# Using k-means to segment customers in the positive class

I have some labeled data (0=didn’t cancel, 1=canceled) that I am creating a model for in my marketing class.

On top of predicting who is likely to cancel, I’d like to explore the possibility of trying different proactive retention strategies. I was thinking of running k-means on the training data where the label=1 and get, say, 4 clusters.

Is this the right way to go about this? I would basically end up with two models and run each customer through the binary classifier, and if it’s predicted to cancel, run the customer through the clustering model.

I’m not sure of this approach because k-means is an unsupervised learning method and I’m sort of helping it by feeding it just the customers in the positive class.

• You may be way too optimistic on k-means ability to process your data in a meaningful way, and of the chances of finding a behavioral pattern, and not finding trivial customers groups (say, old males, old females, young males, young females). Commented Aug 14, 2019 at 5:57

A more statistical way to approach the question would be this. First, can I tell which covariates determine whether or not a customer is likely to cancel? A logistic regression model could do this, for example. If you then suspect that customers who differ on a certain covariate $$X$$ are likely to cancel for different reasons, then you could include an interaction between $$X$$ and other covariates in the logistic regression model.
For example, suppose you're selling newspapers and you observe $$Y =$$ cancellation, $$X =$$ political affiliation (Democrat/Republican/Independent), and a measure of "liberalness" of the news customers read on a 1-10 scale $$Z$$. Then you might run the regression $$Y \sim X + Z$$ and conclude that Democrats are most likely to cancel but that liberalness doesn't really affect cancellation much. But then you might run the regression $$Y \sim X + Z + XZ$$, which includes an interaction term, and realize that higher liberalness decreases cancellation probability for Democrats, doesn't change it for Independents, and increases it for Republicans.