r - 使用 rpart : How to get more variability on predictions?

Question

我正在使用 rpart 包，如下所示:

model <- rpart(totalUSD ~ ., data = df.train)

我注意到超过 80k 行，rpart 将其预测 概括为三个不同的组 ，如下图所示:



我看到几个 configuration options for the rpart method ；但是，我不太了解它们。

有没有办法配置 rpart 以便它创建 个更多的预测(而不是三个) ；不是那么明显的群体，而是介于两者之间的更多层次？

我问的原因是因为我的成本估算器看起来相当简单，因为它只返回三个数字之一!

这是我的数据示例:

structure(list(totalUSD = c(9726.6, 730.14, 750, 200, 60.49, 
310.81, 151.23, 145.5, 3588.13, 400), durationDays = c(730, 724, 
730, 189, 364, 364, 364, 176, 730, 1095), familySize = c(4, 1, 
2, 1, 3, 2, 1, 1, 4, 4), serviceName = c("Service5", 
"Service6", "Service9", "Service4", 
"Service1", "Service2", "Service1", "Service3", 
"Service7", "Service8"), homeLocationGeoLat = c(37.09024, 
10.691803, 37.09024, 35.86166, 55.378051, 35.86166, 51.165691, 
-30.559482, -30.559482, 41.87194), homeLocationGeoLng = c(-95.712891, 
-61.222503, -95.712891, 104.195397, -3.435973, 104.195397, 10.451526, 
22.937506, 22.937506, 12.56738), hostLocationGeoLat = c(55.378051, 
37.09024, 55.378051, 55.378051, 37.09024, 1.352083, 55.378051, 
37.09024, 23.424076, 1.352083), hostLocationGeoLng = c(-3.435973, 
-95.712891, -3.435973, -3.435973, -95.712891, 103.819836, -3.435973, 
-95.712891, 53.847818, 103.819836), geoDistance = c(6838055.10555534, 
4532586.82063172, 6838055.10555534, 7788275.0443749, 6838055.10555534, 
3841784.48282769, 1034141.95021832, 14414898.8246973, 6856033.00945242, 
10022083.1525388)), .Names = c("totalUSD", "durationDays", "familySize", 
"serviceName", "homeLocationGeoLat", "homeLocationGeoLng", "hostLocationGeoLat", 
"hostLocationGeoLng", "geoDistance"), row.names = c(25601L, 6083L, 
24220L, 20235L, 8372L, 456L, 8733L, 27257L, 15928L, 24099L), class = "data.frame")

Answer 1

如果你真的想要一个复杂的树结构，试试这个:

library(rpart)
fit = rpart(totalUSD ~ ., data = df.train, control = rpart.control(cp = 0))

基本上每个拆分都会在 cp = 0 时尝试，无论拆分的改进如何。你得到了一个非常复杂的模型，但你有 80k 观察，所以设置 minsplit或 minbucket到您满意的数字。

我在我正在使用的随机森林实现中使用了这个策略。当心计算时间可能会增加很多。