loo (version 2.5.1)

loo_compare: Model comparison

Description

Compare fitted models based on ELPD.

By default the print method shows only the most important information. Use print(..., simplify=FALSE) to print a more detailed summary.

Usage

loo_compare(x, ...)

# S3 method for default loo_compare(x, ...)

# S3 method for compare.loo print(x, ..., digits = 1, simplify = TRUE)

# S3 method for compare.loo_ss print(x, ..., digits = 1, simplify = TRUE)

Arguments

x

An object of class "loo" or a list of such objects. If a list is used then the list names will be used as the model names in the output. See Examples.

...

Additional objects of class "loo", if not passed in as a single list.

digits

For the print method only, the number of digits to use when printing.

simplify

For the print method only, should only the essential columns of the summary matrix be printed? The entire matrix is always returned, but by default only the most important columns are printed.

Value

A matrix with class "compare.loo" that has its own print method. See the Details section.

Details

When comparing two fitted models, we can estimate the difference in their expected predictive accuracy by the difference in elpd_loo or elpd_waic (or multiplied by \(-2\), if desired, to be on the deviance scale).

When using loo_compare(), the returned matrix will have one row per model and several columns of estimates. The values in the elpd_diff and se_diff columns of the returned matrix are computed by making pairwise comparisons between each model and the model with the largest ELPD (the model in the first row). For this reason the elpd_diff column will always have the value 0 in the first row (i.e., the difference between the preferred model and itself) and negative values in subsequent rows for the remaining models.

To compute the standard error of the difference in ELPD --- which should not be expected to equal the difference of the standard errors --- we use a paired estimate to take advantage of the fact that the same set of \(N\) data points was used to fit both models. These calculations should be most useful when \(N\) is large, because then non-normality of the distribution is not such an issue when estimating the uncertainty in these sums. These standard errors, for all their flaws, should give a better sense of uncertainty than what is obtained using the current standard approach of comparing differences of deviances to a Chi-squared distribution, a practice derived for Gaussian linear models or asymptotically, and which only applies to nested models in any case.

References

Vehtari, A., Gelman, A., and Gabry, J. (2017). Practical Bayesian model evaluation using leave-one-out cross-validation and WAIC. Statistics and Computing. 27(5), 1413--1432. doi:10.1007/s11222-016-9696-4 (journal version, preprint arXiv:1507.04544).

Vehtari, A., Simpson, D., Gelman, A., Yao, Y., and Gabry, J. (2019). Pareto smoothed importance sampling. preprint arXiv:1507.02646

See Also

  • The FAQ page on the loo website for answers to frequently asked questions.

Examples

Run this code
# NOT RUN {
# very artificial example, just for demonstration!
LL <- example_loglik_array()
loo1 <- loo(LL, r_eff = NA)     # should be worst model when compared
loo2 <- loo(LL + 1, r_eff = NA) # should be second best model when compared
loo3 <- loo(LL + 2, r_eff = NA) # should be best model when compared

comp <- loo_compare(loo1, loo2, loo3)
print(comp, digits = 2)

# show more details with simplify=FALSE
# (will be the same for all models in this artificial example)
print(comp, simplify = FALSE, digits = 3)

# can use a list of objects with custom names
# will use apple, banana, and cherry, as the names in the output
loo_compare(list("apple" = loo1, "banana" = loo2, "cherry" = loo3))

# }
# NOT RUN {
# works for waic (and kfold) too
loo_compare(waic(LL), waic(LL - 10))
# }
# NOT RUN {
# }

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