Below, I've used library(mice) to multiply impute 5 datasets from my data.frame popmis. Then, I performed my desired analysis with() all those 5 imputed datasets and finally pool() across those analyses.
Question: Is it possible to replicate the same steps using library(Hmisc)?
library(mice)
library(lme4)
library(broom.mixed)
imp <- mice(popmis, m = 5) # `popmis` is a dataset from `mice`
fit <- with(data = imp, exp = lme4::lmer(popular ~ sex + (1|school)))
pool(fit)
You can use Hmisc::aregImpute:
library(Hmisc)
library(mice)
library(lme4)
library(broom.mixed)
imps <- aregImpute(~pupil + school + popular + sex + texp + teachpop,
data = popmis)$imputed$popular
#> Iteration 1 Iteration 2 Iteration 3 Iteration 4 Iteration 5
fit2 <- lapply(1:5, function(x) lme4::lmer(popular ~ sex + (1|school),
data = within(popmis, popular[is.na(popular)] <- imps[,x])))
pool(fit2)
#> Class: mipo m = 5
#> term m estimate ubar b t dfcom df
#> 1 (Intercept) 5 4.9029688 0.007668137 0.0006677781 0.008469470 1996 358.18134
#> 2 sex 5 0.8569774 0.001215393 0.0002971695 0.001571996 1996 73.99956
#> riv lambda fmi
#> 1 0.1045017 0.09461438 0.09962785
#> 2 0.2934059 0.22684749 0.24692949
Which gives similar results to your code using mice:
imp <- mice(popmis, m = 5) # `popmis` is a dataset from `mice`
fit <- with(data = imp, exp = lme4::lmer(popular ~ sex + (1|school)))
pool(fit)
#> Class: mipo m = 5
#> term m estimate ubar b t dfcom df
#> 1 (Intercept) 5 4.8984082 0.007438482 0.0004775986 0.008011601 1996 549.60298
#> 2 sex 5 0.8543277 0.001177158 0.0009930326 0.002368797 1996 15.55799
#> riv lambda fmi
#> 1 0.07704775 0.07153605 0.07489638
#> 2 1.01230146 0.50305656 0.55661230
Created on 2020-11-08 by the reprex package (v0.3.0)