Functions required for compatibility of brms with emmeans.
Users are not required to call these functions themselves. Instead,
they will be called automatically by the emmeans function
of the emmeans package.
Usage
recover_data.brmsfit(
object,
data,
resp = NULL,
dpar = NULL,
nlpar = NULL,
re_formula = NA,
epred = FALSE,
...
)
emm_basis.brmsfit(
object,
trms,
xlev,
grid,
vcov.,
resp = NULL,
dpar = NULL,
nlpar = NULL,
re_formula = NA,
epred = FALSE,
...
)Arguments
- object
An object of class
brmsfit.- data, trms, xlev, grid, vcov.
Arguments required by emmeans.
- resp
Optional names of response variables. If specified, predictions are performed only for the specified response variables.
- dpar
Optional name of a predicted distributional parameter. If specified, expected predictions of this parameters are returned.
- nlpar
Optional name of a predicted non-linear parameter. If specified, expected predictions of this parameters are returned.
- re_formula
Optional formula containing group-level effects to be considered in the prediction. If
NULL, include all group-level effects; ifNA(default), include no group-level effects.- epred
Logical. If
TRUEcompute predictions of the posterior predictive distribution's mean (seeposterior_epred.brmsfit) while ignoring argumentsdparandnlpar. Defaults toFALSE. If you have specified a response transformation within the formula, you need to setepredtoTRUEfor emmeans to detect this transformation.- ...
Additional arguments passed to emmeans.
Details
In order to ensure compatibility of most brms models with
emmeans, predictions are not generated 'manually' via a design matrix
and coefficient vector, but rather via posterior_linpred.brmsfit.
This appears to generally work well, but note that it produces an `.@linfct`
slot that contains the computed predictions as columns instead of the
coefficients.
Examples
# \dontrun{
fit1 <- brm(time | cens(censored) ~ age * sex + disease + (1|patient),
data = kidney, family = lognormal())
#> Compiling Stan program...
#> Start sampling
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1: Rejecting initial value:
#> Chain 1: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 1: Stan can't start sampling from this initial value.
#> Chain 1: Rejecting initial value:
#> Chain 1: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 1: Stan can't start sampling from this initial value.
#> Chain 1:
#> Chain 1: Gradient evaluation took 2.4e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.24 seconds.
#> Chain 1: Adjust your expectations accordingly!
#> Chain 1:
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#> Chain 1: Elapsed Time: 1.13 seconds (Warm-up)
#> Chain 1: 0.407 seconds (Sampling)
#> Chain 1: 1.537 seconds (Total)
#> Chain 1:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 2).
#> Chain 2: Rejecting initial value:
#> Chain 2: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 2: Stan can't start sampling from this initial value.
#> Chain 2: Rejecting initial value:
#> Chain 2: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 2: Stan can't start sampling from this initial value.
#> Chain 2: Rejecting initial value:
#> Chain 2: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 2: Stan can't start sampling from this initial value.
#> Chain 2: Rejecting initial value:
#> Chain 2: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 2: Stan can't start sampling from this initial value.
#> Chain 2:
#> Chain 2: Gradient evaluation took 1.8e-05 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.18 seconds.
#> Chain 2: Adjust your expectations accordingly!
#> Chain 2:
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#> Chain 2:
#> Chain 2: Elapsed Time: 1.007 seconds (Warm-up)
#> Chain 2: 0.396 seconds (Sampling)
#> Chain 2: 1.403 seconds (Total)
#> Chain 2:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
#> Chain 3:
#> Chain 3: Gradient evaluation took 1.7e-05 seconds
#> Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.17 seconds.
#> Chain 3: Adjust your expectations accordingly!
#> Chain 3:
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#> Chain 3: 0.41 seconds (Sampling)
#> Chain 3: 1.434 seconds (Total)
#> Chain 3:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 4).
#> Chain 4: Rejecting initial value:
#> Chain 4: Log probability evaluates to log(0), i.e. negative infinity.
#> Chain 4: Stan can't start sampling from this initial value.
#> Chain 4:
#> Chain 4: Gradient evaluation took 1.7e-05 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.17 seconds.
#> Chain 4: Adjust your expectations accordingly!
#> Chain 4:
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#> Chain 4:
#> Chain 4: Elapsed Time: 1.015 seconds (Warm-up)
#> Chain 4: 0.274 seconds (Sampling)
#> Chain 4: 1.289 seconds (Total)
#> Chain 4:
summary(fit1)
#> Family: lognormal
#> Links: mu = identity
#> Formula: time | cens(censored) ~ age * sex + disease + (1 | patient)
#> Data: kidney (Number of observations: 76)
#> Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#> total post-warmup draws = 4000
#>
#> Multilevel Hyperparameters:
#> ~patient (Number of levels: 38)
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sd(Intercept) 0.41 0.25 0.02 0.94 1.01 845 1607
#>
#> Regression Coefficients:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> Intercept 2.61 0.98 0.70 4.56 1.00 2415 2877
#> age 0.02 0.02 -0.03 0.07 1.00 1885 2280
#> sexfemale 2.59 1.16 0.33 4.80 1.00 2261 2620
#> diseaseGN -0.45 0.53 -1.50 0.61 1.00 2367 2761
#> diseaseAN -0.54 0.51 -1.55 0.46 1.00 2599 2653
#> diseasePKD 0.57 0.70 -0.83 1.97 1.00 2496 2352
#> age:sexfemale -0.03 0.03 -0.08 0.03 1.00 2244 2393
#>
#> Further Distributional Parameters:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sigma 1.16 0.13 0.93 1.44 1.00 2225 2357
#>
#> Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
#> and Tail_ESS are effective sample size measures, and Rhat is the potential
#> scale reduction factor on split chains (at convergence, Rhat = 1).
# summarize via 'emmeans'
library(emmeans)
#> Welcome to emmeans.
#> Caution: You lose important information if you filter this package's results.
#> See '? untidy'
rg <- ref_grid(fit1)
em <- emmeans(rg, "disease")
summary(em, point.est = mean)
#> disease emmean lower.HPD upper.HPD
#> other 4.13 3.41 4.83
#> GN 3.68 2.99 4.40
#> AN 3.58 2.93 4.24
#> PKD 4.70 3.64 5.80
#>
#> Results are averaged over the levels of: sex
#> Point estimate displayed: mean
#> HPD interval probability: 0.95
# obtain estimates for the posterior predictive distribution's mean
epred <- emmeans(fit1, "disease", epred = TRUE)
summary(epred, point.est = mean)
#> disease emmean lower.HPD upper.HPD
#> other 172.5 63.5 322
#> GN 110.6 38.7 206
#> AN 98.7 41.3 175
#> PKD 336.2 53.9 752
#>
#> Results are averaged over the levels of: censored, sex
#> Point estimate displayed: mean
#> HPD interval probability: 0.95
# model with transformed response variable
fit2 <- brm(log(mpg) ~ factor(cyl), data = mtcars)
#> Compiling Stan program...
#> Start sampling
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1:
#> Chain 1: Gradient evaluation took 1e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.1 seconds.
#> Chain 1: Adjust your expectations accordingly!
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#> Chain 1: 0.027 seconds (Total)
#> Chain 1:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 2).
#> Chain 2:
#> Chain 2: Gradient evaluation took 3e-06 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.03 seconds.
#> Chain 2: Adjust your expectations accordingly!
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#> Chain 2:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
#> Chain 3:
#> Chain 3: Gradient evaluation took 3e-06 seconds
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#> Chain 3:
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 4).
#> Chain 4:
#> Chain 4: Gradient evaluation took 3e-06 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.03 seconds.
#> Chain 4: Adjust your expectations accordingly!
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#> Chain 4:
summary(fit2)
#> Family: gaussian
#> Links: mu = identity
#> Formula: log(mpg) ~ factor(cyl)
#> Data: mtcars (Number of observations: 32)
#> Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#> total post-warmup draws = 4000
#>
#> Regression Coefficients:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> Intercept 3.27 0.05 3.17 3.37 1.00 2785 1847
#> factorcyl6 -0.29 0.08 -0.45 -0.13 1.00 2885 2749
#> factorcyl8 -0.57 0.07 -0.70 -0.44 1.00 3295 2639
#>
#> Further Distributional Parameters:
#> Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sigma 0.17 0.02 0.13 0.22 1.00 3005 2483
#>
#> Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
#> and Tail_ESS are effective sample size measures, and Rhat is the potential
#> scale reduction factor on split chains (at convergence, Rhat = 1).
# results will be on the log scale by default
emmeans(fit2, ~ cyl)
#> cyl emmean lower.HPD upper.HPD
#> 4 3.27 3.16 3.37
#> 6 2.98 2.86 3.11
#> 8 2.70 2.61 2.79
#>
#> Point estimate displayed: median
#> HPD interval probability: 0.95
# log transform is detected and can be adjusted automatically
emmeans(fit2, ~ cyl, epred = TRUE, type = "response")
#> cyl response lower.HPD upper.HPD
#> 4 26.4 23.6 28.9
#> 6 19.7 17.4 22.2
#> 8 14.9 13.6 16.2
#>
#> Point estimate displayed: median
#> Results are back-transformed from the log scale
#> HPD interval probability: 0.95
# }