I have a multi-class classification problem with circular and linear data. For example a feature vector contains 20 circular ($\theta \in [0,2\pi)$) and two linear ($\in (0,1)$) variables. I plan to transform $\theta$ into $(\sin\theta,\cos\theta)$ and apply any traditional classifier (MLP, SVM). In fact my data appear to be linearly separable, so I try Perceptron also. I don't find any specific classifier that handles $\theta$ directly. So, is my planning okay? $\sin\theta$ and $\cos\theta$ seem to be highly correlated. Does that create a problem for a classifier?

I appreciate any help and any reference on this issue.


1 Answer 1


You could try clustering the observations of each $\theta_j$ feature. If there is a preferred direction, $\theta_j^{(1)}$, then you might consider replacing $\theta_j$ with $\theta_j-\theta_j^{(1)}$.

If you have enough data not to overfit, then you could replace $\theta_j$ with the deltas to each of the $m$ most preferred directions, $\theta_j-\theta_j^{(1)}$ to $\theta_j-\theta_j^{(m)}$.

But you say the data appear to be linearly separable. If you can explain that better (to us, or by representing the data appropriately in feature space) then you'll have gone a long way to answering your original question.

  • $\begingroup$ Thanks but I don't really get you answer. My feature vector is $\Xi = (\theta_1,\ldots,\theta_{20}, x_1,x_2)$. I have $n$ such feature vectors corresponding to 10 classes. Is it problematic if I convert $\theta_j$ to $(\sin\theta_j,\cos\theta_j)$ and apply MLP or SVM? Can you suggest some article, because I don't find anything. I don't understand "preferred direction" also. Is that the average of $n$ values of $\theta_j$ component ? Sorry, linear separability is an assumption so far. $\endgroup$
    – Ananda2111
    Nov 23, 2015 at 21:49
  • $\begingroup$ I'm sure it's not problematic at all to convert $\theta_j$ to $(\sin\theta_j,\cos\theta_j)$. Try it and see; you may as well try the preferred direction thing as well. Suppose, for example, that $\theta_1$ consists mostly of angles near $\theta_1^{(1)}=55$ degrees and near $\theta_1^{(2)}=355$ degrees; then an angle of $5$ degrees, say, could be represented by feature values $-50$ (=5-55) and $+10$ (=5-355 mod 360). If plotting the data reveals no such clustering, forget the idea. $\endgroup$
    – Creosote
    Nov 24, 2015 at 21:08

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