我一直在用头撞rpart几天了（尝试为我拥有的这个数据集制作分类树），我认为现在是时候询问生命线了:-)我确信这是我没有看到的愚蠢的事情，但这里是我一直在做什么：
EuropeWater <- read.csv(file=paste("/Users/artessaniccola/Documents/",
"Magic Briefcase/CityTypology/Europe_water.csv",sep=""))
library(rpart)
attach(EuropeWater)
names(EuropeWater)
[1] "City" "waterpercapita_m3" "water_class" "population"
[5] "GDPpercapita" "area_km2" "populationdensity" "climate"
EuropeWater$water_class <- factor(EuropeWater$water_class, levels=1:3,
labels=c("Low", "Medium", "High"))
EuropeWater$climate <- factor(EuropeWater$climate, levels=2:4,
labels=c("Arid", "Warm temperate", "Snow"))
EuropeWater_tree <- rpart(EuropeWater$water_class ~
population+GDPpercapita + area_km2 + populationdensity +
EuropeWater$climate,
data=EuropeWater, method=class)
Error in as.character(x) :
cannot coerce type 'builtin' to vector of type 'character'
对于我的一生，我无法弄清楚错误是什么。
这有效吗？
EuropeWater_tree <- rpart(EuropeWater$water_class ~
population+GDPpercapita + area_km2 + populationdensity + EuropeWater$climate,
data=EuropeWater, method="class")
我认为你应该引用方法类型。
当你使用class代替"class"然后 R 尝试将其自身转换为字符：
as.character(class)
Error in as.character(class) :
cannot coerce type 'builtin' to vector of type 'character'
Because class是一个有类型的函数builtin:
typeof(class)
[1] "builtin"
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