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    <p>The problem you are describing is generally referred to as <a href="http://en.wikipedia.org/wiki/Cluster_analysis" rel="nofollow noreferrer">clustering or cluster analysis</a>. Its can be quite difficult depending on your data set and your analysis constraints. However, in your case it is simple because you have a hard threshold (5 pixels) to use.</p> <p>Aardvarkk has already put out a great solution but it doesn't really demonstrate the process of clustering. Here is a very simple way you could cluster your data and get more or less the same result.</p> <ol> <li>Compute the pairwise distance between your points. For N points this results in an NxN matrix</li> <li>Threshold that matrix</li> <li>Iterate through the rows of the matrix</li> </ol> <p>Each iteration would look like the following:</p> <ul> <li>if <code>i</code> is already clustered, continue</li> <li>if <code>i</code> isn't in a cluster, create a new cluster and assign <code>i</code> to it</li> <li>find all other points that are close to <code>i</code> (columns in row <code>i</code> that are equal to 1)</li> <li>Check to see if any of those points are already in a cluster</li> <li>If yes set <code>i</code> and all points close to <code>i</code> to the minimum cluster id </li> <li>If no set <code>i</code> and all points close to <code>i</code> to <code>i's</code> cluster</li> </ul> <p>Here is a quick example I whipped up:</p> <pre><code>%Generate random points nPts = 300; clustId = zeros(nPts,1); clusterCount = 0; x = randi(3, [1, nPts])*10+ randn(1, nPts); y = randi(3, [1, nPts])*10 + randn(1, nPts); %Compute the distance matrix from http://goo.gl/8mAoC dist = distance([x;y], [x;y]); maxDist = 5; binDist = dist &lt;= maxDist; for i = 1:nPts % if this point is already clustered skip it if clustId(i) ~= 0 continue; end %if the current point isn't in a cluster, create a new cluster and add %the point to that cluster if clustId(i) == 0 clusterCount = clusterCount+1; clustId(i) = clusterCount; fprintf('New point, creating cluster:%d\n', clusterCount); end % get the indices of the points that collide with the i idx = binDist(:,i); % check to see if idx contains points that already belong to another clustered % if it doesn't collisionIdx will be equal to i collisionIdx = idx &amp; clustId&gt;0; %get the smallest cluster from collisionIdx mergeClustId = min(clustId(collisionIdx)); %assing all the original points to that cluster clustId(idx) = mergeClustId; end </code></pre> <p>Simply iterate through the cluster Ids to calculate the centroids:</p> <pre><code>cx = []; cy = []; for i = 1:clusterCount idx = clustId == i; cx(i) = mean(x(idx)); cy(i) = mean(y(idx)); end </code></pre> <p>Then plot the results with: </p> <pre><code>figure; plot(cx,cy,'c.', 'markersize', 50); hold on; plot(x,y,'.'); </code></pre> <p><img src="https://i.stack.imgur.com/jUGTV.png" alt="enter image description here"></p>
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    1. POReducing Local Neighbourhood of pixels
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    1. COThanks for the excellent reply. I wish to ask a few questions: first and foremost how do you get the centroid of your clusters? Secondly I believe yours and aardvark's solutions are much the same the only difference being yours is adaptable to multiple dimensions whereas his isn't (as yours is grounded in cluster analysis of which I have done some work in). My question is given the simplicity of my situation (just an image) is it really necessary to cluster the data and NOT go for the connected-component route? One other thing: I had to adapt your solution to use bsxfun (memory constraints).
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    2. COJust a continuation to give a better explanation of my last point: I don't have the very newest of PC's and the distance function given in the note you made would not run on my system. I just used the bsx function to perform the same calculation (a trick I learnt that allows me to do larger matrix multiplications on my system). Your solution is perfect, its obviously just something I had to change on my end (and will only become a bigger problem for much larger images). Just something to note as why I feel the other solution is more appropriate.
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    3. CO@user901898 I've added the code I used to calculate the cluster centers and the plotting code.
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