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I want to automatically find the "knee" point of the eigenvalue plot. I.e. I have a vector of eigenvalues (sorted from highest to lowest) and I want some heuristic to find the "knee" point.

Is there some heuristic for doing that

I've found the two following proposals so far.

  1. Setting a threshold, say 0.99, or 0.95 and keep m of n eigenvalues when T(m-1) < 0.99 *T(n) <= T(m) where T(m) = sum(i=1:m){lambda(i)}

  2. The knee is located at a point where the radius of curvature is a local minimum. For a curve y = f(x) the curvature is k = y''/(1+(y')^2)^(3/2). Just replace the derivatives with finite differences.

What do you think of these two proposals? How can I implement the second one? I don't understand how to replace the derivatives with the differences

BlackHawk
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  • What have you tried so far? Also check out the other SO sites on maths and programming. – David Brossard Jun 02 '14 at 23:07
  • @DavidBrossard: I've found the two following proposals so far. 1. Setting a threshold, say 0.99, or 0.95 and keep m of n eigenvalues when T(m-1) < 0.99 *T(n) <= T(m) where T(m) = sum(i=1:m){lambda(i)} 2. The knee is located at a point where the radius of curvature. is a local minimum. For a curve y = f(x) the curvature is k = y''/(1+(y')^2)^(3/2). Just replace the derivatives with finite differences. What do you think of these two proposals? How can I implement the second one? I don't understand how to replace the derivatives with the differences. – BlackHawk Jun 03 '14 at 07:46
  • Hi, edit your question with your two proposals. Unfortunately, I cannot help you much but do check out http://math.stackexchange.com/ – David Brossard Jun 03 '14 at 15:19
  • I think you should be able to approximate the derivative numerically by just subtracting the values. `y'(x_i) = y(x_i) - y(x_(i-1))`. `y''(x_i) = y'(x_i) - y'(x_(i-1))`. `_` is subscript. – pbible Aug 28 '14 at 15:16

1 Answers1

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Did you read this paper?

Non-Graphical Solutions for Cattell’s Scree Test https://ppw.kuleuven.be/okp/_pdf/Raiche2013NGSFC.pdf

Pierre
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