Error in eval(expr, envir, enclos) : object not found

17
votes

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
4
rdataframerpart
You don't need the second or third lines of your code. Just do the read.table line then do: fit <- rpart(pre ~ quality, method="class",data=data.train). - Thomas
I tried what you asked me to do, but I still get the same error - Rads
Is there a upper/lower case problem here? I see 'Pre' declared but error is about 'pre'. - ako
No, if instead of all the statements, i just write data.train <- read.table("Assign2.WineComplete.csv",sep=",",header=T) and then fit <- rpart(pre ~ quality, method="class",data=data.train) , i get the same error - Rads

4 Answers

22
votes

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.

14
votes

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

2
votes

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)
0
votes

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.