sde*_*188 5 tree r cart rpart r-caret
数据:我正在使用rsample包中的" attrition "数据集.
问题:使用磨损数据集和rpart库,我可以使用公式"Attrition~OverTime + JobRole"生成树,其中选择OverTime作为第一个分割.但是当我尝试在没有JobRole变量的情况下生长树(即"Attrition~OverTime")时,树不会分裂并仅返回根节点.使用rpart函数以及使用method ="rpart"的插入符号列函数都会发生这种情况.
我对此感到困惑,因为我认为在rpart中实现的CART算法选择了最佳变量以迭代贪婪的方式进行分割,并且没有"向前看"以查看其他变量的存在如何影响其对最佳选择的选择分裂.如果在具有两个解释变量的情况下算法选择OverTime作为值得的第一次拆分,为什么在删除JobRole变量后它不选择OverTime作为值得的第一次拆分？
我在Windows 7中使用R版本3.4.2和RStudio版本1.1.442.
研究:我在这里和这里找到了类似的Stack Overflow问题,但都没有完整的答案.
我可以说,rpart文档似乎在第5页说rpart算法不使用"向前看"规则:
解决这两个问题的一种方法是使用预见规则; 但这些在计算上非常昂贵.相反,rpart使用节点的几种杂质或多样性度量之一.
代码:这是一个代表.任何见解都会很棒 - 谢谢!
suppressPackageStartupMessages(library(rsample))
#> Warning: package 'rsample' was built under R version 3.4.4
suppressPackageStartupMessages(library(rpart))
suppressPackageStartupMessages(library(caret))
suppressPackageStartupMessages(library(dplyr))
#> Warning: package 'dplyr' was built under R version 3.4.3
suppressPackageStartupMessages(library(purrr))
#################################################
# look at data
data(attrition)
attrition_subset <- attrition %>% select(Attrition, OverTime, JobRole)
attrition_subset %>% glimpse()
#> Observations: 1,470
#> Variables: 3
#> $ Attrition <fctr> Yes, No, Yes, No, No, No, No, No, No, No, No, No, N...
#> $ OverTime <fctr> Yes, No, Yes, Yes, No, No, Yes, No, No, No, No, Yes...
#> $ JobRole <fctr> Sales_Executive, Research_Scientist, Laboratory_Tec...
map_dfr(.x = attrition_subset, .f = ~ sum(is.na(.x)))
#> # A tibble: 1 x 3
#> Attrition OverTime JobRole
#> <int> <int> <int>
#> 1 0 0 0
#################################################
# with rpart
attrition_rpart_w_JobRole <- rpart(Attrition ~ OverTime + JobRole, data = attrition_subset, method = "class", cp = .01)
attrition_rpart_w_JobRole
#> n= 1470
#>
#> node), split, n, loss, yval, (yprob)
#> * denotes terminal node
#>
#> 1) root 1470 237 No (0.83877551 0.16122449)
#> 2) OverTime=No 1054 110 No (0.89563567 0.10436433) *
#> 3) OverTime=Yes 416 127 No (0.69471154 0.30528846)
#> 6) JobRole=Healthcare_Representative,Manager,Manufacturing_Director,Research_Director 126 11 No (0.91269841 0.08730159) *
#> 7) JobRole=Human_Resources,Laboratory_Technician,Research_Scientist,Sales_Executive,Sales_Representative 290 116 No (0.60000000 0.40000000)
#> 14) JobRole=Human_Resources,Research_Scientist,Sales_Executive 204 69 No (0.66176471 0.33823529) *
#> 15) JobRole=Laboratory_Technician,Sales_Representative 86 39 Yes (0.45348837 0.54651163) *
attrition_rpart_wo_JobRole <- rpart(Attrition ~ OverTime, data = attrition_subset, method = "class", cp = .01)
attrition_rpart_wo_JobRole
#> n= 1470
#>
#> node), split, n, loss, yval, (yprob)
#> * denotes terminal node
#>
#> 1) root 1470 237 No (0.8387755 0.1612245) *
#################################################
# with caret
attrition_caret_w_JobRole_non_dummies <- train(x = attrition_subset[ , -1], y = attrition_subset[ , 1], method = "rpart", tuneGrid = expand.grid(cp = .01))
attrition_caret_w_JobRole_non_dummies$finalModel
#> n= 1470
#>
#> node), split, n, loss, yval, (yprob)
#> * denotes terminal node
#>
#> 1) root 1470 237 No (0.83877551 0.16122449)
#> 2) OverTime=No 1054 110 No (0.89563567 0.10436433) *
#> 3) OverTime=Yes 416 127 No (0.69471154 0.30528846)
#> 6) JobRole=Healthcare_Representative,Manager,Manufacturing_Director,Research_Director 126 11 No (0.91269841 0.08730159) *
#> 7) JobRole=Human_Resources,Laboratory_Technician,Research_Scientist,Sales_Executive,Sales_Representative 290 116 No (0.60000000 0.40000000)
#> 14) JobRole=Human_Resources,Research_Scientist,Sales_Executive 204 69 No (0.66176471 0.33823529) *
#> 15) JobRole=Laboratory_Technician,Sales_Representative 86 39 Yes (0.45348837 0.54651163) *
attrition_caret_w_JobRole <- train(Attrition ~ OverTime + JobRole, data = attrition_subset, method = "rpart", tuneGrid = expand.grid(cp = .01))
attrition_caret_w_JobRole$finalModel
#> n= 1470
#>
#> node), split, n, loss, yval, (yprob)
#> * denotes terminal node
#>
#> 1) root 1470 237 No (0.8387755 0.1612245)
#> 2) OverTimeYes< 0.5 1054 110 No (0.8956357 0.1043643) *
#> 3) OverTimeYes>=0.5 416 127 No (0.6947115 0.3052885)
#> 6) JobRoleSales_Representative< 0.5 392 111 No (0.7168367 0.2831633) *
#> 7) JobRoleSales_Representative>=0.5 24 8 Yes (0.3333333 0.6666667) *
attrition_caret_wo_JobRole <- train(Attrition ~ OverTime, data = attrition_subset, method = "rpart", tuneGrid = expand.grid(cp = .01))
attrition_caret_wo_JobRole$finalModel
#> n= 1470
#>
#> node), split, n, loss, yval, (yprob)
#> * denotes terminal node
#>
#> 1) root 1470 237 No (0.8387755 0.1612245) *
这是完全有道理的。上面有很多额外的代码，所以我将重复重要的部分。
library(rsample)
library(rpart)
data(attrition)
rpart(Attrition ~ OverTime + JobRole, data=attrition)
n= 1470
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1470 237 No (0.83877551 0.16122449)
2) OverTime=No 1054 110 No (0.89563567 0.10436433) *
3) OverTime=Yes 416 127 No (0.69471154 0.30528846)
6) JobRole=Healthcare_Representative,Manager,Manufacturing_Director,Research_Director 126 11 No (0.91269841 0.08730159) *
7) JobRole=Human_Resources,Laboratory_Technician,Research_Scientist,Sales_Executive,Sales_Representative 290 116 No (0.60000000 0.40000000)
14) JobRole=Human_Resources,Research_Scientist,Sales_Executive 204 69 No (0.66176471 0.33823529) *
15) JobRole=Laboratory_Technician,Sales_Representative 86 39 Yes (0.45348837 0.54651163) *
rpart(Attrition ~ OverTime, data=attrition)
n= 1470
node), split, n, loss, yval, (yprob)
* denotes terminal node
1) root 1470 237 No (0.8387755 0.1612245) *
看一下第一个模型（有两个变量）。就在根下面我们有：
1) root 1470 237 No (0.83877551 0.16122449)
2) OverTime=No 1054 110 No (0.89563567 0.10436433) *
3) OverTime=Yes 416 127 No (0.69471154 0.30528846)
该模型继续拆分节点 3 (OverTime=Yes)，但仅使用 JobRole。由于我们在第二个模型中没有 JobRole，因此 rpart 无法进行其他拆分。但请注意，在节点 2 和 3 处，Attrition=No 是多数类。在节点 3 处，69.5% 的实例为“否”，30.5% 为“是”。因此，对于节点 2 和 3，我们将预测“否”。由于分割两侧的预测相同，因此分割是不必要的并被剪掉。你只需要根节点就可以预测所有实例都是No。
| 归档时间： |
|
| 查看次数： |
277 次 |
| 最近记录： |