Automatically report the results of Bayesian model comparison using the loo package.
Usage
# S3 method for class 'compare.loo'
report(x, include_IC = TRUE, include_ENP = FALSE, ...)Arguments
- x
An object of class brms::loo_compare.
- include_IC
Whether to include the information criteria (IC).
- include_ENP
Whether to include the effective number of parameters (ENP).
- ...
Additional arguments (not used for now).
Value
Objects of class report_text().
Details
The rule of thumb is that the models are "very similar" if |elpd_diff| (the
absolute value of elpd_diff) is less than 4 (Sivula, Magnusson and Vehtari, 2020).
If superior to 4, then one can use the SE to obtain a standardized difference
(Z-diff) and interpret it as such, assuming that the difference is normally
distributed. The corresponding p-value is then calculated as 2 * pnorm(-abs(Z-diff)).
However, note that if the raw ELPD difference is small (less than 4), it doesn't
make much sense to rely on its standardized value: it is not very useful to
conclude that a model is much better than another if both models make very
similar predictions.
Examples
# \donttest{
library(brms)
m1 <- brms::brm(mpg ~ qsec, data = mtcars)
#> Compiling Stan program...
#> Start sampling
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1:
#> Chain 1: Gradient evaluation took 8e-06 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.08 seconds.
#> Chain 1: Adjust your expectations accordingly!
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#> Chain 1: 0.029 seconds (Total)
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#>
#> 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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#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
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#> Chain 3: Gradient evaluation took 4e-06 seconds
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#> Chain 3: Elapsed Time: 0.017 seconds (Warm-up)
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#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 4).
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#> Chain 4: Gradient evaluation took 3e-06 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.03 seconds.
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m2 <- brms::brm(mpg ~ qsec + drat, data = mtcars)
#> Compiling Stan program...
#> Start sampling
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1:
#> Chain 1: Gradient evaluation took 8e-06 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.08 seconds.
#> Chain 1: Adjust your expectations accordingly!
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#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 2).
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#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
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m3 <- brms::brm(mpg ~ qsec + drat + wt, data = mtcars)
#> Compiling Stan program...
#> Start sampling
#>
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1:
#> Chain 1: Gradient evaluation took 8e-06 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.08 seconds.
#> Chain 1: Adjust your expectations accordingly!
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x <- suppressWarnings(brms::loo_compare(
brms::add_criterion(m1, "loo"),
brms::add_criterion(m2, "loo"),
brms::add_criterion(m3, "loo"),
model_names = c("m1", "m2", "m3")
))
report(x)
#> The difference in predictive accuracy, as indexed by Expected Log Predictive
#> Density (ELPD-LOO), suggests that 'm3' is the best model (ELPD = -79.02),
#> followed by 'm2' (diff-ELPD = -13.52 +- 4.40, p = 0.002) and 'm1' (diff-ELPD =
#> -23.28 +- 4.40, p < .001)
report(x, include_IC = FALSE)
#> The difference in predictive accuracy, as indexed by Expected Log Predictive
#> Density (ELPD-LOO), suggests that 'm3' is the best model (ELPD = -79.02),
#> followed by 'm2' (diff-ELPD = -13.52 +- 4.40, p = 0.002) and 'm1' (diff-ELPD =
#> -23.28 +- 4.40, p < .001)
report(x, include_ENP = TRUE)
#> The difference in predictive accuracy, as indexed by Expected Log Predictive
#> Density (ELPD-LOO), suggests that 'm3' is the best model (LOOIC = 158.04, ENP =
#> 4.94), followed by 'm2' (diff-ELPD = -13.52 +- 4.40, p = 0.002, LOOIC = 185.07,
#> ENP = 4.00) and 'm1' (diff-ELPD = -23.28 +- 4.40, p < .001, LOOIC = 204.59, ENP
#> = 2.64)
# }