用“kde”函数估算R中的5-D核密度

2019年8月4日 327次阅读

我想通过在R的“ks”库中使用“kde”函数对5维数据(x,y,z,时间,大小)进行核密度估计.在它的手册中它说它可以做核密度估计1至6维数据(手册第24页：
http://cran.r-project.org/web/packages/ks/ks.pdf).

我的问题是它说超过3个维度我需要指定eval.points.我不知道如何指定评估点,因为没有超过3个维度的示例.例如,如果我想在问题空间中生成常规3D序列数据并将其用作评估点,我该怎么办？
 这是我的数据：

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

这是我正在使用的代码：

library(ks) 
library(rgl)
kern <- read.table(file.choose(), sep=",")
hat <- kde(kern)

它适用于最多3个维度,但对于4维和5维,它表示：需要为超过3个维度指定eval.points.

另外,我想知道如何绘制这些内核？例如,使用z作为条件变量并在3D散点图中绘制x,y,时间,并对不同的大小范围使用不同的颜色

最佳答案 像你一样,我最初找不到一个有用的例子,文档并没有真正描述预期的对象类型.对于你的5d数据集,我尝试设置一个5d网格的点,这些点是从每个维度的10,25,50,70和90百分位数构建的.我的数据集名为“dat”：

evpts <- do.call(expand.grid,  lapply(dat, quantile, prob=c(0.1,.25,.5,.75,.9)) )

然后我将它传递给kde函数,似乎满足了算法.这是否“正确”确实需要检查.没有保证.

> hat <- kde(dat, eval.points= evpts)
> str(hat)
List of 8
 $x          : num [1:31, 1:5] 423 423 205 204 101 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$: NULL
  .. ..$: chr [1:5] "V1" "V2" "V3" "V4" ...
 $eval.points:'data.frame':    3125 obs. of  5 variables:
  ..$V1: Named num [1:3125] 23 118 234 326 415 ...
  .. ..- attr(*, "names")= chr [1:3125] "10%" "25%" "50%" "75%" ...
  ..$V2: Named num [1:3125] 35.2 35.2 35.2 35.2 35.2 ...
  .. ..- attr(*, "names")= chr [1:3125] "10%" "10%" "10%" "10%" ...
  ..$V3: Named num [1:3125] 0.581 0.581 0.581 0.581 0.581 ...
  .. ..- attr(*, "names")= chr [1:3125] "10%" "10%" "10%" "10%" ...
  ..$V4: Named num [1:3125] 43.2 43.2 43.2 43.2 43.2 ...
  .. ..- attr(*, "names")= chr [1:3125] "10%" "10%" "10%" "10%" ...
  ..$V5: Named num [1:3125] 0.749 0.749 0.749 0.749 0.749 ...
  .. ..- attr(*, "names")= chr [1:3125] "10%" "10%" "10%" "10%" ...
  ..- attr(*, "out.attrs")=List of 2
  .. ..$dim     : Named int [1:5] 5 5 5 5 5
  .. .. ..- attr(*, "names")= chr [1:5] "V1" "V2" "V3" "V4" ...
  .. ..$dimnames:List of 5
  .. .. ..$V1: chr [1:5] "V1= 23.0482" "V1=117.8185" "V1=234.2829" "V1=326.1557" ...
  .. .. ..$V2: chr [1:5] "V2= 35.20190" "V2= 59.51319" "V2=149.26953" "V2=211.49194" ...
  .. .. ..$V3: chr [1:5] "V3=0.5806926" "V3=1.1180112" "V3=1.9397874" "V3=2.5830000" ...
  .. .. ..$V4: chr [1:5] "V4= 43.21776" "V4= 71.94553" "V4=129.26484" "V4=151.34103" ...
  .. .. ..$V5: chr [1:5] "V5=0.7487835" "V5=0.7764066" "V5=0.8517871" "V5=0.9190948" ...
 $estimate   : Named num [1:3125] 3.23e-08 5.70e-08 1.01e-08 4.07e-10 6.20e-12 ...
  ..- attr(*, "names")= chr [1:3125] "1" "2" "3" "4" ...
 $H          : num [1:5, 1:5] 5073.879 1010.815 1.211 -651.089 -0.223 ...
 $gridded    : logi FALSE
 $binned     : logi FALSE
 $names      : chr [1:5] "V1" "V2" "V3" "V4" ...
 $w          : num [1:31] 1 1 1 1 1 1 1 1 1 1 ...
 - attr(*, "class")= chr "kde"

我确实找到了一个早期版本的软件包文档,它提供了这个4d执行的工作示例,我认为我的努力基本相同,模数不同：

data(iris)
   ir <- iris[,1:4][iris[,5]=="setosa",]
   H.scv <- Hscv(ir)
   fhat <- kde(ir, H.scv, eval.points=ir)
点赞