Carl Carl - 25 days ago 16
R Question

Using predict with svyglm

I have found some odd behavior with

predict
and the
svyglm
object from the
survey
package. If your newdata in
predict
has a factor/character with one level it spits out error:

Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) :
contrasts can be applied only to factors with 2 or more levels


This error makes sense if I was putting a one level variable as the predictor for a model, but for newdata I don't see the problem.

With regular
glm
this works fine.

MRE:

library(survey)

data(api)

dstrat<-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc)

svymodel <- svyglm(api00~sch.wide,design=dstrat)
# errors
predict(svymodel, data.frame(sch.wide=rep("No",10)))

regmodel <- glm(api00~sch.wide,data=apistrat)
# works
predict(regmodel,data.frame(sch.wide=rep("No",10)))


I find that it works if I hack the levels of the factor, but this shouldn't be necessary:

svymodel <- svyglm(api00~sch.wide,design=dstrat)

predict(svymodel, data.frame(sch.wide=factor(rep("No",10),
levels = c("No","random phrase"))))


Am I misunderstanding something or is this an issue with the
survey
package?

Answer

You aren't putting a factor in newdata; you're putting a character string in. You should put in a factor with the same set of levels as the factor used to fit the model (not some random phrase) -- that's the only way the design matrix makes sense.

predict(svymodel, data.frame(sch.wide=factor(rep("No",10),levels=c("No","Yes"))))

predict.lm recovers the factor levels from the fitted object (I don't remember this being around in 2002, but I might be wrong). You can use that approach to automate:

predict(svymodel, data.frame(sch.wide=factor(rep("No",10),levels=svymodel$xlevels$sch.wide)))

and I'll put that on the list of things to do for the package.

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