Removing neighbours that are too close from eachother in a vector

Hello,

I am trying to solve a problem in which I have, for example, the following vector:

x = [1 8 9 10 15 20 22 25 34]

And i want the output to be like this:

x = [1 9 15 22 34]

So basically analyzing which values are too close from eachother and removing them, while keeping the "middle" one (the one closest to the average of the points removed).

I've tried with

uniquetol(x], 5, 'DataScale', 1)

But uniquetol() only allows to select the hightest or lowest of the values removed. Any ideas/help are welcomed.

[Adam Danz]

Define "too close". What should the threshold be?

How do you plan to deal with ties, for example, if two values are equally close to the mean?

[DGM]

Since you're operating on multiple elements simultaneously, it's unclear what should happen when you have a run of similar values, as your filter may both accept and reject an element in succession.

x = [1 2 3 4];

Assuming a 3-element window:

At step 1, you reject 1 and 3, and you accept 2

At step 2, you reject 2 and 4, and accept 3.

Is this merely determined by the filter direction?

[Rafael Marques]

First of all, thank you for your answers!

The vector is designed to never have ties.

And assuming:

x = [1 2 3 4];

The output should be x=[2] so since both 2 and 3 are closest to average, we pick the smallest number of them. I'm sorry I wasn't clear with my question

Answer 1

[Stephen23]

x = [1,8,9,10,15,20,22,25,34]

x = 1×9

1 8 9 10 15 20 22 25 34

[~,~,y] = uniquetol(x, 5, 'DataScale',1);
z = accumarray(y(:),x(:),[],@(v) v(ceil(end/2)))

z = 5×1

1 9 15 22 34

Answer 2

[DGM]

Here's my guess. The behavior is dependent on whether you can actively accept or reject entries. If it's assumed that values can only be rejected, then:

x = [1 8 9 10 15 20 22 25 34];
%x = [1 9 15 22 34];

wradius = 1; % for a 1x3 window
badrange = 5; % anything less is "too close"

goodvalues = true(size(x));
for k = 1+wradius:numel(x)-wradius
    sidx = k-wradius:k+wradius;
    sample = x(sidx);
    if max(sample)-min(sample) <= badrange
        av = mean(sample);
        [~,idx] = min(abs(sample-av));
        mask = false(1,3);
        mask(idx) = true;
        goodvalues(sidx) = mask & goodvalues(sidx);
    end
end

output = x(goodvalues)

output = 1×5

1 9 15 22 34

I'm only picking the first value that's "closest" to the mean. If there are ties, it's going to keep the leftmost match.

Answer 3

[Image Analyst]

Sounds like you want to use the dbscan clustering algorithm.

So in the above diagram from Wikipedia the red and yellow points are all in one cluster because they are all within a certain distance of at least one other point in the cluster. Point N is not in that cluster - it's in it's own cluster because it is not closer than the specified distance from any of the other points.

I'm attaching an example as applied to images but you could apply it to a 1-D numberline like you have. Once you find the clusters, replace each cluster with the mean of the cluster and then remove duplicates.

dbscan is one of the best clustering algorithms and one you should learn about even if you don't see how it can apply here.