Reputation: 2187
I originally had a data frame composed of 12 columns in N rows. The last column is my class (0 or 1). I had to convert my entire data frame to numeric with
training <- sapply(training.temp,as.numeric)
But then I thought I needed the class column to be a factor column to use the randomforest() tool as a classifier, so I did
training[,"Class"] <- factor(training[,ncol(training)])
I proceed to creating the tree with
training_rf <- randomForest(Class ~., data = trainData, importance = TRUE, do.trace = 100)
But I'm getting two errors:
1: In Ops.factor(training[, "Status"], factor(training[, ncol(training)])) :
<= this is not relevant for factors (roughly translated)
2: In randomForest.default(m, y, ...) :
The response has five or fewer unique values. Are you sure you want to do regression?
I would appreciate it if someone could point out the formatting mistake I'm making.
Thanks!
Upvotes: 10
Views: 13554
Reputation: 2187
So the issue is actually quite simple. It turns out my training data was an atomic vector. So it first had to be converted as a data frame. So I needed to add the following line:
training <- as.data.frame(training)
Problem solved!
Upvotes: 8
Reputation: 2397
First, your coercion to a factor is not working because of syntax errors. Second, you should always use indexing when specifying a RF model. Here are changes in your code that should make it work.
training <- sapply(training.temp,as.numeric)
training[,"Class"] <- as.factor(training[,"Class"])
training_rf <- randomForest(x=training[,1:(ncol(training)-1)], y=training[,"Class"],
importance=TRUE, do.trace=100)
# You can also coerce to a factor directly in the model statement
training_rf <- randomForest(x=training[,1:(ncol(training)-1)], y=as.factor(training[,"Class"]),
importance=TRUE, do.trace=100)
Upvotes: 6