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    <p>This is a difficult problem. I'd even claim it to be impossible to solve in general, because of what I'd call the transitivity problem. Suppose that we have three elements in a set, {A,B,C}. I'll define a simple function isSimilarTo, such that isSimilarTo(A,B) will return a true result if the two inputs are within a specified tolerance of each other. (Note that everything I will say here is meaningful in one dimension as well as in multiple dimensions.) So if two numbers are known to be "similar" to each other, then we will choose to group them together.</p> <p>So suppose we have values {A,B,C} such that isSimilarTo(A,B) is true, and that isSimilarTo(B,C) is also true. Should we decide to group all three together, even though isSimilarTo(A,C) is false?</p> <p>Worse, move to two dimensions. Start with k points equally spaced around the perimeter of a circle. Assume the tolerance is chosen such that any point is within the specified tolerance of its immediate neighbors, but not to any other point. How would you choose to resolve which points are "unique" in the setting?</p> <p>I'll claim that this problem of intransitivity makes the grouping problem not possible to resolve, at least not perfectly, and certainly not in any efficient manner. Perhaps one might try an approach based on a k-means style of aggregation. But this will be quite inefficient, as well, such an approach generally needs to know in advance the number of groups to look for.</p> <p>Having said that, I would still offer a compromise, something that can sometimes work within limits. The trick is found in <a href="http://www.mathworks.com/matlabcentral/fileexchange/8354" rel="noreferrer">Consolidator</a>, as found on the Matlab Central file exchange. My approach was to effectively round the inputs to within the specified tolerance. Having done that, a combination of unique and accumarray allows the aggregation to be done efficiently, even for large sets of data in one or many dimensions.</p> <p>This is a reasonable approach when the tolerance is large enough that when multiple pieces of data belong together, they will be rounded to the same value, with occasional errors made by the rounding step.</p>
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