Scale numbers to be <= 255?

Question

I have cells for whom the numeric value can be anything between 0 and Integer.MAX_VALUE. I would like to color code these cells correspondingly.

If the value = 0, then r = 0. If the value is Integer.MAX_VALUE, then r = 255. But what about the values in between?

I'm thinking I need a function whose limit as x => Integer.MAX_VALUE is 255. What is this function? Or is there a better way to do this?

I could just do (value / (Integer.MAX_VALUE / 255)) but that will cause many low values to be zero. So perhaps I should do it with a log function.

Most of my values will be in the range [0, 10,000]. So I want to highlight the differences there.

Answer 1

The "fairest" linear scaling is actually done like this:

floor(256 * value / (Integer.MAX_VALUE + 1))

Note that this is just pseudocode and assumes floating-point calculations.

If we assume that Integer.MAX_VALUE + 1 is 2^31, and that / will give us integer division, then it simplifies to

value / 8388608

Why other answers are wrong

Some answers (as well as the question itself) suggsted a variation of (255 * value / Integer.MAX_VALUE). Presumably this has to be converted to an integer, either using round() or floor().

If using floor(), the only value that produces 255 is Integer.MAX_VALUE itself. This distribution is uneven.

If using round(), 0 and 255 will each get hit half as many times as 1-254. Also uneven.

Using the scaling method I mention above, no such problem occurs.

Non-linear methods

If you want to use logs, try this:

255 * log(value + 1) / log(Integer.MAX_VALUE + 1)

You could also just take the square root of the value (this wouldn't go all the way to 255, but you could scale it up if you wanted to).

Answer 2

I figured a log fit would be good for this, but looking at the results, I'm not so sure.

However, Wolfram|Alpha is great for experimenting with this sort of thing:

I started with that, and ended up with:

r(x) = floor(((11.5553 * log(14.4266 * (x + 1.0))) - 30.8419) / 0.9687)

Interestingly, it turns out that this gives nearly identical results to Artelius's answer of:

r(x) = floor(255 * log(x + 1) / log(2^31 + 1)

IMHO, you'd be best served with a split function for 0-10000 and 10000-2^31.