使用 rpart 对新因子（分类）变量进行预测

Question

我正在使用 R 练习机器学习。我正在使用 rpart 方法进行训练。数据是来自 UCI 的成人数据集。 Link如下

http://archive.ics.uci.edu/ml/datasets/Adult

#Get the data    
adultData <- read.table("adult.data", header = FALSE, sep = ",")
adultName <- read.csv("adult.name", header = TRUE, sep = ",", stringsAsFactors = FALSE)
names(adultData) <- names(adultName)

为了简化练习，我只select几个属性并且将数据集减少到20%而已

selected <- c("age", "education", "marital.status", "relationship", "sex", "hours.per.week", "salary")
adultData <- subset(adultData, select = selected)
trainIndex = createDataPartition(adultData$salary, p=0.20, list=FALSE)
training = adultData[ trainIndex, ]

使用 "rpart" 拟合模型大约需要一分钟（使用 "gbm" 或 "rf" 会更慢）

set.seed(33833)
modFit <- train(salary ~ ., method = "rpart", data=training)

问题来自我对新数据值的预测。我创建了一个新的数据框

a <- data.frame(age = 40, education = "Bachelors", marital.status = "Divorced", relationship = "Wife", sex = "Female", hours.per.week = 40)
predict(modFit, newdata = a)

它 returns 一个错误 "education has a new level"。

我知道问题出在那些分类（因子）变量上。不知何故，他们不承认 "Bachelors" 是他们已经拥有的一个因素，而是一个新字符串（新因素）。

Answer 1

问题源于数据清理不当

下载数据后，我发现了 R 中因子的一个常见问题：标签具有额外的 space，因此，当您调用标签时（例如，在您的示例中为 "Bachelors"），系统无法识别它，因为该级别具有额外的因素 space:

“单身汉”

你可以通过调用因子的水平来看到这一点：levels(education)

您可以通过将 strip.white 参数设置为 TRUE

来删除读取调用中的白色 spaces

如果你以标准方式上传数据集，你可以看到因子的标签有额外的space

# Not Run 
#  adultData <- read.csv2("AdultDataRenamed.csv", header = TRUE)

# levels(adultData$education)

 # [1] " 10th"         " 11th"         " 12th"         " 1st-4th"     
 # [5] " 5th-6th"      " 7th-8th"      " 9th"          " Assoc-acdm"  
 # [9] " Assoc-voc"    " Bachelors"    " Doctorate"    " HS-grad"     
# [13] " Masters"      " Preschool"    " Prof-school"  " Some-college"

如果你上传数据集 strip.white = TRUE，你可以看到因子的标签没有多余的 space

# Not Run 
# adultData <- read.csv2("AdultDataRenamed.csv", header = TRUE, strip.white = TRUE)

# levels(adultData$education)

 # [1] "10th"         "11th"         "12th"         "1st-4th"      "5th-6th"     
 # [6] "7th-8th"      "9th"          "Assoc-acdm"   "Assoc-voc"    "Bachelors"   
# [11] "Doctorate"    "HS-grad"      "Masters"      "Preschool"    "Prof-school" 
# [16] "Some-college"

我通过上传我已重命名的干净数据集重现了该示例

# Not Run 
# adultData <- read.csv2("AdultDataRenamed.csv", header = TRUE, strip.white = TRUE)

数据集太宽，无法在此处发布；它可以很容易地从上面的指令中复制出来 link。我的干净数据集可以从这里下载 http://www.insular.it/?wpdmact=process&did=OC5ob3RsaW5r

经常看数据

dim(adultData)
head(adultData)
str(adultData)

调用你需要的库

library(rpart)
library(caret)

我选择了与您选择的相同的属性，并且我已将数据集减少到仅 40%（这对于训练是可接受的）

selected <- c("age", "education", "marital.status", "relationship", "sex", "hours.per.week", "salary")
adultData <- subset(adultData, select = selected)
trainIndex = createDataPartition(adultData$salary, p=0.40, list=FALSE)
training = adultData[ trainIndex, ]

我还添加了一个测试集

test = adultData[ -trainIndex, ]

模型拟合

set.seed(33833)
modFit <- train(salary ~ ., method = "rpart", data=training)

总体准确度

prediction <- predict(modFit, newdata=test)

tab <- table(prediction, test$salary)

sum(diag(tab))/sum(tab)

使用 caret 包进行更好的测试

rpartPred<-predict(modFit,test)

confusionMatrix(rpartPred,test$salary)

绘制模型（不是很清楚）

library(rattle)

fancyRpartPlot(modFit$finalModel)

备选

library(partykit)

finalModel <-as.party(modFit$finalModel)
plot(finalModel)

根据您指定的新数据值进行预测

a <- data.frame(age = 40, education = "Bachelors", marital.status = "Divorced", relationship = "Wife", sex = "Female", hours.per.week = 40)

predict(modFit, newdata = a)

使用 rpart 对新因子（分类）变量进行预测

Prediction using rpart on new factor (categorical) variables

r

machine-learning

prediction

rpart

categorical-data