2D coordinate normalization

Question

I need to implement a function which normalizes coordinates. I define normalize as (please suggest a better term if Im wrong):

Mapping entries of a data set from their natural range to values between 0 and 1.

Now this was easy in one dimension:

    static List<float> Normalize(float[] nums)
    {
        float max = Max(nums);
        float min = Min(nums);
        float delta = max - min;

        List<float> li = new List<float>();
        foreach (float i in nums)
        {
            li.Add((i - min) / delta);
        }
        return li;
    }

I need a 2D version as well and that one has to keep the aspect ratio intact. But Im having some troubles figuring out the math.

Although the code posted is in C# the answers need not to be.

Thanks in advance. :)

Answer 1

If I understand correctly, you want to map points to the square [(0,0), (1,0), (0,1), (1,1)]. Let max_xy be the maximum of both the x and the y values, and similarly for min_xy. Use those values to calculate delta. Compute two offsets, one for x and one for y.

Since you're dividing by the same delta, the aspect ratio is preserved.

offset = min == 0f ? 0 : -min is unnecessarily long by the way. What is -0?

Answer 2

Simple idea: Find out which dimension is bigger and normalize in this dimension. The second dimension can be computed by using the ratio. This way the ratio is kept and your values are between 0 and 1.

Answer 3

It seems you want each vector (1D, 2D or ND) to have length <= 1.
If that's the only requirement, you can just divide each vector by the length of the longest one.

double max = maximum (|vector| for each vector in 'data');
foreach (Vector v : data) {
    li.add(v / max);
}

That will make the longest vector in result list to have length 1.

But this won't be equivalent of your current code for 1-dimensional case, as you can't find minimum or maximum in a set of points on the plane. Thus, no delta.