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Home/ Questions/Quick solution to fix the error: error: `data` and `reference` should be factors with the same levels
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Mila Maire
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Mila Maire
Asked: May 12, 20222022-05-12T09:43:30+00:00 2022-05-12T09:43:30+00:00In: r

Quick solution to fix the error: error: `data` and `reference` should be factors with the same levels

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I get the error message:

Error in confusionMatrix.default(pred, testing$Final) : the data and reference factors must have the same number of levels

Has anyone ever faced this problem? How to troubleshoot the “ error: `data` and `reference` should be factors with the same levels.”
But the problem is appearing when I try to operate the following program

EnglishMarks <- read.csv("E:/Subject Wise Data/EnglishMarks.csv", 
 header=TRUE)
 inTrain<-createDataPartition(y=EnglishMarks$Final,p=0.7,list=FALSE)
 training<-EnglishMarks[inTrain,]
 testing<-EnglishMarks[-inTrain,]
 predictionsTree <- predict(treeFit, testdata)
 confusionMatrix(predictionsTree, testdata$catgeory)
 modFit<-train(Final~UT1+UT2+HalfYearly+UT3+UT4,method="lm",data=training)
 pred<-format(round(predict(modFit,testing))) 
 confusionMatrix(pred,testing$Final)
> str(pred)
 chr [1:148] "85" "84" "87" "65" "88" "84" "82" "84" "65" "78" "78" "88" "85" 
 "86" "77" ...
 > str(testing$Final)
 int [1:148] 88 85 86 70 85 85 79 85 62 77 ...
 
 > levels(pred)
 NULL
 > levels(testing$Final)
 NULL

error in confusion matrix
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    lyytutoria Expert
    2022-06-02T02:06:27+00:00Added an answer on June 2, 2022 at 2:06 am

    The cause: This error occured due to data argument was not casted as factor.

    Solution: You can try this my following suggestion:

    confusionMatrix(pred,as.factor(testing$Final))
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  2. Paul Cohen
    2022-05-25T19:39:28+00:00Added an answer on May 25, 2022 at 7:39 pm

    table(pred) and table(testing$Final) table(pred) and table(testing$Final) will show you that at least one number is not predicted in the testing set. Never present in pred This is why there are “different levels”. 

    This trick worked for me.

    table(factor(pred, levels=min(test):max(test)), 
     factor(test, levels=min(test):max(test)))

    It should provide exactly the same confusion matrix that you get from the function.

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