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I'm wondering if there is library support in python (such as sklearn) for doing KNN on a data set that has a custom distance matrix (positive definite) for each data point (x is a query point, $x_i$ is a data set point): $$ d(x,x_i) = (x-x_i)^TQ_i(x-x_i) $$

I know that for a fixed positive definite matrix for all data points, this is a metric that I can transform into $$ Q = A^TA \ \ \ \ \ \ \ d(x,x_i) = (Ax - Ax_i)^T(Ax - Ax_i) $$ Which I can compute via normal KNN by first transforming the input space via multiplying $A$.

My problem of having a separate matrix for each data point came up because I have a covariance around the neighbourhood of each point. KNN can then be interpreted as what are the most likely neighbourhoods this query point lies in. If a neighbourhood doesn't vary along a dimension then we should penalize difference along that dimension highly in terms of increasing distance.

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    $\begingroup$ You can define custom metrics in sklearn stackoverflow.com/q/21052509/58737 $\endgroup$ Apr 20, 2019 at 11:29
  • $\begingroup$ Can Q_i be calculated from x_i? $\endgroup$ Apr 20, 2019 at 16:58
  • $\begingroup$ @PedroHenriqueMonforte No, that is part of the problem. If so I could use the metric keyword in sklearn. One idea I have is to append the $Q_i$ to $x_i$ and then use a custom metric function to extract each part. The problem is that I don't think this is an actual metric (triangle inequality doesn't make sense here), so k-d trees or ball trees wouldn't necessarily work. $\endgroup$
    – LemonPi
    Apr 20, 2019 at 18:12

1 Answer 1

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As pointed out by @Pratik Deoghare you can create a custom metric on sklearn kNN and you can see how in the link he provided.

But you want a function to that is different for each $x_i$, that is not a metric in the mathematical sense, but I can see how that could benefit the algorithm either way.

the function you pass as a metric (see how in the link) could be defined as

def creatmydist(AllA):
    Alist = AllA
    def mydist(x, y):
        nonlocal Alist
        if x[-1] == 0: i = y[-1]
        else: i=x[-1]        
        A = Alist[i]
        x = np.dot(A,x[0:len(x)-1])
        y = np.dot(A,y[0:len(y)-1])
        return np.dot(x-y,x-y)
    return mydist

Where AllA should be a list with all $A = (A^T)^{-1}Q$ and every $x_i$ should have as last element it's index $i$

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  • $\begingroup$ I considered doing this (comment on the original question), but I don't think k-d trees or ball trees would work without a proper metric. I'll give it a test and update when I have some time! $\endgroup$
    – LemonPi
    Apr 20, 2019 at 21:35

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