Вопрос

I cannot understand what is going wrong here.

data.train <- read.table("Assign2.WineComplete.csv",sep=",",header=T)
# Building decision tree
Train <- data.frame(residual.sugar=data.train$residual.sugar,
                total.sulfur.dioxide=data.train$total.sulfur.dioxide, 
                alcohol=data.train$alcohol,
                quality=data.train$quality)
Pre <- as.formula("pre ~ quality")

fit <- rpart(Pre, method="class",data=Train)

I am getting the following error :

Error in eval(expr, envir, enclos) : object 'pre' not found
Это было полезно?

Решение

Don't know why @Janos deleted his answer, but it's correct: your data frame Train doesn't have a column named pre. When you pass a formula and a data frame to a model-fitting function, the names in the formula have to refer to columns in the data frame. Your Train has columns called residual.sugar, total.sulfur, alcohol and quality. You need to change either your formula or your data frame so they're consistent with each other.

And just to clarify: Pre is an object containing a formula. That formula contains a reference to the variable pre. It's the latter that has to be consistent with the data frame.

Другие советы

This can happen if you don't attach your dataset.

I think I got what I was looking for..

data.train <- read.table("Assign2.WineComplete.csv",sep=",",header=T)
fit <- rpart(quality ~ ., method="class",data=data.train)
plot(fit)
text(fit, use.n=TRUE)
summary(fit)

i use colname(train) = paste("A", colname(train)) and it turns out to the same problem as yours.

I finally figure out that randomForest is more stingy than rpart, it can't recognize the colname with space, comma or other specific punctuation.

paste function will prepend "A" and " " as seperator with each colname. so we need to avert the space and use this sentence instead:

colname(train) = paste("A", colname(train), sep = "")

this will prepend string without space.

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