绘制R中的高维内核密度

Question

我对R中的核密度估计有一个问题。我有一个5维数据，它包括（x，y，z）位置，发生的时间和某些事件的大小（例如地震）（I＆＃） 39;附上数据集）。我在R中编写了以下代码，以便找到5D内核密度估计：

library(ks) 
library(rgl)
kern <- read.table(file.choose(), sep=",")
evpts <- do.call(expand.grid,lapply(kern,quantile, prob=c(.1,.15,.2,.25,.3,.35,.4,.45,.5,.55,.6,.65,.7,.75,.8,.85,.9,.95)))
hat <- kde(kern, eval.points= evpts)
str(hat)

现在，我想要可视化核密度估计。我更喜欢在一个图中显示关于所有5个维度的内核（通过对点使用不同的颜色或大小）或者至少分别关于三个维度。你对我有什么建议吗？

以下是数据：

       x          y          z            time         size
422.697323  164.19886   2.457419    8.083796636  0.83367586
423.008236  163.32434   0.5551326   37.58477455  0.893893903
204.733908  218.36365   1.9397874   37.88324312  0.912809449
203.963056  218.4808    0.3723791   43.21775903  0.926406005
100.727581  46.60876    1.4022341   49.41510519  0.782807523
453.335182  244.25521   1.6292517   51.73779175  0.903910803
134.909462  210.96333   2.2389119   53.13433521  0.896529401
135.300562  212.02055   0.6739541   67.55073745  0.748783521
258.237117  134.29735   2.1205291   76.34032587  0.735699304
341.305271  149.26953   3.718958    94.33975483  0.849509216
307.138925  59.60571    0.6311074   106.9636715  0.987923188
307.76875   58.91453    2.6496741   113.8515307  0.802115718
415.025535  217.17398   1.7155688   115.7464603  0.875580325
414.977687  216.73327   1.7107369   115.9776948  0.767143582
311.006135  173.24378   2.7819572   120.8079566  0.925380118
310.116929  174.28122   4.3318722   129.2648401  0.776528535
347.260911  37.34946    3.5155427   136.7851291  0.851787115
351.317624  33.65703    0.5806926   138.7349284  0.909723017
4.471892    59.42068    1.4062959   139.0543783  0.967270976
5.480223    59.72857    2.7326106   139.2114277  0.987787428
199.513023  21.53302    2.5163259   143.5895625  0.864164659
198.718031  23.50163    0.4801849   147.2280466  0.741587333
26.650517   35.2019     0.8246514   150.4876506  0.744788202
25.089379   90.47825    0.8700944   152.1944046  0.777252476
26.307439   88.41552    2.4422487   155.9090026  0.952215177
234.282901  236.11422   1.8115261   155.9658144  0.776284654
235.052948  236.77437   1.9644963   156.6900297  0.944285448
23.048202   98.6261     3.4573048   159.7700912  0.773057491
21.516695   98.05431    2.5029284   160.8202997  0.978779087
213.936324  151.87013   3.1042192   161.0612489  0.80499513
277.887935  197.25753   1.3659279   163.673142   0.758978575
277.239746  197.54001   2.2109361   166.2629868  0.775325157

Answer 1

根据时间四分位数将“大小”分为4个颜色类，根据“时间”四分位数将4个线类型分为4个颜色类。它不使用内核密度信息：

require(MASS)
png();parcoord(dat, col=cut(dat$size, quantile(dat$size, c(0,.25,.5,.75,1)), include.lowest=TRUE), lty= as.numeric(cut(dat$time, quantile(dat$time, c(0,.25,.5,.75,1)), include.lowest=TRUE))); dev.off()