0
$\begingroup$

I was running into a situation in which my input feature experience a very large variation in term of magnitude.

Particularly, consider feature 1 belong to group 1 and feature 2 3 4 belong to group 2,

Like this picture below

enter image description here

I was really worried that in this case feature 1 might dominate feature 2,3,4 (group 2) because its corresponding value is so large (I was trying to train this data set on a neural network).

In this situation, what would be the appropriate scaling strategy ? Update: I know for sure that the value of feature 1 is an integer that is uniform on the interval [22,42] But for feature 2 ,3 ,4 I do not have any insight

Thank you for your enthusiast !

$\endgroup$
0
$\begingroup$

You should check the distribution of each feature and scale them accordingly, but in any case you should aim to roughly the same interval of values for every feature. For example, if f1 has the standard distribution and f2 is close to the uniform one, then you can scale f1 to N(0,1) and f2 to U(-1,1). In other words, try to have maximum, minimum and mean values for all features as close as possible, while keeping their original distributions.

| improve this answer | |
$\endgroup$
0
$\begingroup$

You should normalize every column individually. It will work just fine. Sum up the column and divide every element of that column by sum of that column. But as your feature 2,3,4 are of very small scale you should consider some transformation like log transformation as you might encounter numerical underflow.

| improve this answer | |
$\endgroup$
  • $\begingroup$ What is log transformation for feature 2 3 4 ? Could you explain a little bit clearer ? Thank you very much $\endgroup$ – Tuong Nguyen Minh Jun 19 at 13:36
  • $\begingroup$ Instead of x, use log(1+x) or something like it. It will prevent numerical underflow $\endgroup$ – SrJ Jun 19 at 13:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.