r - 通过r中的for循环搜索最佳随机森林参数

Question

大家好，我正在尝试通过for循环搜索最佳参数。然而，结果实在令我困惑。下面的代码应该提供相同的结果，因为参数“mtry”是相同的。

       gender Partner   tenure Churn
3521     Male      No 0.992313   Yes
2525.1   Male      No 4.276666    No
567      Male     Yes 2.708050    No
8381   Female      No 4.202127   Yes
6258   Female      No 0.000000   Yes
6569     Male     Yes 2.079442    No
27410  Female      No 1.550804   Yes
6429   Female      No 1.791759   Yes
412    Female     Yes 3.828641    No
4655   Female     Yes 3.737670    No

<小时/>

RFModel = randomForest(Churn ~ .,
                     data = ggg,
                     ntree = 30,
                     mtry = 2,
                     importance = TRUE,
                     replace = FALSE)
print(RFModel$confusion)

    No Yes class.error
No   4   1         0.2
Yes  1   4         0.2

<小时/>

for(i in c(2)){
   RFModel = randomForest(Churn ~ .,
                     data = Trainingds,
                     ntree = 30,
                     mtry = i,
                     importance = TRUE,
                     replace = FALSE)
   print(RFModel$confusion)
}

     No Yes class.error
No   3   2         0.4
Yes  2   3         0.4

<小时/>

1. Code1 和 code2 应提供相同的输出。

Answer 1

每次您都会得到略有不同的结果，因为算法中内置了随机性。为了构建每棵树，算法对数据帧进行重新采样，并随机选择 mtry 列来根据重新采样的数据帧构建树。如果您希望使用相同参数(例如 mtry、ntree)构建的模型每次都给出相同的结果，则需要设置随机种子。

例如，让我们运行 randomForest 10 次，并检查每次运行的均方误差的平均值。请注意，平均 mse 每次都不同:

library(randomForest)

replicate(10, mean(randomForest(mpg ~ ., data=mtcars)$mse))

[1] 5.998530 6.307782 5.791657 6.125588 5.868717 5.845616 5.427208 6.112762 5.777624 6.150021

如果运行上面的代码，您将获得另外 10 个与上面的值不同的值。

如果您希望能够重现使用相同参数(例如 mtry 和 ntree)运行的给定模型的结果，那么您可以设置随机种子。例如:

set.seed(5)
mean(randomForest(mpg ~ ., data=mtcars)$mse)

[1] 6.017737

如果使用相同的种子值，您将得到相同的结果，否则会得到不同的结果。使用较大的 ntree 值会减少但不会消除模型运行之间的可变性。

更新:当我使用您提供的数据示例运行代码时，并不总是每次都能得到相同的结果。即使使用 replace=TRUE(这会导致在不进行替换的情况下对数据帧进行采样)，每次选择用于构建树的列也可能不同:

> randomForest(Churn ~ .,
+              data = ggg,
+              ntree = 30,
+              mtry = 2,
+              importance = TRUE,
+              replace = FALSE)

Call:
 randomForest(formula = Churn ~ ., data = ggg, ntree = 30, mtry = 2,      importance = TRUE, replace = FALSE) 
               Type of random forest: classification
                     Number of trees: 30
No. of variables tried at each split: 2

        OOB estimate of  error rate: 30%
Confusion matrix:
    No Yes class.error
No   3   2         0.4
Yes  1   4         0.2
> randomForest(Churn ~ .,
+              data = ggg,
+              ntree = 30,
+              mtry = 2,
+              importance = TRUE,
+              replace = FALSE)

Call:
 randomForest(formula = Churn ~ ., data = ggg, ntree = 30, mtry = 2,      importance = TRUE, replace = FALSE) 
               Type of random forest: classification
                     Number of trees: 30
No. of variables tried at each split: 2

        OOB estimate of  error rate: 20%
Confusion matrix:
    No Yes class.error
No   4   1         0.2
Yes  1   4         0.2
> randomForest(Churn ~ .,
+              data = ggg,
+              ntree = 30,
+              mtry = 2,
+              importance = TRUE,
+              replace = FALSE)

Call:
 randomForest(formula = Churn ~ ., data = ggg, ntree = 30, mtry = 2,      importance = TRUE, replace = FALSE) 
               Type of random forest: classification
                     Number of trees: 30
No. of variables tried at each split: 2

        OOB estimate of  error rate: 30%
Confusion matrix:
    No Yes class.error
No   3   2         0.4
Yes  1   4         0.2

以下是使用内置 iris 数据框得出的一组类似结果:

> randomForest(Species ~ ., data=iris, ntree=30, mtry=2, importance = TRUE,
+              replace = FALSE)

Call:
 randomForest(formula = Species ~ ., data = iris, ntree = 30,      mtry = 2, importance = TRUE, replace = FALSE) 
               Type of random forest: classification
                     Number of trees: 30
No. of variables tried at each split: 2

        OOB estimate of  error rate: 3.33%
Confusion matrix:
           setosa versicolor virginica class.error
setosa         50          0         0        0.00
versicolor      0         47         3        0.06
virginica       0          2        48        0.04
> randomForest(Species ~ ., data=iris, ntree=30, mtry=2, importance = TRUE,
+              replace = FALSE)

Call:
 randomForest(formula = Species ~ ., data = iris, ntree = 30,      mtry = 2, importance = TRUE, replace = FALSE) 
               Type of random forest: classification
                     Number of trees: 30
No. of variables tried at each split: 2

        OOB estimate of  error rate: 4.67%
Confusion matrix:
           setosa versicolor virginica class.error
setosa         50          0         0        0.00
versicolor      0         47         3        0.06
virginica       0          4        46        0.08
> randomForest(Species ~ ., data=iris, ntree=30, mtry=2, importance = TRUE,
+              replace = FALSE)

Call:
 randomForest(formula = Species ~ ., data = iris, ntree = 30,      mtry = 2, importance = TRUE, replace = FALSE) 
               Type of random forest: classification
                     Number of trees: 30
No. of variables tried at each split: 2

        OOB estimate of  error rate: 6%
Confusion matrix:
           setosa versicolor virginica class.error
setosa         50          0         0        0.00
versicolor      0         47         3        0.06
virginica       0          6        44        0.12

您还可以查看每个模型运行生成的树，它们通常会有所不同。例如，假设我运行以下代码三次，将结果存储在对象 m1、m2 和 m3 中。

randomForest(Churn ~ .,
             data = ggg,
             ntree = 30,
             mtry = 2,
             importance = TRUE,
             replace = FALSE)

现在让我们看看每个模型对象的前四棵树，我已将其粘贴在下面。输出是一个列表。您可以看到每个模型运行的第一棵树都是不同的。第二棵树对于前两次模型运行是相同的，但对于第三次模型运行是不同的，依此类推。

check.trees = lapply(1:4, function(i) {
  lapply(list(m1=m1,m2=m2,m3=m3), function(model) getTree(model, i, labelVar=TRUE))
  })

check.trees

[[1]]
[[1]]$m1
  left daughter right daughter split var split point status prediction
1             2              3   Partner    1.000000      1       <NA>
2             4              5    gender    1.000000      1       <NA>
3             0              0      <NA>    0.000000     -1         No
4             0              0      <NA>    0.000000     -1        Yes
5             6              7    tenure    2.634489      1       <NA>
6             0              0      <NA>    0.000000     -1        Yes
7             0              0      <NA>    0.000000     -1         No

[[1]]$m2
  left daughter right daughter split var split point status prediction
1             2              3    gender    1.000000      1       <NA>
2             0              0      <NA>    0.000000     -1        Yes
3             4              5    tenure    1.850182      1       <NA>
4             0              0      <NA>    0.000000     -1        Yes
5             0              0      <NA>    0.000000     -1         No

[[1]]$m3
  left daughter right daughter split var split point status prediction
1             2              3    tenure    2.249904      1       <NA>
2             0              0      <NA>    0.000000     -1        Yes
3             0              0      <NA>    0.000000     -1         No


[[2]]
[[2]]$m1
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             0              0      <NA>           0     -1        Yes
3             0              0      <NA>           0     -1         No

[[2]]$m2
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             0              0      <NA>           0     -1        Yes
3             0              0      <NA>           0     -1         No

[[2]]$m3
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             4              5    gender           1      1       <NA>
3             0              0      <NA>           0     -1         No
4             0              0      <NA>           0     -1        Yes
5             0              0      <NA>           0     -1         No


[[3]]
[[3]]$m1
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             4              5    gender           1      1       <NA>
3             0              0      <NA>           0     -1         No
4             0              0      <NA>           0     -1        Yes
5             0              0      <NA>           0     -1        Yes

[[3]]$m2
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             0              0      <NA>           0     -1        Yes
3             0              0      <NA>           0     -1         No

[[3]]$m3
  left daughter right daughter split var split point status prediction
1             2              3    tenure    2.129427      1       <NA>
2             0              0      <NA>    0.000000     -1        Yes
3             0              0      <NA>    0.000000     -1         No


[[4]]
[[4]]$m1
  left daughter right daughter split var split point status prediction
1             2              3    tenure    1.535877      1       <NA>
2             0              0      <NA>    0.000000     -1        Yes
3             4              5    tenure    4.015384      1       <NA>
4             0              0      <NA>    0.000000     -1         No
5             6              7    tenure    4.239396      1       <NA>
6             0              0      <NA>    0.000000     -1        Yes
7             0              0      <NA>    0.000000     -1         No

[[4]]$m2
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             0              0      <NA>           0     -1        Yes
3             0              0      <NA>           0     -1         No

[[4]]$m3
  left daughter right daughter split var split point status prediction
1             2              3   Partner           1      1       <NA>
2             0              0      <NA>           0     -1        Yes
3             0              0      <NA>           0     -1         No