0

我想在 3d 核密度估计中绘制特定 %-contour 的等值面。然后，我想知道哪些点在那个 3d 形状内。

我将展示我接近 2d 的情况来说明我的问题（从R 模仿的代码 - 如何在特定的 Contour 中找到点和如何绘制等高线，以显示 95% 的值落在 R 和 ggplot2 中的位置）。

library(MASS)
library(misc3d)
library(rgl)
library(sp)

# Create dataset
set.seed(42)
Sigma <- matrix(c(15, 8, 5, 8, 15, .2, 5, .2, 15), 3, 3)
mv <- data.frame(mvrnorm(400, c(100, 100, 100),Sigma))

### 2d ###
# Create kernel density
dens2d <- kde2d(mv[, 1], mv[, 2], n = 40)
# Find the contour level defined in prob
dx <- diff(dens2d$x[1:2])
dy <- diff(dens2d$y[1:2])
sd <- sort(dens2d$z)
c1 <- cumsum(sd) * dx * dy 
prob <- .5
levels <- sapply(prob, function(x) { 
  approx(c1, sd, xout = 1 - x)$y
})

# Find which values are inside the defined polygon
ls <- contourLines(dens2d, level = levels)
pinp <- point.in.polygon(mv[, 1], mv[, 2], ls[[1]]$x, ls[[1]]$y)

# Plot it
plot(mv[, 1], mv[, 2], pch = 21, bg = "gray")
contour(dens2d, levels = levels, labels = prob,
        add = T, col = "red")
points(mv[pinp == 1, 1], mv[pinp == 1, 2], pch = 21, bg = "orange")

二维情况 因此，50% 的轮廓是使用 approx 定义的，轮廓是使用 contourLines 创建的，然后 point.in.polygon 找到该轮廓内的点。

我想做同样的事情，但在 3d 情况下。这就是我所管理的：

### 3d ###
# Create kernel density
dens3d <- kde3d(mv[,1], mv[,2], mv[,3], n = 40)

# Find the contour level defined in prob
dx <- diff(dens3d$x[1:2])
dy <- diff(dens3d$y[1:2])
dz <- diff(dens3d$z[1:2])
sd3d <- sort(dens3d$d)
c3d <- cumsum(sd3d) * dx * dy * dz
levels <- sapply(prob, function(x) {
  approx(c3d, sd3d, xout = 1 - x)$y
})

# Find which values are inside the defined polygon
# # No idea

# Plot it
points3d(mv[,1], mv[,2], mv[,3], size = 2)
box3d(col = "gray")
contour3d(dens3d$d, level = levels, x = dens3d$x, y = dens3d$y, z = dens3d$z, #exp(-12)
          alpha = .3, color = "red", color2 = "gray", add = TRUE)
title3d(xlab = "x", ylab = "y", zlab = "z")

3D情况

所以，我还没有走多远。

我意识到我在 3d 案例中定义关卡的方式是不正确的，我猜问题出在其中，c3d <- cumsum(sd3d) * dx * dy * dz但老实说我不知道​​如何进行。

而且，一旦正确定义了 3d 轮廓，我将不胜感激有关如何处理该轮廓内的哪些点的任何提示。

非常感谢！

编辑：根据 user2554330 的建议，我将编辑我的问题以添加测试代码，将他或她的建议与我在此处发布的建议进行比较。（我确实意识到使用轮廓作为新数据点的推断的目的不在最初的问题中，我为此修改道歉。）

另外，我在下面的评论中有点草率。这两种方法在 2D 情况下的表现如何取决于样本有多大。在样本 n = 48 左右时，来自 user2554330 的方法捕获了大约 69% 的人口（而我发布的方法捕获了大约 79%），但是在样本 n = 400 左右时，user2554330 的方法捕获了大约 79%（对 83% ）。

# Load libraries
library(MASS)
library(misc3d)
library(rgl)
library(sp)
library(oce)
library(akima)

# Create dataset
set.seed(42)
tn <- 1000 # number in pop
Sigma <- matrix(c(15, 8, 5, 8, 15, .2, 5, .2, 15), 3, 3)
mv <- data.frame(mvrnorm(tn, c(100, 100, 100),Sigma)) # population

prob <- .8 # rather than .5
simn <- 100 # number of simulations
pinp <- rep(NA, simn)
cuts <- pinp
sn <- 48 # sample size, at n = 400 user2554330 performs better

### 2d scenario
for (isim in 1:simn) {

  # Sample
  smv <- mv[sample(1:tn, sn), ]

  # Create kernel density
  dens2d <- kde2d(smv[, 1], smv[, 2], n = 40,
                  lims = c(min(smv[, 1]) - abs(max(smv[, 1]) - min(smv[, 1])) / 2,
                           max(smv[, 1]) + abs(max(smv[, 1]) - min(smv[, 1])) / 2,
                           min(smv[, 2]) - abs(max(smv[, 2]) - min(smv[, 2])) / 2,
                           max(smv[, 2]) + abs(max(smv[, 2]) - min(smv[, 2])) / 2))



  # Approach based on https://stackoverflow.com/questions/30517160/r-how-to-find-points-within-specific-contour
  # Find the contour level defined in prob
  dx <- diff(dens2d$x[1:2])
  dy <- diff(dens2d$y[1:2])
  sd <- sort(dens2d$z)
  c1 <- cumsum(sd) * dx * dy 
  levels <- sapply(prob, function(x) { 
    approx(c1, sd, xout = 1 - x)$y
  })
  # Find which values are inside the defined polygon
  ls <- contourLines(dens2d, level = levels)
  # Note below that I check points from "population"
  pinp[isim] <- sum(point.in.polygon(mv[, 1], mv[, 2], ls[[1]]$x, ls[[1]]$y)) / tn



  # Approach based on user2554330
  # Find the estimated density at each observed point
  sdatadensity<- bilinear(dens2d$x, dens2d$y, dens2d$z, 
                    smv[,1], smv[,2])$z
  # Find the contours
  levels2 <- quantile(sdatadensity, probs = 1- prob, na.rm = TRUE)
  # Find within
  # Note below that I check points from "population"
  datadensity <- bilinear(dens2d$x, dens2d$y, dens2d$z, 
                    mv[,1], mv[,2])$z
  cuts[isim] <- sum(as.numeric(cut(datadensity, c(0, levels2, Inf))) == 2, na.rm = T) / tn

}

summary(pinp)
summary(cuts)

> summary(pinp)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0030  0.7800  0.8205  0.7950  0.8565  0.9140 
> summary(cuts)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.5350  0.6560  0.6940  0.6914  0.7365  0.8120

我还尝试使用以下代码查看 user2554330 的方法在 3D 情况下的表现：

# 3d scenario
for (isim in 1:simn) {

  # Sample
  smv <- mv[sample(1:tn, sn), ]

  # Create kernel density
  dens3d <- kde3d(smv[,1], smv[,2], smv[,3], n = 40,
                  lims = c(min(smv[, 1]) - abs(max(smv[, 1]) - min(smv[, 1])) / 2,
                           max(smv[, 1]) + abs(max(smv[, 1]) - min(smv[, 1])) / 2,
                           min(smv[, 2]) - abs(max(smv[, 2]) - min(smv[, 2])) / 2,
                           max(smv[, 2]) + abs(max(smv[, 2]) - min(smv[, 2])) / 2,
                           min(smv[, 3]) - abs(max(smv[, 3]) - min(smv[, 3])) / 2,
                           max(smv[, 3]) + abs(max(smv[, 3]) - min(smv[, 3])) / 2))

  # Approach based on user2554330
  # Find the estimated density at each observed point
  sdatadensity <- approx3d(dens3d$x, dens3d$y, dens3d$z, dens3d$d, 
                          smv[,1], smv[,2], smv[,3])
  # Find the contours
  levels <- quantile(sdatadensity, probs = 1 - prob, na.rm = TRUE)
  # Find within
  # Note below that I check points from "population"
  datadensity <- approx3d(dens3d$x, dens3d$y, dens3d$z, dens3d$d, 
                          mv[,1], mv[,2], mv[,3])
  cuts[isim] <- sum(as.numeric(cut(datadensity, c(0, levels, Inf))) == 2, na.rm = T) / tn

}

summary(cuts)

> summary(cuts)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.1220  0.1935  0.2285  0.2304  0.2620  0.3410

我更愿意定义轮廓，使得指定的概率是（接近）捕获从同一总体中提取的未来数据点的概率，即使样本 n 相对较小（即 < 50）也是如此。

4

1 回答 1

1

我不会尝试找出轮廓内的点，而是尝试评估每个点的密度，并根据该值与轮廓水平的比较为点着色。对于边界附近的几个点，它可能会做出不同的决定，但应该非常接近。

要进行该评估，您可以oce::approx3d在密度估计上使用该函数。

我要做的另一件事是根据观察到的密度的分位数选择轮廓，而不是尝试模拟估计密度的 3-d 积分。

这是执行所有这些操作的代码：

library(MASS)
library(misc3d)
library(rgl)
library(oce)
#> Loading required package: testthat
#> Loading required package: gsw

# Create dataset
set.seed(42)
Sigma <- matrix(c(15, 8, 5, 8, 15, .2, 5, .2, 15), 3, 3)
mv <- data.frame(mvrnorm(400, c(100, 100, 100),Sigma))

### 3d ###
# Create kernel density
dens3d <- kde3d(mv[,1], mv[,2], mv[,3], n = 40)

# Find the estimated density at each observed point
datadensity <- approx3d(dens3d$x, dens3d$y, dens3d$z, dens3d$d, 
                        mv[,1], mv[,2], mv[,3])

# Find the contours
prob <- .5
levels <- quantile(datadensity, probs = prob, na.rm = TRUE)

# Plot it
colours <- c("gray", "orange")
cuts <- cut(datadensity, c(0, levels, Inf))
for (i in seq_along(levels(cuts))) {
  gp <- as.numeric(cuts) == i
  spheres3d(mv[gp,1], mv[gp,2], mv[gp,3], col = colours[i], radius = 0.2)
}
box3d(col = "gray")
contour3d(dens3d$d, level = levels, x = dens3d$x, y = dens3d$y, z = dens3d$z, #exp(-12)
          alpha = .1, color = "red", color2 = "gray", add = TRUE)
title3d(xlab = "x", ylab = "y", zlab = "z")

这是制作的情节：

截屏

于 2020-02-01T14:38:39.913 回答