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loo_compare() returns elpd_diff with the opposite sign for psis_loo_ss objects #390

Description

@florence-bockting

What happened

loo_compare() returns elpd_diff with the opposite sign for subsampled
psis_loo_ss objects. It also returns a matrix, not a data.frame.

What I expected

In loo_compare() the difference is comparison-model minus reference-model,
and the output is a data.frame.

Minimal reproducible example

library(loo)

set.seed(4711)
N <- 500
S <- 500
data <- data.frame(y = rnorm(N, mean = 0, sd = 1))

# two models: model1 is well centred, model2 is biased by 0.3
draws1 <- matrix(rnorm(S, 0.0, 0.05), ncol = 1, dimnames = list(NULL, "mu"))
draws2 <- matrix(rnorm(S, 0.3, 0.05), ncol = 1, dimnames = list(NULL, "mu"))

llfun <- function(data_i, draws, ...) {
  dnorm(data_i$y, mean = draws[, "mu"], sd = 1, log = TRUE)
}

l1 <- loo(llfun, data = data, draws = draws1, r_eff = rep(1, N))
l2 <- loo(llfun, data = data, draws = draws2, r_eff = rep(1, N))
loo_compare(list(model1 = l1, model2 = l2))

set.seed(4711)
s1 <- loo_subsample(llfun, data = data, draws = draws1,
                    observations = 200, loo_approximation = "plpd",
                    r_eff = rep(1, N))
set.seed(4711)
s2 <- loo_subsample(llfun, data = data, draws = draws2,
                    observations = s1, loo_approximation = "plpd",
                    r_eff = rep(1, N))
loo_compare(list(model1 = s1, model2 = s2))

Actual output

> loo_compare(list(model1 = l1, model2 = l2))
  model elpd_diff se_diff p_worse diag_diff diag_elpd
 model1       0.0     0.0      NA
 model2     -25.8     6.6    1.00

> loo_compare(list(model1 = s1, model2 = s2))
       elpd_diff se_diff subsampling_se_diff
model1  0.0       0.0     0.0
model2 25.7       6.6     0.0

The return types also differ:

full: class compare.loo/data.frame                  | rownames 1,2
ss  : class compare.loo_ss/compare.loo/matrix/array | rownames model1,model2

Environment

  • OS: Ubuntu 24.04.4 LTS
  • Package version: loo 2.10.0
  • R version: 4.6.1 (2026-06-24)

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