如何使用'ks'和'rgl'从R中内置的3D内核密度图中提取值 [英] How to extract values from a 3D kernel density plot built in R using 'ks' and 'rgl'

问题描述

我一直在使用 ks包和 rgl包来生成3D内核密度估计值和这些估计值的3D图。第一部分工作得很好（下面的简要示例）。我无法确定的是，是否有可能首先提取用于构建内核的给定xyz位置的内核值。换句话说，提取3D图中点的值，类似于光栅包中用于2D曲面的提取命令。有没有人有做这样的事情的经验，可以指引我正确的方向？非常感谢。 -DJ

I've been using the 'ks' package along with the 'rgl' package to produce 3D kernel density estimates and 3D plots of these. This first part has worked out fine (brief example below). What I can't figure out is if it's possible to extract the values of the kernels for the given xyz locations used to build the kernels in the first place. In other words, extract the values for points in a 3D plot, akin to the extract command used for 2D surfaces in the 'raster' package. Does anyone have experience doing something like this that can point me in the right direction? Thanks much. -DJ

library("rgl")
library("ks")

# call the plug-in bandwidth estimator
H.pi <- Hpi(b,binned=TRUE) ## b is a matrix of x,y,z points

# calculate the kernel densities
fhat2 <- kde(b, H=H.pi)

#plot the 50% and 95% kernels in gray and blue
plot(fhat2,cont=c(50,95),colors=c("gray","blue"),drawpoints=TRUE
    ,xlab="", ylab="", zlab="",size=2, ptcol="white", add=FALSE, box=TRUE, axes=TRUE) 




#Structure of fhat2. Original df consists of ~6000 points.  

List of 9
 $ x          : num [1:6173, 1:3] -497654 -497654 -497929 -498205 -498205 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : chr [1:6173] "50" "57" "70" "73" ...
  .. ..$ : chr [1:3] "x" "max_dep" "y"

$ eval.points:List of 3
  ..$ : num [1:51] -550880 -546806 -542733 -538659 -534586 ...
  ..$ : num [1:51] -7.9 -4.91 -1.93 1.06 4.05 ...
  ..$ : num [1:51] -376920 -374221 -371522 -368823 -366124 ...

$ estimate   : num [1:51, 1:51, 1:51] 0 0 0 0 0 ...

$ H          : num [1:3, 1:3] 3.93e+07 -2.97e+03 8.95e+06 -2.97e+03 2.63e+01 ...

$ gridtype   : chr [1:3] "linear" "linear" "linear"

$ gridded    : logi TRUE

$ binned     : logi FALSE

$ names      : chr [1:3] "x" "max_dep" "y"

$ w          : num [1:6173] 1 1 1 1 1 1 1 1 1 1 ...
 - attr(*, "class")= chr "kde"

推荐答案

尝试一下

## from ?plot.kde
library(ks)
library(MASS)
 data(iris)

 ## trivariate example
 fhat <- kde(x=iris[,1:3])

## this indicates the orientation
image(fhat$eval.points[[1]], fhat$eval.points[[2]], apply(fhat$estimate, 1:2, sum))
points(fhat$x[,1:2])

library(raster)

## convert to RasterBrick from raw array
## with correct orientation relative to that of ?base::image
b <- brick(fhat$estimate[,ncol(fhat$estimate):1,], 
    xmn = min(fhat$eval.points[[1]]), xmx = max(fhat$eval.points[[1]]), ymn = min(fhat$eval.points[[2]]), ymx = max(fhat$eval.points[[2]]), 
    transpose = TRUE)

## check orientation
plot(calc(b, sum))
points(fhat$x[,1:2])

现在我们很高兴，因为栅格功率很好。

Now we are happy because raster power is good.

plot(b)

## note this is a matrix with nrows = nrow(fhat$x), ncols = nlayers(b)
extract(b, fhat$x[,1:2])

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