反距离加权插值

问题描述

我想计算重量的距离的东西互惠如反距离加权插值。

I would like to compute a weight as reciprocal of a distance for something like inverse distance weighting interpolation.

double wgt = 0, wgt_tmp, result = 0;
for (int i = 0; i < num; i++) {
   wgt_tmp = 1.0/dist[i];
   wgt += wgt_tmp;
   result += wgt_tmp * values[i];
}
results /= wgt;

不过的距离可以 0 ，我需要做的重量适合计算。如果只有一个距离测距[I] 是 0 ，我想它的对应值值[I] 是主导。如果有几个距离是 0 ，我想有自己的价值观为贡献同样的结果。此外，即使测距[I] 不为零，但非常小，我想有一个合理的标准来检查和处理。任何想法如何实现它？

However the distance can be 0 and I need to make the weight suitable for computation. If there is only one distance dist[i] is 0, I would like its corresponding value values[i] to be dominant. If there are several distances are 0, I would like to have their values to contribute equally to the result. Also even if dist[i] is not zero but very small, I would like to have a reasonable criterion to check it and deal with it. Any idea how to implement it?

推荐答案

我不明白，除了分段任何方式 - 你需要一个不同的功能比小距离倒数距离。最简单的事情是只砍掉顶部：

I don't see any way besides piecewise - you need a different function than reciprocal distance for small distances. The simplest thing would be to just chop off the top:

modified_dist[i] = dist[i] < MIN_DIST ? MIN_DIST : dist[i]

但你可以替换的东西仍然在下降，如果你想，像（MIN_DIST + DIST [I]）/ 2 。

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