Reputation: 301
OK, so another newbie question related to the Titanic Competition:
I am trying to run a Random Forest prediction against my test data. All my work has been done on combined test and training data.
I have now split the 2 to testdata and trainingdata
I have the following code:
trainingdata <- droplevels(data.combined[1:891,])
testdata <- droplevels(data.combined[892:1309,])
fitRF <- randomForest(as.factor(Survived) ~ Pclass + Sex + Age + SibSp
+ Parch + Fare + Embarked
+ new.title + family.size + FamilyID2,
data=trainingdata,
importance =T,
ntree=2000)
varImpPlot(fitRF)
#All works up to this point
Prediction <- predict(fitRF, testdata)
#This line above generates error
submit <- data.frame(PassengerID = data.combined$PassengerId, Survived
= Prediction)
write.csv(submit, file="14072017_1_RF", row.names = F)
When I run the Prediction line I get the following error:
> Prediction <- predict(fitRF, testdata)
Error in predict.randomForest(fitRF, testdata) :
New factor levels not present in the training data
When i run str(testdata) and str(trainingdata) I can see 2 factors that no longer match
Trainingdata
$ Parch : Factor w/ 7 levels
Testdata
$ Parch : Factor w/ 8
Trainingdata
$ FamilyID2 : Factor w/ 22
Testdata
$ FamilyID2 : Factor w/ 18
Is it these differences that are causing my error to occur? And if so, how do I resolve this?
Many Thanks
Additional Information: I have removed Parch and FamilyID2 from the RandomForest creation line, and the code now works, so it is definitely those 2 variables that are causing the issue with mismatched levels.
Upvotes: 1
Views: 3692
Reputation: 374
Fellow newbie here, I was just toying around with Titanic these days. I think it doesn´t make sense to have the Parch variable as a factor, so maybe make it numeric and that may solve the problem:
train$Parch <- as.numeric(train$Parch)
Otherwise, the test data has 2 obs with the value of 9 for Parch, which are not present in the train data:
> table(train$Parch)
0 1 2 3 4 5 6
678 118 80 5 4 5 1
> table(test$Parch)
0 1 2 3 4 5 6 9
324 52 33 3 2 1 1 2
>
Alternatively, if you need the variable to be a factor, then you could just add another level to it:
train$Parch <- as.factor(train$Parch) # in my data, Parch is type int
train$Parch
levels(train$Parch) <- c(levels(train$Parch), "9")
train$Parch # now Parch has 7 levels
table(train$Parch) # level 9 is empty
Upvotes: 3