多维网格的逆插值

问题描述

我正在做一个插入样本数据{(x_i,y_i)}的项目，其中x_i的输入域位于4D空间并输出y_i 位于 3D 空间.我需要为两个方向生成两个查找表.我设法生成了 4D ->3D 表.但是 3D ->4D 一个是棘手的.样本数据不在规则的网格点上，也不是一对一的映射.有没有已知的方法来治疗这种情况?我在网上做了一些搜索，但我发现的只是 3D ->3D 映射，不适合这种情况.谢谢！

I am working on a project of interpolating sample data {(x_i,y_i)} where the input domain for x_i locates in 4D space and output y_i locates in 3D space. I need generate two look up tables for both directions. I managed to generate the 4D -> 3D table. But the 3D -> 4D one is tricky. The sample data are not on regular grid points, and it is not one to one mapping. Is there any known method to treat this situation? I did some search online, but what I found is only for 3D -> 3D mapping, which are not suitable for this case. Thank you!

回答 Spektre 的问题:

To answer the questions of Spektre:

X(3D) ->Y(4D) 就是这种情况 1X ->nY

我想生成一个表，对于任何给定的 X，我们都可以找到 Y 的值.样本数据没有占据X的所有域.但是没关系，我们只需要样本数据域内的点的准确性.例如，我们有像 {(x1,x2,x3) ->(y1,y2,y3,y4)} 这样的样本数据.可能我们还有一个样本数据 {(x1,x2,x3) ->(y1_1,y2_1,y3_1,y4_1)}.但是没关系.我们需要一个空间X中任意(a,b,c)的表格，它对应一个(e,f,g,h) 在空格 Y 中.可能有不止一种选择，但我们只需要一种.(对不起，如果有符号混淆)

I want to generate a table that for any given X, we can find the value for Y. The sample data is not occupy all the domain of X. But it's fine, we only need accuracy for point inside the domain of sample data. For example, we have sample data like {(x1,x2,x3) ->(y1,y2,y3,y4)}. It is possible we also have a sample data {(x1,x2,x3) -> (y1_1,y2_1,y3_1,y4_1)}. But it is OK. We need a table for any (a,b,c) in space X, it corresponds to ONE (e,f,g,h) in space Y. There might be more than one choice, but we only need one. (Sorry for the symbol confusing if any)

一种可能的处理方法:由于我已经建立了从Y->X的平滑映射，我可以使用牛顿法或任何其他方法来反向搜索点y 对于任何给定的 x.但它不够准确，而且耗时.因为我需要对表中的每个点都做搜索，误差是模型误差与搜索误差之和.

One possible way to deal with this: Since I have already established a smooth mapping from Y->X, I can use Newton's method or any other method to reverse search the point y for any given x. But it is not accurate enough, and time consuming. Because I need do search for each point in the table, and the error is the sum of the model error with the search error.

所以我想知道可以直接找到一个映射来插入样本数据，而不是在 3 中进行这种搜索.

So I want to know it is possible to find a mapping directly to interpolate the sample data instead of doing such kind of search in 3.

推荐答案

1. 您正在寻找投影/映射

正如你提到的，你有投影 X(3D) ->Y(4D) 在你的情况下不是一对一的，那么它是 (1 X -> n Y) 还是 (n X -> 1 Y) 或 (n X -> m Y) ?

as you mentioned you have projection X(3D) -> Y(4D) which is not one to one in your case so what case it is (1 X -> n Y) or (n X -> 1 Y) or (n X -> m Y) ?

您要使用查找表

我假设您只想为给定的 Y 生成所有 X 非 (1 to 1) 映射的问题是您可以仅当有查找表时才使用查找表

I assume you just want to generate all X for given Y the problem with non (1 to 1) mappings is that you can use lookup table only if it has

• 所有有效点
• 或映射具有某种几何或数学对称性(例如X和Y空间中的点之间的距离相似，并且映射是连续的)

• all valid points
• or mapping has some geometric or mathematic symmetry (for example distance between points in X and Yspace is similar,and mapping is continuous)

您无法在通用映射点之间进行插值，所以问题是您想到的是哪种映射/投影?

You can not interpolate between generic mapped points so the question is what kind of mapping/projection you have in mind?

首先是 1->1 投影/映射插值

1. 如果您的 X->Y 投影映射适合插值

1. if your X->Y projection mapping is suitable for interpolation

然后对于 3D->4D 使用 tri-linear 插值.找到最近的 8 个点(每个点在其轴上形成网格超立方体)并在所有 4 维

then for 3D->4D use tri-linear interpolation. Find closest 8 points (each in its axis to form grid hypercube) and interpolate between them in all 4 dimensions

如果您的 X 投影映射适合插值

if your X<-Y projection mapping is suitable for interpolation

然后对于 4D->3D 使用 quatro-linear 插值.找到最近的 16 个点(每个点都在其轴上形成网格超立方体)并在所有 3 维度上在它们之间进行插值.

then for 4D->3D use quatro-linear interpolation. Find closest 16 points (each in its axis to form grid hypercube) and interpolate between them in all 3 dimensions.

现在 1->n 或 n->m 投影/映射

Now what about 1->n or n->m projections/mappings

这完全取决于我一无所知的投影/映射属性.尝试提供数据集示例，最好添加一些图像.

That solely depends on the projection/mapping properties which I know nothing of. Try to provide an example of your datasets and adding some image would be best.

[edit1] 1 X

我仍然会使用quatro-linear 插值.您仍然需要搜索您的 Y 表，但如果您将其分组为 4D 网格，那么它应该很容易.

I still would use quatro-linear interpolation. You still will need to search your Y table but if you group it like 4D grid then it should be easy enough.

1. 在 Y-table 中找到 16 个最接近输入 Y 点的点

1. find 16 closest points in Y-table to your input Y point

这些点应该是所有轴的每个 +/- 方向上最接近您的 Y 的点.在 3D 中，它看起来像这样:

These points should be the closest points to your Y in each +/- direction of all axises. In 3D it looks like this:

• 红点是你的输入 Y 点
• 蓝色点是找到的最近点(网格)，它们不需要像图像上那样对称.

请不要让我绘制有意义的 4D 示例:)(至少对于清醒的头脑)

Please do not want me to draw 4D example that make sense :) (at least for sober mind)

插值

找到对应的X点.如果每个点有多个然后选择一个与其他点更接近的点......现在你应该有 16 X 点和 16+1 Y 点.然后从 Y 点开始，您只需要计算从输入 Y 点沿线的距离.这些距离用作线性插值的参数.将它们规范化为 where

find corresponding X points. If there is more then one per point chose the closer one to the others ... Now you should have 16 X points and 16+1 Y points. Then from Y points you need just to calculate the distance along lines from your input Y point. These distances are used as parameter for linear interpolations. Normalize them to <0,1> where

• 0 表示左"点，1 表示右"点
• 0.5 表示正中

您将需要在每个 Y 域维度中的标量距离.现在只需计算沿线性插值的所有 X 点，直到在 X 域中得到相应的红点.

You will need this scalar distance in each of Y-domain dimension. Now just compute all the X points along the linear interpolations until you get the corresponding red point in X-domain.

使用三线性插值(3D)有4+2+1=7线性插值(如图像).对于quatro-linear 插值(4D)，有8+4+2+1=15 线性插值.

With tri-linear interpolation (3D) there are 4+2+1=7 linear interpolations (as on image). For quatro-linear interpolation (4D) there are 8+4+2+1=15 linear interpolations.

线性插值

X = X0 + (X1-X0)*t

• X 为插值点
• X0,X1 分别是左"、右"点
• t 是距离参数
• X is interpolated point
• X0,X1 are the 'left','right' points
• t is the distance parameter <0,1>

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