从 rpart.object 取回原始名称

Question

我已经保存了使用 R 中的 rpart 包创建的模型。我正在尝试从这些保存的模型中检索一些信息；具体来自 rpart.object。虽然文档 - rpart doc - 很有帮助，但仍有一些事情不清楚：

如何找出哪些变量是分类变量，哪些变量是数字变量？目前，我所做的是参考 splits 矩阵中的 'index' 列。我注意到仅对于数字变量，条目不是整数。有更简洁的方法吗？
csplit 矩阵指的是分类变量可以使用整数取的各种值，即 R 将原始名称映射为整数。有没有办法访问此映射？对于前。如果我的原始变量，比如说，Country 可以取任何值 France, Germany, Japan 等，csplit 矩阵让我知道某个拆分是基于 Country == 1, 2。在这里，rpart 分别用 1, 2 替换了对 France, Germany 的引用。如何从模型文件中获取原始名称 - France, Germany, Japan - ？另外，我怎么知道名称和整数之间的映射是什么？

Answer 1

通常，terms 组件会包含此类信息。 See ?rpart::rpart.object.

fit <- rpart::rpart(Kyphosis ~ Age + Number + Start, data = kyphosis)
fit$terms  # notice that the attribute dataClasses has the information
attr(fit$terms, "dataClasses")
#------------
 Kyphosis       Age    Number     Start 
 "factor" "numeric" "numeric" "numeric"

该示例的结构中没有 csplit 节点，因为 none 的 hte 变量是因子。你可以很容易地制作一个：

> fit <- rpart::rpart(Kyphosis ~ Age + factor(findInterval(Number,c(0,4,6,Inf))) + Start, data = kyphosis)
> fit$csplit
     [,1] [,2] [,3]
[1,]    1    1    3
[2,]    1    1    3
[3,]    3    1    3
[4,]    1    3    3
[5,]    3    1    3
[6,]    3    3    1
[7,]    3    1    3
[8,]    1    1    3
> attr(fit$terms, "dataClasses")
                                     Kyphosis 
                                     "factor" 
                                          Age 
                                    "numeric" 
factor(findInterval(Number, c(0, 4, 6, Inf))) 
                                     "factor" 
                                        Start 
                                    "numeric"

整数只是因子变量的值，因此 "mapping" 与从 as.numeric() 到 levels() 的因子相同。如果我试图构建 fit$csplit-矩阵的字符矩阵版本，用因子变量中的水平名称代替，这将是成功的途径之一：

> kyphosis$Numlev <- factor(findInterval(kyphosis$Number, c(0, 4, 6, Inf)), labels=c("low","med","high"))
> str(kyphosis)
'data.frame':   81 obs. of  5 variables:
 $ Kyphosis: Factor w/ 2 levels "absent","present": 1 1 2 1 1 1 1 1 1 2 ...
 $ Age     : int  71 158 128 2 1 1 61 37 113 59 ...
 $ Number  : int  3 3 4 5 4 2 2 3 2 6 ...
 $ Start   : int  5 14 5 1 15 16 17 16 16 12 ...
 $ Numlev  : Factor w/ 3 levels "low","med","high": 1 1 2 2 2 1 1 1 1 3 ...
> fit <- rpart::rpart(Kyphosis ~ Age +Numlev + Start, data = kyphosis)
> Levels <- fit$csplit
> Levels[] <- levels(kyphosis$Numlev)[Levels]
> Levels
     [,1]   [,2]   [,3]  
[1,] "low"  "low"  "high"
[2,] "low"  "low"  "high"
[3,] "high" "low"  "high"
[4,] "low"  "high" "high"
[5,] "high" "low"  "high"
[6,] "high" "high" "low" 
[7,] "high" "low"  "high"
[8,] "low"  "low"  "high"

评论回复：如果只有模型就用str()看一下。我在我创建的示例中看到一个 "ordered" 叶子，它的因子标签存储在名为 "xlevels":

的属性中

$ ordered            : Named logi [1:3] FALSE FALSE FALSE
  ..- attr(*, "names")= chr [1:3] "Age" "Numlev" "Start"
 - attr(*, "xlevels")=List of 1
  ..$ Numlev: chr [1:3] "low" "med" "high"
 - attr(*, "ylevels")= chr [1:2] "absent" "present"
 - attr(*, "class")= chr "rpart"

从 rpart.object 取回原始名称

Getting back original names from rpart.object

r

rpart