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I am trying to get something like a confusion matrix for different classes, but without training a model.

The idea is to use some kind of distance between classes.

The data set is like this, just for better understanding

Class1

A    B    C ...  YZ
500  3    2 ...  43
500  15   1 ...  2
300  23   9 ...  68

Class 2

A    B    C ...  YZ
100  56   2 ...  58
100  35   6 ...  34
300  23   9 ...  1

...
Class N

The classes have different row number.

I have only one idea so far - Calculate means for each column separately (one resulting row) and measure distances between these rows

What is the right way to calculate the similarity of classes? Maybe a python library for that?

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  • $\begingroup$ As you are anew contributor here is a hint: if an answer actually answers the question, you should mark it as "accepted" so that it is removed from the list of unanswered questions. $\endgroup$ – cdalitz Nov 27 '20 at 14:18
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Your suggestion of summing up the (squared) distances between the class means actually is an established method and is the trace of the between scatter matrix:

$$S_B = \sum_{i=1}^{C} n_i (\vec{\mu}_i - \vec{\mu})\cdot (\vec{\mu}_i - \vec{\mu})^T$$

It is however more useful to set this value in relation to the spread inside the classes, or the trace of the within scatter matrix:

$$S_W = \sum\limits_{i=1}^{C} \left(\sum\limits_{\vec{x}\in\omega_i}(\vec{x} - \vec{\mu}_i)\cdot (\vec{x} - \vec{\mu}_i)^T\right)$$

Measures for class separability are thus

$$J_1 = \frac{trace(S_B)}{trace(S_W)} \quad\mbox{ or }\quad J_2 = trace\Big(S_W^{-1} \cdot S_B\Big)$$

where trace stands for the sum of the diagonal elements, which represent the variances.

The normalization with $trace(S_W)$ makes the measure scale invariant and measures how large the distance between the classes is compared to the distance within the classes.

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  • $\begingroup$ Thank you very much for your answer! I really appreciate your attention to my question. Could you please explain me the difference between the approaches, why one is better than another? $\endgroup$ – Statsnewbie Nov 27 '20 at 14:11
  • $\begingroup$ Which approaches do you mean? $J_1$ and $J_2$? These do not make a great difference, I think. Or do you mean the normalization with $trace(S_W)$ compared to your approach that only measures $trace(S_B)$? I have added an explanation for this normalization. $\endgroup$ – cdalitz Nov 27 '20 at 14:15
  • $\begingroup$ Thank you for help, your answer makes me happy 😊 $\endgroup$ – Statsnewbie Nov 27 '20 at 14:56

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