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無法使用“ party”包在R中實現決策樹。 怎么做?

[英]Can't implement Decision tree in R using 'party' package. How to do it?

我正在嘗試使用“ party”包在R中構造決策樹,

我正在遵循http://www.rdatamining.com/examples/decision-tree上提到的方法

他們使用“ party”包顯示了決策樹。

我的數據集類似於示例中顯示的虹膜數據集。 這是指向我的數據集副本的鏈接。 https://drive.google.com/file/d/0B6cqWmwsEk20TXQyMnVlbGppcTQ/edit?usp=sharing

這是我嘗試過的代碼。 我使用read.csv命令加載了數據,並將其提供給dat3變量。

library(party)
> str(dat3)
'data.frame':   1000 obs. of  4 variables:
 $ Road_Type              : num  2 3 3 1 1 1 3 3 1 3 ...
 $ Light_Conditions       : num  2 3 3 3 3 3 3 3 3 3 ...
 $ Road_Surface_Conditions: num  1 2 2 2 2 2 2 2 2 2 ...
 $ Accident_Severity      : chr  "three" "three" "three" "three" ...
> dat3$Accident_Severity<-as.factor(dat3$Accident_Severity)
> str(dat3)
'data.frame':   1000 obs. of  4 variables:
 $ Road_Type              : num  2 3 3 1 1 1 3 3 1 3 ...
 $ Light_Conditions       : num  2 3 3 3 3 3 3 3 3 3 ...
 $ Road_Surface_Conditions: num  1 2 2 2 2 2 2 2 2 2 ...
 $ Accident_Severity      : Factor w/ 3 levels "one","three",..: 2 2 2 2 3 2 2 2 3 2 ...
> mytree<- ctree(Accident_Severity ~ Road_Type + Light_Conditions + Road_Surface_Conditions, data=dat3)
> print(mytree)

     Conditional inference tree with 1 terminal nodes

Response:  Accident_Severity 
Inputs:  Road_Type, Light_Conditions, Road_Surface_Conditions 
Number of observations:  1000 

1)*  weights = 1000 
> 

如您所見,構建的樹沒有節點,當我以圖形方式繪制該樹時,結果也不會像沒有構建樹那樣不理想。 我不確定我在做什么錯。

我認為數據中沒有足夠的信息來執行0.95的顯着性水平。 查看表格拆分:

> with( dat3, table(Accident_Severity, Light_Conditions, Road_Type))
, , Road_Type = 1

                 Light_Conditions
Accident_Severity   1   2   3
            one     0   2   4
            three   2 157 158
            two     0  14  35

, , Road_Type = 2

                 Light_Conditions
Accident_Severity   1   2   3
            one     0   0   0
            three   1  17  11
            two     0   0   0

, , Road_Type = 3

                 Light_Conditions
Accident_Severity   1   2   3
            one     0   2   2
            three   3 269 251
            two     0  38  34

因此,我認為沒有顯而易見的分歧。 該函數認為它已經被充分分割。 如果降低最低標准,則會產生分裂:

 mytree<- ctree(Accident_Severity ~ Road_Type + Light_Conditions + Road_Surface_Conditions, 
                  data=dat3, control=ctree_control(  mincriterion =0.50) )
 print(mytree)
#----------------------
     Conditional inference tree with 4 terminal nodes

Response:  Accident_Severity 
Inputs:  Road_Type, Light_Conditions, Road_Surface_Conditions 
Number of observations:  1000 

1) Light_Conditions <= 2; criterion = 0.653, statistic = 4.043
  2) Road_Surface_Conditions <= 1; criterion = 0.9, statistic = 6.742
    3)*  weights = 193 
  2) Road_Surface_Conditions > 1
    4)*  weights = 312 
1) Light_Conditions > 2
  5) Road_Type <= 1; criterion = 0.792, statistic = 5.187
    6)*  weights = 197 
  5) Road_Type > 1
    7)*  weights = 298 

plot(mytree)

在此處輸入圖片說明

如果在變量名稱周圍使用factor(),則它們將作為非普通變量進行處理:

 mytree2 <- ctree(Accident_Severity ~ factor(Road_Type) + factor(Light_Conditions) + factor(Road_Surface_Conditions), 
                   data=dat3, control=ctree_control(  mincriterion =0.50) )
  print(mytree2)
#------------------------
     Conditional inference tree with 2 terminal nodes

Response:  Accident_Severity 
Inputs:  factor(Road_Type), factor(Light_Conditions), factor(Road_Surface_Conditions) 
Number of observations:  1000 

1) factor(Road_Type) == {1, 3}; criterion = 0.635, statistic = 6.913
  2)*  weights = 971 
1) factor(Road_Type) == {2}
  3)*  weights = 29 

在此處輸入圖片說明

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