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I have a dataset which contains only categorical data i.e.A,B,C,D (like factors) for each predictor. There are 10 predictors and the dependent variable is binary, 0,1.

UPDATE: MY predictors are answers for multiple choice questions for a questionnaire. So each predictor only takes on categorical values, i.e. X_1 can be A,B,C or D, X_2 can be A,B,C,D,E,F,G or H.

Is it feasible to fit a logistic regression over this dataset? Ideally, if I can fit a logistic regression the data, I will then use it for prediction over a set of test data, which again contains only categorical data.

What are the pitfalls that I should look out for?

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    $\begingroup$ Yes, you should be able to. I would watch out for how you're grouping/binning the levels of each predictor to improve credibility and homogeneity. I would also look out for missing data. Lastly, because you're fitting to a logistic regression, you will need to have three separate datasets - one for model fitting, the second to select the logistic probability/value for which you have 0 vs 1 separation, and the third for model validation. $\endgroup$ – Frank H. Nov 3 '15 at 16:54
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    $\begingroup$ Yes, you can train a logistic regression model on categorical data. Each feature will be basically on/off which actually simplifies the things. It depends though on implementation how it handles such features. $\endgroup$ – Vladislavs Dovgalecs Nov 3 '15 at 16:54
  • $\begingroup$ Hi @Frank.H, you mentioned that I will need 3 mutually exclusive datasets. I understand the first and third are for training and testing the model. What is the second one for? I thought if the probability computed from the logistic regression is greater than 0.5, then the response variable should be 1. If p <0.5, then response variable should be be 0. And regarding binning levels of each predictor, since all my predictors have values like A,B,C and number of levels for each predictor is different, can I just useas.factor for all the predictor variables ? $\endgroup$ – mynameisJEFF Nov 3 '15 at 17:17
  • $\begingroup$ @mynameisJEFF See: ats.ucla.edu/stat/r/dae/logit.htm $\endgroup$ – rightskewed Nov 3 '15 at 17:28
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Yes of course you can. Just be aware of the nature of your categorical data - is it ordered or unordered?

If ordered (e.g. small, medium, large) you might want a single feature X1 with values like (1, 1, 3, 2, 3, 1, ...) where 1 represents small, 2 represents medium, etc.

If unordered (e.g. red, blue, green) you'll want multiple features like X1 = (0, 0, 1, 0) representing "is red?", X2 = (1, 0, 0, 1) representing "is blue?" and so forth.

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  • $\begingroup$ Hi, I am not understanding the part where the data is unordered. Can you elaborate on how you create multiple features? $\endgroup$ – mynameisJEFF Nov 3 '15 at 17:06
  • $\begingroup$ There's a lot of discussion about this if you search for it. But suppose your training data contains a feature like (red, blue, blue, green, red). Then you'll want to train your logistic regression model using three features. X1 = IsRed? = (1, 0, 0, 0, 1), X2 = IsBlue? = (0, 1, 1, 0, 0), X3 = IsGreen = (0, 0, 0, 1, 0) where 1s represent "yes" or "true" and 0s represent "no" or "false". In other words, you create a binary vector for each unique class (i.e. category). $\endgroup$ – Ben Nov 3 '15 at 17:20
  • $\begingroup$ Upon reading this again, it looks like you're searching for implementation details (i.e. R code) yes? If so, you should include the R tag or possibly consider posting in StackOverflow. $\endgroup$ – Ben Nov 3 '15 at 17:23
  • $\begingroup$ No, I am not looking for implementation. I just didn't get how you split unordered values into multiple features. That's why I include more details in the quesiton. $\endgroup$ – mynameisJEFF Nov 4 '15 at 0:31
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Yes, this is doable.

The (potentially) unseen pitfall is that your model may require a great deal more data than you expect. A general rule of thumb for logistic regression is that you need at least $15$ observations in the less commonly occurring category (i.e., either $0$s or $1$s) for each variable in the model (cf., here). You may think that you have just $2$ variables (viz., X_1 and X_2), and thus, you will be OK as long as you have at least $30$ 'successes' and $30$ 'failures'. However, there is a subtle inconsistency between how we interpret your variables and how a statistical model will use them. You will quite naturally think of X_1 as a single variable, but the model will treat it as $3$. Likewise, the model will treat X_2 as $7$ (!) additional variables, not one. More specifically, you are using the number of levels minus one ($4-1=3$ and $8-1=7$) in your model for every categorical variable you add. The upshot of this is that you want to have at least $150$ 'successes' and $150$ 'failures' ($N>300$) in your dataset to fit a model with just your X_1 and X_2 variables.

A related issue is that you want to be sure there are sufficient data in each of those levels. Obviously, if no one chose X_2 = G, you won't be able to estimate anything about the effect of that level of X_2, but you will also have a problem if some did choose G, but everyone who did has Y = 1. That would lead to the problem of separation. Moreover, if you want to fit the interaction, you will need sufficient data in every combination of levels ($32$, in your case). To read more about these topics, you may want to peruse some of our threads categorized under and .

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Of course it is possible.

You just need to transform your categorical variables into binary variables and to remove each time one item. For instance, if the variable X takes two values A and B, you need to create the variable which is equal to 1 if X == A and to 0 otherwise. Since X == A implies X != B, you'll have a collinearity in your model if you add the variable which is equal to 1 if X == B and to 0 otherwise.

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    $\begingroup$ Hi, if my categorical variable X can take values A,B,C, do I still need to make them binary ? Can't I just use as.factor(X) ? $\endgroup$ – mynameisJEFF Nov 3 '15 at 17:04
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    $\begingroup$ You do not need to make your categorical variables to only have two levels. They can have as many as you can justify statistically and meaningfully. $\endgroup$ – Frank H. Nov 3 '15 at 17:07
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    $\begingroup$ as.factor(X) is just an R function which makes it easy to transform categorical variables into a set of binary variables. $\endgroup$ – PAC Nov 4 '15 at 9:18

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