Subject:
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Re: On a scale of 0 to 10?
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 26 Mar 2003 17:41:19 GMT
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Viewed:
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2312 times
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In lugnet.off-topic.geek, David Eaton writes:
> I've got a function that represents a quality rating, based on 2 parameters,
> which are on a scale of 0 to 1:
>
> (d-k)*k => quality (q)
>
> Notice that this yields a sort of wacky result, insofar as the quality can
> vary between -1 and 0.25, not -1 and 1 or -1 and 0 or 0 and 1. Icky. Anyway,
> what I'd *like* to do is translate the resulting quality rating (q) onto a
> scale (s) of 0 to 10. In other words, for q = -1, you get a 0, and for a q
> of .25, you get a 10.
>
> Problem. Assuming a linear distribution for 0.25 => 10, and -1 => 0, we get:
>
> s = 8q + 8
>
> Well, that's all well and good, except for when q = 0, we get s = 8, and I'd
> rather that a quality of 0 translated to 5 rather than 8, since an 8 implies
> that such a quality is really good, and in reality, a quality of 0
> represents 'middle of the road'.
>
> So. I figured why not assume a quadratic solution? IE: s = Aq^2 + Bq + C,
> with known points (-1,0), (0,5), and (0.25,10). However. This seems to yield:
>
> s = 12q^2 + 17q + 5
>
> So, what's wrong with that? Well, for some values of q (like q = -0.72), s
> results in a *negative* value (in this case -1.0192). ACK! That means that
> *some* things that have better quality ratings actually have *LOWER* 0-to-10
> rankings!
>
> I assume I want a logarithmic scale function instead... How do I set that up
> & solve for it? Or do I want something totally different?
>
> DaveE
How about a double-linear solution? from -1 to 0, translate linearly from 0 to
5. From 0 to .25, translate linearly from 5 to 10. You won't have a single
equation, but that seems to accomplish what you want, right?
Adrian
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Message has 1 Reply:
 | | Re: On a scale of 0 to 10?
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| (...) Thought about that... If necessary I guess, but I'm pretty sure there's a way to do it. I think what I need is: s = Ae^(q + B) + C And at the moment, I've solved: C = -Ae^(B - 1) A = 10 / (e^(B + .25) - e^(B - 1)) And I'm scratching my head (...) (24 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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Message is in Reply To:
| | On a scale of 0 to 10?
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| I've got a function that represents a quality rating, based on 2 parameters, which are on a scale of 0 to 1: (d-k)*k => quality (q) Notice that this yields a sort of wacky result, insofar as the quality can vary between -1 and 0.25, not -1 and 1 or (...) (24 years ago, 26-Mar-03, to lugnet.off-topic.geek)
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