I\'m kind of stuck here, I guess it\'s a bit of a brain teaser. If I have numbers in the range between 0.5 to 1 how can I normalize it to be between 0 to 1?
Thanks for a
Others have provided you the formula, but not the work. Here's how you approach a problem like this. You might find this far more valuable than just knowning the answer.
To map [0.5, 1]
to [0, 1]
we will seek a linear map of the form x -> ax + b
. We will require that endpoints are mapped to endpoints and that order is preserved.
Method one: The requirement that endpoints are mapped to endpoints and that order is preserved implies that 0.5
is mapped to 0
and 1
is mapped to 1
a * (0.5) + b = 0 (1)
a * 1 + b = 1 (2)
This is a simultaneous system of linear equations and can be solved by multiplying equation (1)
by -2
and adding equation (1)
to equation (2)
. Upon doing this we obtain b = -1
and substituting this back into equation (2)
we obtain that a = 2
. Thus the map x -> 2x - 1
will do the trick.
Method two: The slope of a line passing through two points (x1, y1)
and (x2, y2)
is
(y2 - y1) / (x2 - x1).
Here we will use the points (0.5, 0)
and (1, 1)
to meet the requirement that endpoints are mapped to endpoints and that the map is order-preserving. Therefore the slope is
m = (1 - 0) / (1 - 0.5) = 1 / 0.5 = 2.
We have that (1, 1)
is a point on the line and therefore by the point-slope form of an equation of a line we have that
y - 1 = 2 * (x - 1) = 2x - 2
so that
y = 2x - 1.
Once again we see that x -> 2x - 1
is a map that will do the trick.
× 2 − 1
should do the trick
You could always use clamp or saturate within your math to make sure your final value is between 0-1. Some saturate at the end, but I've seen it done during a computation, too.
To add another generic answer.
If you want to map the linear range [A..B] to [C..D], you can apply the following steps:
Shift the range so the lower bound is 0. (subract A from both bounds:
[A..B] -> [0..B-A]
Scale the range so it is [0..1]. (divide by the upper bound):
[0..B-A] -> [0..1]
Scale the range so it has the length of the new range which is D-C. (multiply with D-C):
[0..1] -> [0..D-C]
Shift the range so the lower bound is C. (add C to the bounds):
[0..D-C] -> [C..D]
Combining this to a single formula, we get:
(D-C)*(X-A)
X' = ----------- + C
(B-A)
In your case, A=0.5, B=1, C=0, D=1 you get:
(X-0.5)
X' = ------- = 2X-1
(0.5)
Note, if you have to convert a lot of X to X', you can change the formula to:
(D-C) C*B - A*D
X' = ----- * X + ---------
(B-A) (B-A)
It is also interesting to take a look at non linear ranges. You can take the same steps, but you need an extra step to transform the linear range to a nonlinear range.
Lazyweb answer: To convert a value x
from [minimum..maximum]
to [floor..ceil]
:
General case:
normalized_x = ((ceil - floor) * (x - minimum))/(maximum - minimum) + floor
To normalize to [0..255]:
normalized_x = (255 * (x - minimum))/(maximum - minimum)
To normalize to [0..1]:
normalized_x = (x - minimum)/(maximum - minimum)
Subtract 0.5 (giving you a new range of 0 - 0.5) then multiply by 2.
double normalize( double x )
{
// I'll leave range validation up to you
return (x - 0.5) * 2;
}