# How to normalize a range of values taking the difference into consideration

Given a set of numbers

e.g. {1, 2, 3, 4, 5} or { 50, 100} or {50000, 50001}

I want to normalize these into a range with a min and max e.g. 2 >= x <= 50

My current algorithm is $$((range_{max} - range_{min}) / (x_{max} - x_{min})) * (x - x_{min}) + range_{min}$$

This does result is numbers within the range however a set of numbers like {50000,500001} will result in 50000 = 2 and 50000 = 50 which is too skewed. In this case I would like a result still in the range but with 2 numbers closer together e.g. 2 & 3 or 30 & 31 .

What formula could I use to do this? I'm guessing I need to use $\log(x_{max} / x_{min})$ somewhere but I'm not sure how to work it into the equation.

-
That depends a lot on the domain of what your $x$ can be and also on what you want to do later with the result. –  Tunococ Aug 23 '12 at 1:33
You could try pretending that zero is also in your data set, i.e. replace $x_\min$ with $0$. –  Rahul Aug 23 '12 at 1:36
x could be any value. I'm generating a scatter chart from the data and the normalized value will be the radius of the point. –  Chris Herring Aug 23 '12 at 1:44
Using 0 in the data set might work better, will have a think about it. It would however produce a bias towards larger numbers in the range which I would need to tweak as a bias towards smaller numbers is better for my purposes. –  Chris Herring Aug 23 '12 at 1:47
Then pretend that some very big number is in your given set. –  Gerry Myerson Aug 23 '12 at 4:01

If what you're trying to do is pick radii of points for visualization, I would just use $$\max\left(2, 50\sqrt{x/x_\max}\right).$$ This accomplishes two things: the area of the point becomes proportional to the value, and extremely small values get clamped so their points are not too small to be visible.
That is quite nice. Only issue is that it can be biased towards a larger radii for smaller data sets however I could use $\ 2*(x/x_{min})$ when $\ 2*(x_{max}/x_{min}) <= 50$ as well or @Gerry Myerson suggestion above. Thanks. –  Chris Herring Aug 23 '12 at 5:21
However don't think this will work when $\ x < 0$ –  Chris Herring Aug 23 '12 at 5:47
1. If the radii are too big then you should replace $50$ with a smaller value. 2. Yes, this won't work when the data values are negative. –  Rahul Aug 23 '12 at 17:55
I think what you should think about first is whether you want the results to represent the relative magnitudes of the data, or just their relative differences. That is, do you consider $\{1,2,3\}$ to be the same as $\{5,10,15\}$ and $\{500,1000,1500\}$? If so, then you're going to have to do something special about negative values. On the other hand, if you want to treat $\{-5,0,5\}$ the same as $\{5,10,15\}$, then it'll also be the same as $\{4995,5000,5005\}$, and you can't complain about the latter getting mapped to too wide a range. –  Rahul Aug 23 '12 at 18:40