Reputation: 1042
`glmRob()` does not predict when given `newdata` argument
The following is some slightly modified code from the glmRob() examples. When given the newdata argument, predict.glmRob() errors out. Am I doing something wrong?
suppressMessages(library(robust))
data(breslow.dat)
bres.rob <- glmRob(sumY ~ Age10 + Base4 * Trt, family = poisson(), data = breslow.dat)
predict(bres.rob, newdata = breslow.dat)
Error in NextMethod("predict"): no method to invoke
devtools::session_info()
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Created on 2020-12-14 by the reprex package (v0.3.0)
Worth noting that the exact same thing does work with regular glm().
suppressMessages(library(robust))
data(breslow.dat)
bres.glm <- glm(sumY ~ Age10 + Base4 * Trt, family = poisson(), data = breslow.dat)
predict(bres.glm, newdata = breslow.dat)
#> 1 2 3 4 5 6 7 8
#> 2.957756 2.933407 2.704879 3.015431 3.913226 3.250763 2.979112 4.101214
#> 9 10 11 12 13 14 15 16
#> 3.360129 2.863352 3.955120 3.257157 2.912460 3.741554 4.459110 3.668917
#> 17 18 19 20 21 22 23 24
#> 3.034205 5.093412 3.131601 2.906475 2.930414 2.671553 3.110244 3.174723
#> 25 26 27 28 29 30 31 32
#> 3.873096 3.134184 2.644211 3.507451 3.917288 3.375050 2.641300 2.675629
#> 33 34 35 36 37 38 39 40
#> 2.592602 2.854897 3.163672 2.890335 2.625822 3.756825 3.201280 2.557211
#> 41 42 43 44 45 46 47 48
#> 2.784067 2.988840 3.585320 3.060731 3.448097 2.484164 3.182476 2.577124
#> 49 50 51 52 53 54 55 56
#> 5.757692 3.003209 3.274328 3.308657 3.525533 3.268830 2.863768 2.857114
#> 57 58 59
#> 2.805090 2.891444 2.892553
Created on 2020-12-14 by the reprex package (v0.3.0)
Upvotes: 3
Views: 235
Answers (1)
Reputation: 317
Just take out the newdata argument since you're running it on the same data.
From the robust docs here
newdata - optionally, a data frame in which to look for variables with which to predict. If omitted, the fitted linear predictors are used.
Edited to add
One of the listed authors replied and recommended using the package "robustbase" (whom he's the maintainer of) as opposed to "robust" because the methods used in the former are more modern, as well as backed by more extensive testing and examples.
Here's my sample code for a quick comparison between the two. Note that the 'r' in robustbase's glmrob is lowercase.
library(robustbase)
library(robust)
data(breslow.dat)
## Comparison of methods.
bres.robustbase <- glmrob(sumY ~ Age10 + Base4 * Trt, family = poisson, data = breslow.dat)
predict(bres.robustbase, newdata = breslow.dat)
bres.robust <- glmRob(sumY ~ Age10 + Base4 * Trt, family = poisson(), data = breslow.dat)
predict(bres.robust)
## predict handling test data on robustbase's glmrob object.
lastIndex = round(nrow(breslow.dat)*0.7)
train = breslow.dat[1:lastIndex,]
test = breslow.dat[(lastIndex + 1):nrow(breslow.dat),]
bres.robustbase <- glmrob(sumY ~ Age10 + Base4 * Trt, family = poisson, data = train)
predict(bres.robustbase, newdata = test)
Notice that you'll get different coefficient estimates, and therefore predictions, between glmrob and glmRob. Deciding which one's more accurate is beyond my knowledge at the moment (hopefully not for you), but I've followed up with our author to see if he can give a high-level explanation that can be posted here.
Upvotes: 2
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