1042
  1045    1045
  1082    1082
  1125    1125
  1372    1372
  1440    1440
  1463    1463
  1598    1598
  1687    1687
  3778    3778
  4147    4147
  4709    4709
  5009    5009
  5452    5452
  6058    6058

It seems to be perhaps putting the 3rd column (i.e. intensity) for some of the rows

Walter Roberson 2017-1-9

在 MATLAB Online 中打开

aveA(k,:) = max(A(ia{k},:), 1);

In the case where there was only a single row, max by default works along the columns.

But I suspect you want something closer to

    [~, maxidx] = max(A(ia{k},3));
    aveA(k,:) = A(ik{k(maxidx)},:);

Jason 2017-1-10

编辑：Jason 2017-1-10

在 MATLAB Online 中打开

hmmm, I feel Im nearly there but not quite....

[C,ia] = uniquetol(double(A), 3, 'ByRows', true, ...
    'OutputAllIndices', true, 'DataScale', DS);
%Average each group of points that are within tolerance 
length(ia)
for k = 1:length(ia)
    k
    [~, maxidx] = max(A(ia{k},3))
    aveA(k,:) = A(ia{k(maxidx)},:);
end

gives me

Index exceeds matrix dimensions.
Error in AfunctionATS_QC>pushbutton62_Callback (line 4194)
    aveA(k,:) = A(ia{k(maxidx)},:);

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Answer 2

Image Analyst 2017-1-6

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The way I find spikes in spectra that look like that is to first flatten the hump with adapthisteq(). Then maybe some noise reduction if needed. Then imregionalmax() or simple thresholding. Then if there are still small noise specs, call bwareaopen(). At this point you should have fairly large round spots. Finally call regionprops() to get the centroids of those spots.

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Jason 2017-1-6

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So my FFT image as displayed (fLog )is via f = fftshift(fft2(IM2)); fabs=abs(f); fLog = log(1 + abs(f));

typical values are:

3449   14.0344   13.9478   14.5760   14.7319   14.7613   14.0889   13.8858   14.0256   13.8655   14.3519   14.6125   14.4053   14.5912   13.9406   14.4580   14.6937   14.6081   14.2810
6524   13.6782   13.1208   14.0463   12.8391   13.6880   14.5778   14.0849   13.8230   13.0803   13.9648   13.6392   14.9358   13.0224   13.1800   15.0674   14.7089   14.4756   14.4593
2441   14.0987   13.8662   14.9372   13.7176   14.4019   14.2667   15.1542   14.4959   14.4358   14.4857   14.3052   14.6665   13.9729   13.2541   14.8806   13.9349   14.7924   14.8968
4770   14.1713   14.4296   14.3623   14.3190   14.2337   14.3327   13.6597   14.2302   13.5637   13.5564   13.7410   14.6453   13.8705   13.6382   12.6438   13.7971   14.3293   14.0198
1876   14.1564   14.9214   12.9028   14.8500   14.1732   14.1156   14.1639   14.6839   13.0326   14.8514   14.2794   14.0046   14.4100   14.1975   14.2900   14.8544   13.5223   14.6178
9660   14.2728   14.1810   12.8452   13.8589   14.9243   14.2886   13.8030   14.5397   14.5398   14.0764   14.7958   12.8295   14.2470   13.4628   14.3374   14.1477   14.7460   13.5973
3420   14.3576   14.0068   13.7979   14.8972   14.3773   13.8775   13.4591   13.7093   14.0612   14.7619   13.8294   14.2826   13.4302   14.6212   14.6125   13.7437   14.0563   15.0035
4081   12.1394   14.4322   13.9806   11.5343   12.6346   13.5777   13.3499   14.6571   13.3027   14.2555   13.0297   12.6997   12.9625   14.0681   14.6287   14.4310   15.0224   14.7864

When I perform

J = adapthisteq(fLog)

the resultant image is just 1 everywhere.

Image Analyst 2017-1-6

编辑：Image Analyst 2017-1-6

fft_filter.m

You must not be doing it correctly. Try varying some of the parameters.

For what it's worth, attached is a demo where I take the log of the spectrum to sort of flatten it, and then threshold and filter.

People could help more if you'd attach your original image.

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Answer 3

Kelly Kearney 2017-1-10

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If you simply want to isolate the points, perhaps my findclump function could be of help:

x = [...
    75
    75
   140
   268
   333
   333];
y = [...
   204
   205
   315
    93
   203
   204];
z = rand(size(x));
[xyunq, ixy] = findclump([x y], 10);
zmax = accumarray(ixy, z, [max(ixy) 1], @max);
[~,row] = ismember([(1:max(ixy))' zmax], [ixy z], 'rows');

Original points:

[x y z]
ans =
           75          204      0.25999
           75          205      0.78686
          140          315      0.51158
          268           93      0.56253
          333          203      0.68479
          333          204     0.092397

Clumped points, keeping one with highest z value:

[x(row) y(row) z(row)]
ans =
           75          205      0.78686
          140          315      0.51158
          268           93      0.56253
          333          203      0.68479