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From: Oleg Komarov on 16 Mar 2010 06:49
"Satish " <dhungana(a)kth.se> wrote in message <hnnbh1$cun$1(a)fred.mathworks.com>...
> dpb <none(a)non.net> wrote in message <hnm43d$7u9$1(a)news.eternal-september.org>...
>
> > Of course you can; it's simply a linear transformation -- you have two
> > points (min/max values) you want to scale to another range. That boils
> > down to solving two equations in two unknowns...
> >
> > --
>
> Hi dpb,
>
> Thanks , Finally i made it with linear transformation using following formula.
>
> value=((y(i) + abs(min(y)))/(max(y)-min(y)))*100
>
> If you have also any other formula than let me know.

you can rearrange your formula to:
value = (y + abs(min(y))) / range(y) * 100

Note that you don't need to loop here.

Oleg
From: ImageAnalyst on 16 Mar 2010 08:19
Why are you taking the absolute value of the min? Why not simply use
the regular, usual normalization equation:
normalizedY = (Y - min(Y)) / range(Y) * 100
Don't bother taking the abs().
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