Scaling A Pair Of Numbers

[Maverick22]

Situation:

I have a pair real floating point numbers from (-∞, +∞) which I compare using relational operators ( <, >, <=,>=,==)

For the sake of being visually pleasing, I need to find a way to represent these number scaled down to [-100,+100]

I cant do modulus division because if the numbers are say (1001, 1) the (mod 100) of each number would be 1. Which does me no good relationally

Regular Division doesn't work either...

Can someone help me figure this out? Thanks!

[Ganchrow]

Situation:

I have a pair real floating point numbers from (-ꝏ,ꝏ) which I compare using relational operators ( <, >, <=,>=,==)

For the sake of being visually pleasing, I need to find a way to represent these number scaled down to [-100,100]

I cant do modulus division because if the numbers are say (1001,1) the (mod 100) of each number would be 1. Which doest me no good.

Regular Division doesn't work either...

Can someone help me figure this out? Thanks!

I can't read the characters in your specified domain but I'll just assume they're meant to be (-∞ , +∞).

That said, there's no linear bijective mapping (i.e., both 1-1 and onto) between the set of reals and any a given subset with finite endpoints.

But if you really needed the bijectivity and you were OK with a non-linear mapping you could try the inverse of the logit() function translated about the origin and scalar multiplied to yield the desired range of (-100, +100). Specifically:

F(x) = 200 * invlg(x) - 100
F(x) = 200 1 + e^-x - 100

Where e^X is the base of the natural logarithm raised to the X power (=exp() in Excel.)

This would map:

±∞ → ±100
±35 → ~±99.999999999985789145284798
±2.944438979 → ~±90
±1.945910149 → ~±75
±1.098612289 → ~±50
±0.510825624 → ~±25
0 → 0

Which could clearly wreck havoc on widely ranging floating points values.

If you wanted to scale it a bit differently so that, say, ±x₀ mapped to ±y₀ you'd need to include a scale factor (call it σ) such that:

σ = -x₀ / ln( 200 y₀+100 - 1)
F(x;σ) = 200 1 + e^-x/σ - 100

So if you wanted to guarantee a mapping of 100,000 to ~90 (say), you would use:

σ ≈ 33,962.3272

Yielding:

±∞ → ±100
±100,000 → ~±90
±66,088 → ~±75
±37,311 → ~±50
±17,349 → ~±25
0 → 0

If you needed a linear transformation and you could specify limits on your domain endpoints (call them a & b) then the mapping of ±x₀ ∈ [a, b] → ±y₀ ∈ [-100, 100] is just be the line segment with endpoints (-100, a) and (+100, b), which is:

F(x) = 100*(a + b - 2x) a - b

[unusialsusp5]

what an idiotic question about nothing

[Maverick22]

I can't read the characters in your specified domain but I'll just assume they're meant to be (-∞ , +∞).

That said, there's no linear bijective mapping (ie., 1-1 and onto) between the set of reals and any a given finite subset.

But if you really needed the bijectivity and you were OK with a non-linear mapping you could try the inverse of the logit() function translated about the origin and scalar multiplied to yield the desired range of (-100, +100). Specifically:

F(x) = 200 * invlg(x) - 100
F(x) = 200 1 + e^-x - 100

Where e^X is the base of the natural logarithm raised to the X power (=exp() in Excel.)

This would map:

±∞ → ±100
±35 → ~±99.999999999985789145284798
±2.944438979 → ~±90
±1.945910149 → ~±75
±1.098612289 → ~±50
±0.510825624 → ~±25
0 → 0

Which could clearly wreck havoc on widely ranging floating points values.

If you wanted to scale it a bit differently so that, say, ±x₀ mapped to ±y₀ you'd need to include a scale factor (call it σ) such that:

σ = -x₀ / ln( 200 y₀+100 - 1)

F(x;σ) = 200 1 + e^-x/σ - 100

So if you wanted to guarantee a mapping of 100,000 to ~90 (say), you would use:

σ ≈ 33,962.3272

Yielding:

±∞ → ±100
±100,000 → ~±90
±66,088 → ~±75
±37,311 → ~±50
±17,349 → ~±25
0 → 0

If you needed a linear transformation and you could specify limits on your domain endpoints (call them a & b) then the mapping of ±x₀ ∈ [a, b] → ±y₀ ∈ [-100, 100] is just be the line segment with endpoints (-100, a) and (+100, b), which is:

F(x) = 100*(a + b - 2x) a - b

Phew... what a read. I will try to code this up today, and let you know how it goes. I had passing thoughts yesterday of using log/invlog, or e^x/ln. Let's me know that i'm not a complete dunce

I'll paste the function(s) when i finish, will be later tonight.

Thanks Ganch!

what an idiotic question about nothing

And you... if you need your first post of the day, don't do it in my threads...
and if you don't understand what I am asking... you should just keep your mouth shut.

[Wrecktangle]

Mav, you may want to try logarithmic regression as a transform. It can map your negative to positive Infinity numbers to -1 / +1 which you can then multiply by 100.

In another direction: is it my imagination or are we suffering more idiots lately in the Tank?

[sharpcat]

Mav, you may want to try logarithmic regression as a transform. It can map your negative to positive Infinity numbers to -1 / +1 which you can then multiply by 100.

In another direction: is it my imagination or are we suffering more idiots lately in the Tank?

I think the misspelling of "unusual" in the username "unusialsusp5" is a good indicator of one who may be out of his habitat.

[unusialsusp5]

maybe the misspelling was intentional. congratulations on your mathematical expertise. how much money did you waste going to school. i don't believe your astute math knowledge could help you handicap sports out of a paper bag.

[durito]

maybe the misspelling was intentional. congratulations on your mathematical expertise. how much money did you waste going to school. i don't believe your astute math knowledge could help you handicap sports out of a paper bag.

You'd be very wrong.

[djiddish98]

Ha, handicapping without math

[unusialsusp5]

you probably believe we went to the moon in 1969 too.

[Maverick22]

Mav, you may want to try logarithmic regression as a transform. It can map your negative to positive Infinity numbers to -1 / +1 which you can then multiply by 100.

In another direction: is it my imagination or are we suffering more idiots lately in the Tank?

My regression skills are nil... Gonna need someone to handhold me on that

But i will see if i can google my way into it.

Thanks!

[Flying Dutchman]

My regression skills are nill... Gonna need someone to handhold me on that

But i will see if i can google my way into it.

Thanks!

Maverick, you can do it in Excel. Give me a call or send me the data in e-mail and we'll sort it.

[Indecent]

If you needed a linear transformation and you could specify limits on your domain endpoints (call them a & b) then the mapping of ±x₀ ∈ [a, b] → ±y₀ ∈ [-100, 100] is just be the line segment with endpoints (-100, a) and (+100, b), which is:

F(x) = 100*(a + b - 2x) a - b

Wow, I'm embarrassed to share how convoluted my solution to this was.

It's always nice to have a one liner replace crappy code, thanks for the reminder to consider the obvious (and easiest!) solution.