我尝试做的事情，应该是没有什么比二维，线性插值但目前我不找到正确的方法。为了说明这个问题有点简单：没有与3000x3000像素的大小，我有画如绘图区一个水平线。要做到这一点我画点或短线从每个像素位置到下一个像素的位置，然后形成一条直线。

I try to do a thing that should be nothing more than a two-dimensional, linear interpolation but currently I fail finding the correct approach. To describe the problem a bit simplified: there is a drawing area with a size of 3000x3000 pixels where I have to draw e.g. a horizontal line. To do that I'm drawing dots or short lines from every pixel position to the next pixel position which then forms a line.

现在的校正已被应用到整个事情，其中​​校正信息可在发现此（例如简化的）4乘4阵列，其中的每个单元包含一对校正后描述的值的坐标。因此，中性阵列（没有校正）是这样的：

Now a correction has to be applied to the whole thing where correction information can be found in a (for this example simplified) 4 by 4 array, where every element contains a pair of coordinates describing the values after correction. So a neutral array (with no correction) would look like this:

0,0      1000,0      2000,0      3000,0
0,1000   1000,1000   2000,1000   3000,1000
0,2000   1000,2000   2000,2000   3000,2000
0,3000   1000,3000   2000,3000   3000,3000

描述修正

一个真正的校正表将包含其他坐标做：

A real correction table would contain other coordinates describing the correction to be done:

因此​​，作为输入数据我点的坐标上，而不校正线时，字段值而不进行校正和校正数据。但是，我怎么能计算出点线现在应用修正值，以它这样一个扭曲的线条绘制像显示在右侧如果图片？我目前的做法有两个独立的线性插值X和Y不工作，有Y位置跳到上一个单元格边框，但并不顺利小区内的改变。

So as input data I have the coordinates of points on the line without correction, the fields values without correction and the correction data. But how can I calculate the lines points now applying the correction values to it so that a distorted line is drawn like shown in right side if the image? My current approach with two separate linear interpolations for X and Y does not work, there the Y-position jumps on a cells border but does not change smoothly within a cell.

所以...任何想法怎么可以这样做？

So...any ideas how this could be done?

推荐答案

您必须同意对插值方法第一。我建议无论是双线性或重心插值。在我的previous帖子之一我可视化这两种方法之间的差异。

You have to agree on an interpolation method first. I would suggest either bilinear or barycentric interpolation. In one of my previous posts I visualized the difference between both methods.

我会专注于双线性插值。我们希望将任何一点的小区内的修正点。因此，所有的点可以单独转化

I'll concentrate on the bilinear interpolation. We want to transform any point within a cell to its corrected point. Therefore, all points could be transformed separately.

我们需要插补参数 U 和 v 的点（X，Y ）。因为我们有一个轴对齐网格，这是pretty的简单：

We need the interpolation parameters u and v for the point (x, y). Because we have an axis-aligned grid, this is pretty simple:

u = (x - leftCellEdge) / (rightCellEdge - leftCellEdge)
v = (y - bottomCellEdge) / (topCellEdge - bottomCellEdge)

我们可以重建点通过双线性插值：

We could reconstruct the point by bilinear interpolation:

p2       p4
   x----x
   |  o |
   x----x
p1       p3

o = (1 - u) * ((1 - v) * p1 + v * p2) + u * ((1 - v) * p3 + v * p4)

现在，相同的公式可以用于校正的点。如果您使用原始分 P1 到 P4 ，你会得到裸线点。如果通过 P4 使用校正细胞分 P1 ，你会得到纠正线点。

Now, the same formula can be used for the corrected points. If you use the original points p1 through p4, you'll get the uncorrected line point. If you use the corrected cell points for p1 through p4, you'll get the corrected line point.