我使用 RPart 来构建决策树。没有问题,我正在这样做。但是,我需要了解(或计算)树被分割了多少次?我的意思是,树有多少条规则(if-else 语句)? 例如:
X
- -
if (a<9)- - if(a>=9)
Y H
-
if(b>2)-
Z
有3条规则。
当我写总结(模型)时:
摘要(model_dt)
Call:
rpart(formula = Alert ~ ., data = train)
n= 18576811
CP nsplit rel error xerror xstd
1 0.9597394 0 1.00000000 1.00000000 0.0012360956
2 0.0100000 1 0.04026061 0.05290522 0.0002890205
Variable importance
ip.src frame.protocols tcp.flags.ack tcp.flags.reset frame.len
20 17 17 17 16
ip.ttl
` 12
Node number 1: 18576811 observations, complexity param=0.9597394
predicted class=yes expected loss=0.034032 P(node) =1
class counts: 632206 1.79446e+07
probabilities: 0.034 0.966
left son=2 (627091 obs) right son=3 (17949720 obs)
Primary splits:
ip.src splits as LLLLLLLRRRLLRR ............ LLRLRLRRRRRRRRRRRRRRRR
improve=1170831.0, (0 missing)
ip.dts splits as LLLLLLLLLLLLLLLLLLLRLLLLLLLLLLL, improve=1013082.0, (0 missing)
tcp.flags.ctl < 1.5 to the right, improve=1007953.0, (2645 missing)
tcp.flags.syn < 1.5 to the right, improve=1007953.0, (2645 missing)
frame.len < 68 to the right, improve= 972871.3, (30 missing)
Surrogate splits:
frame.protocols splits as LLLLLLLLLLLLLLLLLLLRLLLLLLLLLLL, agree=0.995, adj=0.841, (0 split)
tcp.flags.ack < 1.5 to the right, agree=0.994, adj=0.836, (0 split)
tcp.flags.reset < 1.5 to the right, agree=0.994, adj=0.836, (0 split)
frame.len < 68 to the right, agree=0.994, adj=0.809, (0 split)
ip.ttl < 230.5 to the right, agree=0.987, adj=0.612, (0 split)
Node number 2: 627091 observations
predicted class=no expected loss=0.01621615 P(node) =0.03375666
class counts: 616922 10169
probabilities: 0.984 0.016
Node number 3: 17949720 observations
predicted class=yes expected loss=0.0008514896 P(node) =0.9662433
class counts: 15284 1.79344e+07
probabilities: 0.001 0.999
如果有人帮助我理解,我将不胜感激
真诚的 伊雷
有几种方法可以通过了解如何返回树对象 (
?rpart.object
) 来实现此目的。
我将按照
kyphosis
中的第一个示例展示在 R 中使用 ?rpart
数据集的两种方法:
fit <- rpart(Kyphosis ~ Age + Number + Start, data = kyphosis)
> tail(fit$cptable[, "nsplit"], 1)
3
4
> unname(tail(fit$cptable[, "nsplit"], 1)) ## or
[1] 4
来自
cptable
,其中包含给定大小的树的成本复杂性信息
> fit$cptable
CP nsplit rel error xerror xstd
1 0.17647059 0 1.0000000 1.000000 0.2155872
2 0.01960784 1 0.8235294 1.176471 0.2282908
3 0.01000000 4 0.7647059 1.176471 0.2282908
据我所知,该表的最后一行将引用当前最大的树。如果根据 CP 将树修剪到特定大小,则该矩阵的最后一行将包含该大小的树的信息:
> fit2 <- prune(fit, cp = 0.02)
> fit2$cptable
CP nsplit rel error xerror xstd
1 0.1764706 0 1.0000000 1.000000 0.2155872
2 0.0200000 1 0.8235294 1.176471 0.2282908
第二个选项是计算拟合模型的
<leaf>
分量的 var
列中 frame
的出现次数:
> fit$frame
var n wt dev yval complexity ncompete nsurrogate yval2.V1 yval2.V2
1 Start 81 81 17 1 0.17647059 2 1 1.00000000 64.00000000
2 Start 62 62 6 1 0.01960784 2 2 1.00000000 56.00000000
4 <leaf> 29 29 0 1 0.01000000 0 0 1.00000000 29.00000000
5 Age 33 33 6 1 0.01960784 2 2 1.00000000 27.00000000
10 <leaf> 12 12 0 1 0.01000000 0 0 1.00000000 12.00000000
11 Age 21 21 6 1 0.01960784 2 0 1.00000000 15.00000000
22 <leaf> 14 14 2 1 0.01000000 0 0 1.00000000 12.00000000
23 <leaf> 7 7 3 2 0.01000000 0 0 2.00000000 3.00000000
3 <leaf> 19 19 8 2 0.01000000 0 0 2.00000000 8.00000000
yval2.V3 yval2.V4 yval2.V5 yval2.nodeprob
1 17.00000000 0.79012346 0.20987654 1.00000000
2 6.00000000 0.90322581 0.09677419 0.76543210
4 0.00000000 1.00000000 0.00000000 0.35802469
5 6.00000000 0.81818182 0.18181818 0.40740741
10 0.00000000 1.00000000 0.00000000 0.14814815
11 6.00000000 0.71428571 0.28571429 0.25925926
22 2.00000000 0.85714286 0.14285714 0.17283951
23 4.00000000 0.42857143 0.57142857 0.08641975
3 11.00000000 0.42105263 0.57894737 0.23456790
该值 - 1 是分割数。为了进行计数,我们可以使用:
> grepl("^<leaf>$", as.character(fit$frame$var))
[1] FALSE FALSE TRUE FALSE TRUE FALSE TRUE TRUE TRUE
> sum(grepl("^<leaf>$", as.character(fit$frame$var))) - 1
[1] 4
我使用的正则表达式可能有点矫枉过正,但它意味着检查以 (
^
) 开头并以 ($
) "<leaf>"
结尾的字符串,即,这是整个字符串。我使用 grepl()
将
var
列上的匹配项作为逻辑向量返回,我们可以对
TRUE
求和并从中减去 1。由于
var
存储为因子,因此我在
grepl()
调用中将其转换为字符向量。您还可以使用
grep()
来返回匹配项的索引并使用
length()
来对它们进行计数:
> grep("^<leaf>$", as.character(fit$frame$var))
[1] 3 5 7 8 9
> length(grep("^<leaf>$", as.character(fit$frame$var))) - 1
[1] 4
rpart.plot
包来计算行数:
library(rpart.plot)
> fit <- rpart(Kyphosis ~ Age + Number + Start, data = kyphosis)
> data.frame(rpart.rules(fit))
Kyphosis Var.2 Var.3 Var.4 Var.5 Var.6 Var.7 Var.8 Var.9 Var.10 Var.11 Var.12 Var.13
4 0.00 when Start >= 15
10 0.00 when Start is 9 to 15 & Age < 55
22 0.14 when Start is 9 to 15 & Age >= 111
23 0.57 when Start is 9 to 15 & Age is 55 to 111
3 0.58 when Start < 9
> nrow(data.frame(rpart.rules(tree)))
5
由于我的结果是5,而另一个答案是4,也许我没有完全数清你想要的是什么? 我想这往往就是我想要的。