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3.1、随机森林之随机森林实例_随机森林error in yhat[keep] <- ans$ypred : nas are not

作者：知新_RL | 2024-02-22 21:00:04

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随机森林error in yhat[keep] <- ans$ypred : nas are not allowed in subscripte

随机森林实例

Markdown脚本及数据集：http://pan.baidu.com/s/1bnY6ar9

实例一、用随机森林对鸢尾花数据进行分类


#1、加载数据并查看
data("iris")
summary(iris)


##   Sepal.Length    Sepal.Width     Petal.Length    Petal.Width   
##  Min.   :4.300   Min.   :2.000   Min.   :1.000   Min.   :0.100  
##  1st Qu.:5.100   1st Qu.:2.800   1st Qu.:1.600   1st Qu.:0.300  
##  Median :5.800   Median :3.000   Median :4.350   Median :1.300  
##  Mean   :5.843   Mean   :3.057   Mean   :3.758   Mean   :1.199  
##  3rd Qu.:6.400   3rd Qu.:3.300   3rd Qu.:5.100   3rd Qu.:1.800  
##  Max.   :7.900   Max.   :4.400   Max.   :6.900   Max.   :2.500  
##        Species  
##  setosa    :50  
##  versicolor:50  
##  virginica :50  
##                 
##                 
##

str(iris)


## 'data.frame':    150 obs. of  5 variables:
##  $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
##  $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
##  $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
##  $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
##  $ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...


#2、创建训练集和测试集数据
set.seed(2001)
library(caret)

## Loading required package: lattice

## Loading required package: ggplot2

## Warning: package 'ggplot2' was built under R version 3.2.3


index <- createDataPartition(iris$Species, p=0.7, list=F)
train_iris <- iris[index, ]
test_iris <- iris[-index, ]
 
#3、建模 
library(randomForest)

## randomForest 4.6-12

## Type rfNews() to see new features/changes/bug fixes.


## 
## Attaching package: 'randomForest'


## The following object is masked from 'package:ggplot2':
## 
##     margin


model_iris <- randomForest(Species~., data=train_iris, ntree=50, nPerm=10, mtry=3, proximity=T, importance=T)
 
#4、模型评估
model_iris


## 
## Call:
##  randomForest(formula = Species ~ ., data = train_iris, ntree = 50,      nPerm = 10, mtry = 3, proximity = T, importance = T) 
##                Type of random forest: classification
##                      Number of trees: 50
## No. of variables tried at each split: 3
## 
##         OOB estimate of  error rate: 4.76%
## Confusion matrix:
##            setosa versicolor virginica class.error
## setosa         35          0         0  0.00000000
## versicolor      0         32         3  0.08571429
## virginica       0          2        33  0.05714286

str(model_iris)


## List of 19
##  $ call           : language randomForest(formula = Species ~ ., data = train_iris, ntree = 50,      nPerm = 10, mtry = 3, proximity = T, importance = T)
##  $ type           : chr "classification"
##  $ predicted      : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
##   ..- attr(*, "names")= chr [1:105] "5" "7" "8" "11" ...
##  $ err.rate       : num [1:50, 1:4] 0.0513 0.0758 0.0741 0.0435 0.0505 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : NULL
##   .. ..$ : chr [1:4] "OOB" "setosa" "versicolor" "virginica"
##  $ confusion      : num [1:3, 1:4] 35 0 0 0 32 2 0 3 33 0 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : chr [1:3] "setosa" "versicolor" "virginica"
##   .. ..$ : chr [1:4] "setosa" "versicolor" "virginica" "class.error"
##  $ votes          : matrix [1:105, 1:3] 1 1 1 1 1 1 1 1 1 1 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : chr [1:105] "5" "7" "8" "11" ...
##   .. ..$ : chr [1:3] "setosa" "versicolor" "virginica"
##   ..- attr(*, "class")= chr [1:2] "matrix" "votes"
##  $ oob.times      : num [1:105] 15 23 22 16 17 11 20 20 17 19 ...
##  $ classes        : chr [1:3] "setosa" "versicolor" "virginica"
##  $ importance     : num [1:4, 1:5] 0 0 0.3417 0.34918 -0.00518 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##   .. ..$ : chr [1:5] "setosa" "versicolor" "virginica" "MeanDecreaseAccuracy" ...
##  $ importanceSD   : num [1:4, 1:4] 0 0 0.04564 0.04711 0.00395 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##   .. ..$ : chr [1:4] "setosa" "versicolor" "virginica" "MeanDecreaseAccuracy"
##  $ localImportance: NULL
##  $ proximity      : num [1:105, 1:105] 1 1 1 1 1 1 1 1 1 1 ...
##   ..- attr(*, "dimnames")=List of 2
##   .. ..$ : chr [1:105] "5" "7" "8" "11" ...
##   .. ..$ : chr [1:105] "5" "7" "8" "11" ...
##  $ ntree          : num 50
##  $ mtry           : num 3
##  $ forest         :List of 14
##   ..$ ndbigtree : int [1:50] 11 5 9 9 9 9 9 11 11 9 ...
##   ..$ nodestatus: int [1:17, 1:50] 1 -1 1 1 1 -1 -1 1 -1 -1 ...
##   ..$ bestvar   : int [1:17, 1:50] 4 0 4 3 3 0 0 1 0 0 ...
##   ..$ treemap   : int [1:17, 1:2, 1:50] 2 0 4 6 8 0 0 10 0 0 ...
##   ..$ nodepred  : int [1:17, 1:50] 0 1 0 0 0 2 3 0 3 2 ...
##   ..$ xbestsplit: num [1:17, 1:50] 0.8 0 1.65 5.25 4.85 0 0 6.05 0 0 ...
##   ..$ pid       : num [1:3] 1 1 1
##   ..$ cutoff    : num [1:3] 0.333 0.333 0.333
##   ..$ ncat      : Named int [1:4] 1 1 1 1
##   .. ..- attr(*, "names")= chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##   ..$ maxcat    : int 1
##   ..$ nrnodes   : int 17
##   ..$ ntree     : num 50
##   ..$ nclass    : int 3
##   ..$ xlevels   :List of 4
##   .. ..$ Sepal.Length: num 0
##   .. ..$ Sepal.Width : num 0
##   .. ..$ Petal.Length: num 0
##   .. ..$ Petal.Width : num 0
##  $ y              : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
##   ..- attr(*, "names")= chr [1:105] "5" "7" "8" "11" ...
##  $ test           : NULL
##  $ inbag          : NULL
##  $ terms          :Classes 'terms', 'formula' length 3 Species ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width
##   .. ..- attr(*, "variables")= language list(Species, Sepal.Length, Sepal.Width, Petal.Length, Petal.Width)
##   .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
##   .. .. ..- attr(*, "dimnames")=List of 2
##   .. .. .. ..$ : chr [1:5] "Species" "Sepal.Length" "Sepal.Width" "Petal.Length" ...
##   .. .. .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##   .. ..- attr(*, "term.labels")= chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##   .. ..- attr(*, "order")= int [1:4] 1 1 1 1
##   .. ..- attr(*, "intercept")= num 0
##   .. ..- attr(*, "response")= int 1
##   .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv> 
##   .. ..- attr(*, "predvars")= language list(Species, Sepal.Length, Sepal.Width, Petal.Length, Petal.Width)
##   .. ..- attr(*, "dataClasses")= Named chr [1:5] "factor" "numeric" "numeric" "numeric" ...
##   .. .. ..- attr(*, "names")= chr [1:5] "Species" "Sepal.Length" "Sepal.Width" "Petal.Length" ...
##  - attr(*, "class")= chr [1:2] "randomForest.formula" "randomForest"


pred <- predict(model_iris, train_iris)
mean(pred==train_iris[, 5])

## [1] 1


#5、预测
pred_iris <- predict(model_iris, test_iris)
table(pred_iris, test_iris[, 5])


##             
## pred_iris    setosa versicolor virginica
##   setosa         15          0         0
##   versicolor      0         13         2
##   virginica       0          2        13

mean(pred_iris==test_iris[, 5])

## [1] 0.9111111


library(gmodels)
CrossTable(pred_iris, test_iris[, 5])


## 
##  
##    Cell Contents
## |-------------------------|
## |                       N |
## | Chi-square contribution |
## |           N / Row Total |
## |           N / Col Total |
## |         N / Table Total |
## |-------------------------|
## 
##  
## Total Observations in Table:  45 
## 
##  
##              | test_iris[, 5] 
##    pred_iris |     setosa | versicolor |  virginica |  Row Total | 
## -------------|------------|------------|------------|------------|
##       setosa |         15 |          0 |          0 |         15 | 
##              |     20.000 |      5.000 |      5.000 |            | 
##              |      1.000 |      0.000 |      0.000 |      0.333 | 
##              |      1.000 |      0.000 |      0.000 |            | 
##              |      0.333 |      0.000 |      0.000 |            | 
## -------------|------------|------------|------------|------------|
##   versicolor |          0 |         13 |          2 |         15 | 
##              |      5.000 |     12.800 |      1.800 |            | 
##              |      0.000 |      0.867 |      0.133 |      0.333 | 
##              |      0.000 |      0.867 |      0.133 |            | 
##              |      0.000 |      0.289 |      0.044 |            | 
## -------------|------------|------------|------------|------------|
##    virginica |          0 |          2 |         13 |         15 | 
##              |      5.000 |      1.800 |     12.800 |            | 
##              |      0.000 |      0.133 |      0.867 |      0.333 | 
##              |      0.000 |      0.133 |      0.867 |            | 
##              |      0.000 |      0.044 |      0.289 |            | 
## -------------|------------|------------|------------|------------|
## Column Total |         15 |         15 |         15 |         45 | 
##              |      0.333 |      0.333 |      0.333 |            | 
## -------------|------------|------------|------------|------------|
## 
##

实例二、用坦泰尼克号乘客是否存活数据应用到随机森林算法中

在随机森林算法的函数randomForest()中有两个非常重要的参数，而这两个参数又将影响模型的准确性，它们分别是mtry和ntree。一般对mtry的选择是逐一尝试，直到找到比较理想的值，ntree的选择可通过图形大致判断模型内误差稳定时的值。 randomForest包中的randomForest(formula, data, ntree, nPerm, mtry, proximity, importace)函数：随机森林分类与回归。ntree表示生成决策树的数目（不应设置太小，默认为 500）；nPerm表示计算importance时的重复次数，数量大于1给出了比较稳定的估计，但不是很有效（目前只实现了回归）；mtry表示选择的分裂属性的个数；proximity表示是否生成邻近矩阵，为T表示生成邻近矩阵；importance表示输出分裂属性的重要性。

下面使用坦泰尼克号乘客是否存活数据应用到随机森林算法中，看看模型的准确性如何。


#1、加载数据并查看：同时读取训练样本和测试样本集
train <- read.table("F:\\R\\Rworkspace\\RandomForest/train.csv", header=T, sep=",")
test <- read.table("F:\\R\\Rworkspace\\RandomForest/test.csv", header=T, sep=",")
#注意：训练集和测试集数据来自不同的数据集，一定要注意测试集和训练集的factor的levels相同，否则，在利用训练集训练的模型对测试集进行预测时，会报错！！！
 
str(train)


## 'data.frame':    891 obs. of  8 variables:
##  $ Survived: int  0 1 1 1 0 0 0 0 1 1 ...
##  $ Pclass  : int  3 1 3 1 3 3 1 3 3 2 ...
##  $ Sex     : Factor w/ 2 levels "female","male": 2 1 1 1 2 2 2 2 1 1 ...
##  $ Age     : num  22 38 26 35 35 NA 54 2 27 14 ...
##  $ SibSp   : int  1 1 0 1 0 0 0 3 0 1 ...
##  $ Parch   : int  0 0 0 0 0 0 0 1 2 0 ...
##  $ Fare    : num  7.25 71.28 7.92 53.1 8.05 ...
##  $ Embarked: Factor w/ 4 levels "","C","Q","S": 4 2 4 4 4 3 4 4 4 2 ...

str(test)


## 'data.frame':    418 obs. of  7 variables:
##  $ Pclass  : int  3 3 2 3 3 3 3 2 3 3 ...
##  $ Sex     : Factor w/ 2 levels "female","male": 2 1 2 2 1 2 1 2 1 2 ...
##  $ Age     : num  34.5 47 62 27 22 14 30 26 18 21 ...
##  $ SibSp   : int  0 1 0 0 1 0 0 1 0 2 ...
##  $ Parch   : int  0 0 0 0 1 0 0 1 0 0 ...
##  $ Fare    : num  7.83 7 9.69 8.66 12.29 ...
##  $ Embarked: Factor w/ 3 levels "C","Q","S": 2 3 2 3 3 3 2 3 1 3 ...


#从上可知：训练集数据共891条记录，8个变量，Embarked因子水平为4；测试集数据共418条记录，7个变量，Embarked因子水平为3；训练集中存在缺失数据；Survived因变量为数字类型，测试集数据无因变量
 
#2、数据清洗
#1）调整测试集与训练基地因子水平
levels(train$Embarked)

## [1] ""  "C" "Q" "S"

levels(test$Embarked)

## [1] "C" "Q" "S"


levels(test$Embarked) <- levels(train$Embarked)
 
#2)把因变量转化为因子类型
train$Survived <- as.factor(train$Survived)
 
#3)使用rfImpute()函数补齐训练集的缺失值NA
library(randomForest)
train_impute <- rfImpute(Survived~., data=train)


## ntree      OOB      1      2
##   300:  16.39%  7.83% 30.12%
## ntree      OOB      1      2
##   300:  16.50%  8.93% 28.65%
## ntree      OOB      1      2
##   300:  16.72%  8.74% 29.53%
## ntree      OOB      1      2
##   300:  16.50%  8.56% 29.24%
## ntree      OOB      1      2
##   300:  17.28%  9.47% 29.82%


#4)补齐测试集的缺失值：对待测样本进行预测,发现待测样本中存在缺失值，这里使用多重插补法将缺失值补齐
summary(test)


##      Pclass          Sex           Age            SibSp       
##  Min.   :1.000   female:152   Min.   : 0.17   Min.   :0.0000  
##  1st Qu.:1.000   male  :266   1st Qu.:21.00   1st Qu.:0.0000  
##  Median :3.000                Median :27.00   Median :0.0000  
##  Mean   :2.266                Mean   :30.27   Mean   :0.4474  
##  3rd Qu.:3.000                3rd Qu.:39.00   3rd Qu.:1.0000  
##  Max.   :3.000                Max.   :76.00   Max.   :8.0000  
##                               NA's   :86                      
##      Parch             Fare         Embarked
##  Min.   :0.0000   Min.   :  0.000    :102   
##  1st Qu.:0.0000   1st Qu.:  7.896   C: 46   
##  Median :0.0000   Median : 14.454   Q:270   
##  Mean   :0.3923   Mean   : 35.627   S:  0   
##  3rd Qu.:0.0000   3rd Qu.: 31.500           
##  Max.   :9.0000   Max.   :512.329           
##                   NA's   :1


#可是看出测试集数据存在缺失值NA,Age和Fare的数据有NA
 
#多重插补法填充缺失值：
library(mice)

## Loading required package: Rcpp

## mice 2.25 2015-11-09

imput <- mice(data=test, m=10)


## 
##  iter imp variable
##   1   1  Age  Fare
##   1   2  Age  Fare
##   1   3  Age  Fare
##   1   4  Age  Fare
##   1   5  Age  Fare
##   1   6  Age  Fare
##   1   7  Age  Fare
##   1   8  Age  Fare
##   1   9  Age  Fare
##   1   10  Age  Fare
##   2   1  Age  Fare
##   2   2  Age  Fare
##   2   3  Age  Fare
##   2   4  Age  Fare
##   2   5  Age  Fare
##   2   6  Age  Fare
##   2   7  Age  Fare
##   2   8  Age  Fare
##   2   9  Age  Fare
##   2   10  Age  Fare
##   3   1  Age  Fare
##   3   2  Age  Fare
##   3   3  Age  Fare
##   3   4  Age  Fare
##   3   5  Age  Fare
##   3   6  Age  Fare
##   3   7  Age  Fare
##   3   8  Age  Fare
##   3   9  Age  Fare
##   3   10  Age  Fare
##   4   1  Age  Fare
##   4   2  Age  Fare
##   4   3  Age  Fare
##   4   4  Age  Fare
##   4   5  Age  Fare
##   4   6  Age  Fare
##   4   7  Age  Fare
##   4   8  Age  Fare
##   4   9  Age  Fare
##   4   10  Age  Fare
##   5   1  Age  Fare
##   5   2  Age  Fare
##   5   3  Age  Fare
##   5   4  Age  Fare
##   5   5  Age  Fare
##   5   6  Age  Fare
##   5   7  Age  Fare
##   5   8  Age  Fare
##   5   9  Age  Fare
##   5   10  Age  Fare


Age <- data.frame(Age=apply(imput$imp$Age, 1, mean))
Fare <- data.frame(Fare=apply(imput$imp$Fare, 1, mean))
 
#添加行标号：
test$Id <- row.names(test)
Age$Id <- row.names(Age)
Fare$Id <- row.names(Fare)
 
#替换缺失值：
test[test$Id %in% Age$Id, 'Age'] <- Age$Age
test[test$Id %in% Fare$Id, 'Fare'] <- Fare$Fare
summary(test)


##      Pclass          Sex           Age            SibSp       
##  Min.   :1.000   female:152   Min.   : 0.17   Min.   :0.0000  
##  1st Qu.:1.000   male  :266   1st Qu.:22.00   1st Qu.:0.0000  
##  Median :3.000                Median :26.19   Median :0.0000  
##  Mean   :2.266                Mean   :29.41   Mean   :0.4474  
##  3rd Qu.:3.000                3rd Qu.:36.65   3rd Qu.:1.0000  
##  Max.   :3.000                Max.   :76.00   Max.   :8.0000  
##      Parch             Fare         Embarked      Id           
##  Min.   :0.0000   Min.   :  0.000    :102    Length:418        
##  1st Qu.:0.0000   1st Qu.:  7.896   C: 46    Class :character  
##  Median :0.0000   Median : 14.454   Q:270    Mode  :character  
##  Mean   :0.3923   Mean   : 35.583   S:  0                      
##  3rd Qu.:0.0000   3rd Qu.: 31.472                              
##  Max.   :9.0000   Max.   :512.329


#从上可知：测试数据集中已经没有了NA值。
 
#3、选着随机森林的mtry和ntree值
#1）选着mtry
(n <- length(names(train)))

## [1] 8


library(randomForest)
for(i in 1:n) {
  model <- randomForest(Survived~., data=train_impute,  mtry=i)
  err <- mean(model$err.rate)
  print(err)
}


## [1] 0.2100028
## [1] 0.1889116
## [1] 0.1776607
## [1] 0.1902606
## [1] 0.1960938
## [1] 0.1953451
## [1] 0.1951303
## [1] 0.2018745


#从上可知：mtry=2或者mtry=3时，模型内评价误差最小，故确定参数mtry=2或者mtry=3
 
#2)选着ntree
set.seed(2002)
model <- randomForest(Survived~., data=train_impute, mtry=2, ntree=1000)
plot(model)


#从上图可知：ntree在400左右时，模型内误差基本稳定，故取ntree=400
 
#4、建模
model_fit <- randomForest(Survived~., data=train_impute, mtry=2, ntree=400, importance=T)
 
#5、模型评估
model_fit


## 
## Call:
##  randomForest(formula = Survived ~ ., data = train_impute, mtry = 2,      ntree = 400, importance = T) 
##                Type of random forest: classification
##                      Number of trees: 400
## No. of variables tried at each split: 2
## 
##         OOB estimate of  error rate: 16.61%
## Confusion matrix:
##     0   1 class.error
## 0 500  49  0.08925319
## 1  99 243  0.28947368


#查看变量的重要性
(importance <- importance(x=model_fit))


##                  0         1 MeanDecreaseAccuracy MeanDecreaseGini
## Pclass   16.766454 28.241508             32.16125         33.15984
## Sex      46.578191 76.145306             72.42624        100.74843
## Age      19.882605 24.586274             30.52032         60.85186
## SibSp    19.070707  2.834303             18.95690         16.11720
## Parch    10.366140  8.380559             13.18282         12.28725
## Fare     18.649672 20.967558             29.43262         66.31489
## Embarked  7.904436 11.479919             14.18780         12.68924


#绘制变量的重要性图
varImpPlot(model_fit)


#从上图可知：模型中乘客的性别最为重要，接下来的是Pclass,age,Fare和Fare,age,Pclass。
 
#6、预测
#1）对训练集数据预测：
train_pred <- predict(model_fit, train_impute)
mean(train_pred==train_impute$Survived)

## [1] 0.9135802

table(train_pred, train_impute$Survived)


##           
## train_pred   0   1
##          0 535  63
##          1  14 279


#模型的预测精度在90%以上
 
#2)对测试集数据预测：
test_pred <- predict(model_fit, test[, 1:7])
head(test_pred)


## 1 2 3 4 5 6 
## 0 0 0 0 1 0 
## Levels: 0 1

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3.1、随机森林之随机森林实例_随机森林error in yhat[keep] <- ans$ypred : nas are not

随机森林

junjun

2016年2月8日

随机森林实例

实例一、用随机森林对鸢尾花数据进行分类

实例二、用坦泰尼克号乘客是否存活数据应用到随机森林算法中