获得最靠近网格点的点

Question

I have a one dimensional gird. its spacing is a floating point. I have a point with floating point coordinate as well. I need to find its distance to the closest grid point.
For example:

            0.12
             |
             *
 |---------|---------|---------|---------|---------|
 0        0.1       0.2       0.3       0.4       0.5

The result would be -0.02 since the closest point is behind it.
However if it was

                -0.66
                  |
                  *
 |---------|---------|---------|---------|---------|
-1       -0.8      -0.6      -0.4      -0.2        0

The result will be 0.06 . As you can see its in floating point and can be negative.
I tried the following:

float spacing = ...;
float point = ...;

while(point >= spacing) point -= spacing;
while(point < 0) point += spacing;

if(std::abs(point - spacing) < point) point -= spacing;

It works, but I'm sure there is a way without loops

Answer 1

Let us first compute the nearest points on the left and right as follows:

leftBorder = spacing * floor(point/spacing);
rightBorder = leftBorder + spacing;

Then the distance is straightforward:

if ((point - leftBorder) < (rightBorder - point))
    distance = leftBorder - point;
else
    distance = rightBorder - point;

Note that, we could find the nearest points alternatively by ceiling:

rightBorder = spacing * ceil(point/spacing);
leftBorder = rightBorder - spacing;

Answer 2

std::vector<float> spacing = ...;
float point = ...;
float result;

Since you say the spacing isn't (linear), I would cache the sums:

std::vector<float> sums(1, 0.0);
float sum=0;
for(int i=0; i<spacing.size(); ++i)
    sums.push_back(sum+=spacing[i]);
//This only needs doing once.
//sums needs to be in increasing order.  

Then do a binary search to find the point to the left:

std::vector<float>::iterator iter;
iter = std::lower_bound(sums.begin(), sums.end(), point);

Then find the result from there:

if (iter+1 == sums.end())
    return point-*iter;
else {
    float midpoint = (*iter + *(iter+1))/2;
    if (point < midpoint)
        result = point - *iter;
    else
        result = *(iter+1) - point;
}

[EDIT] Don't I feel silly. You said the spacing wasn't constant. I interpreted that as not-linear. But then your sample code is linear, just not a compile-time constant. My bad. I'll leave this answer as a more general solution, though your (linear) question is solvable much faster.

Answer 3

Here is my first blush attempt, note that this is not tested at all.

float remainder = fmod(point, spacing); // This is the fractional difference of the spaces
int num_spaces = point/spacing;  // This is the number of "spaces" down you are, rounded down


// If our fractional part is greater than half of the space length, increase the number of spaces.
// Not sure what you want to do when the point is equidistant to both grid points
if(remainder > .5 * spacing) 
{
  ++num_spaces;
}

float closest_value = num_spaces*spacing;
float distance = closest_value - point;

Answer 4

You should just round the number using this:

float spacing = ...;
float point = ...;
(point > 0.0) ? floor(point + spacing/2) : ceil(point - spacing/2);

Answer 5

更常见的是，对于任意间距，尺寸和距离度量（度量），您正在寻找的结构将是Voronoi图。