I have an SFrame and a model:

train_data,test_data = products.random_split(.8, seed=0)
selected_words_model = graphlab.logistic_classifier.create(train_data,

After computing the accuracy of the model with `selected_words_model.evaluate(test_data) I'm asked "What is the accuracy majority class classifier on this task?" Yet I don't even know what this "means accuracy majority class classifier", shouldn't it be "accuracy of the majority class classifier" ?

Here is my attempt.

All these materials come from this coursera ML fundations course exercise.


2 Answers 2


I suspect you are right that there is a missing "of the," and that the "majority class classifier" is the classifier that predicts the majority class for every input. Such a classifier is useful as a baseline model, and is particularly important when using accuracy as your metric. This matches what your notebook comments in the next bullet, so that's likely what was intended.

  • $\begingroup$ Thank you ! That seems the most simple classification method ;) I still have one doube : how should I compare different learned models with the baseline approach where we are just predicting the majority class ? I tried a selected_words_model.show(view='Evaluation') but it still might depend on the threshold ? $\endgroup$ Sep 5, 2019 at 14:33

Naive classifiers, such as picking at random or always picking the majority class, serve as a braindead baseline to compare your model against. Hopefully your model can out-perform a naive model that simply randomly picks the response from the input rows, or doesn't look at the predictors at all!

It is also a warning about using metrics like accuracy for unbalanced datasets, ex. a dataset with response 90% 0s will give a majority classifier a 90% accuracy.

The majority classifier is in fact the optimal Bayes classifier assuming a prior based on the training class proportion, without looking at the predictor variables (so $Y$ and $X$ are independent):

$$ \hat y = \operatorname{argmax}_y P(y \mid \mathbf x) = \operatorname{argmax}_y P(y) $$


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