1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615
|
---
title: "Using Leave-one-out cross-validation for large data"
author: "Mans Magnusson, Paul Bürkner, Aki Vehtari and Jonah Gabry"
date: "`r Sys.Date()`"
output:
html_vignette:
toc: yes
params:
EVAL: !r identical(Sys.getenv("NOT_CRAN"), "true")
---
<!--
%\VignetteEngine{knitr::rmarkdown}
%\VignetteIndexEntry{Using Leave-one-out cross-validation for large data}
-->
```{r settings, child="children/SETTINGS-knitr.txt"}
```
```{r, child="children/SEE-ONLINE.txt", eval = if (isTRUE(exists("params"))) !params$EVAL else TRUE}
```
# Introduction
This vignette demonstrates how to do leave-one-out cross-validation for large
data using the __loo__ package and Stan. There are two approaches covered:
LOO with subsampling and LOO using approximations to posterior distributions.
Some sections from this vignette are excerpted from the papers
* Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2020). Leave-One-Out Cross-Validation for Model Comparison in Large Data.
Proceedings of the 23rd International Conference on Artificial Intelligence and Statistics (AISTATS), in PMLR 108. [arXiv preprint arXiv:2001.00980](https://arxiv.org/abs/2001.00980).
* Magnusson, M., Andersen, M., Jonasson, J. & Vehtari, A. (2019). Bayesian leave-one-out cross-validation for large data. Proceedings of the 36th International Conference on Machine Learning, in PMLR 97:4244-4253
[online](http://proceedings.mlr.press/v97/magnusson19a.html),
[arXiv preprint arXiv:1904.10679](https://arxiv.org/abs/1904.10679).
* 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. Links: [published](https://link.springer.com/article/10.1007/s11222-016-9696-4) | [arXiv preprint](https://arxiv.org/abs/1507.04544).
* Vehtari, A., Simpson, D., Gelman, A., Yao, Y., and Gabry, J. (2024).
Pareto smoothed importance sampling. *Journal of Machine Learning Research*,
25(72):1-58. [PDF](https://jmlr.org/papers/v25/19-556.html)
which provide important background for understanding the methods implemented in
the package.
# Setup
In addition to the __loo__ package, we'll also be using __rstan__:
```{r setup, message=FALSE}
library("rstan")
library("loo")
set.seed(4711)
```
# Example: Well water in Bangladesh
We will use the same example as in the vignette [_Writing Stan programs for use
with the loo package_](http://mc-stan.org/loo/articles/loo2-with-rstan.html).
See that vignette for a description of the problem and data.
The sample size in this example is only $N=3020$, which is not large enough to
_require_ the special methods for large data described in this vignette, but
is sufficient for demonstration purposes in this tutorial.
## Coding the Stan model
Here is the Stan code for fitting the logistic regression model, which
we save in a file called `logistic.stan`:
```
// Note: some syntax used in this program requires RStan >= 2.26 (or CmdStanR)
// To use an older version of RStan change the line declaring `y` to:
// int<lower=0,upper=1> y[N];
data {
int<lower=0> N; // number of data points
int<lower=0> P; // number of predictors (including intercept)
matrix[N,P] X; // predictors (including 1s for intercept)
array[N] int<lower=0,upper=1> y; // binary outcome
}
parameters {
vector[P] beta;
}
model {
beta ~ normal(0, 1);
y ~ bernoulli_logit(X * beta);
}
```
Importantly, unlike the general approach recommended in
[_Writing Stan programs for use with the loo package_](http://mc-stan.org/loo/articles/loo2-with-rstan.html),
we do _not_ compute the log-likelihood for each observation in the
`generated quantities` block of the Stan program. Here we are assuming we have a
large data set (larger than the one we're actually using in this demonstration)
and so it is preferable to instead define a function in R to compute the
log-likelihood for each data point when needed rather than storing all of the
log-likelihood values in memory.
The log-likelihood in R can be coded as follows:
```{r llfun_logistic}
# we'll add an argument log to toggle whether this is a log-likelihood or
# likelihood function. this will be useful later in the vignette.
llfun_logistic <- function(data_i, draws, log = TRUE) {
x_i <- as.matrix(data_i[, which(grepl(colnames(data_i), pattern = "X")), drop=FALSE])
logit_pred <- draws %*% t(x_i)
dbinom(x = data_i$y, size = 1, prob = 1/(1 + exp(-logit_pred)), log = log)
}
```
The function `llfun_logistic()` needs to have arguments `data_i` and `draws`.
Below we will test that the function is working by using the `loo_i()` function.
## Fitting the model with RStan
Next we fit the model in Stan using the **rstan** package:
```{r, eval=FALSE}
# Prepare data
url <- "http://stat.columbia.edu/~gelman/arm/examples/arsenic/wells.dat"
wells <- read.table(url)
wells$dist100 <- with(wells, dist / 100)
X <- model.matrix(~ dist100 + arsenic, wells)
standata <- list(y = wells$switch, X = X, N = nrow(X), P = ncol(X))
# Compile
stan_mod <- stan_model("logistic.stan")
# Fit model
fit_1 <- sampling(stan_mod, data = standata, seed = 4711)
print(fit_1, pars = "beta")
```
```
mean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat
beta[1] 0.00 0 0.08 -0.15 -0.05 0.00 0.06 0.16 1933 1
beta[2] -0.89 0 0.10 -1.09 -0.96 -0.89 -0.82 -0.69 2332 1
beta[3] 0.46 0 0.04 0.38 0.43 0.46 0.49 0.54 2051 1
```
Before we move on to computing LOO we can now test that the log-likelihood
function we wrote is working as it should. The `loo_i()` function is a helper
function that can be used to test a log-likelihood function on a single observation.
```{r, eval=FALSE}
# used for draws argument to loo_i
parameter_draws_1 <- extract(fit_1)$beta
# used for data argument to loo_i
stan_df_1 <- as.data.frame(standata)
# compute relative efficiency (this is slow and optional but is recommended to allow
# for adjusting PSIS effective sample size based on MCMC effective sample size)
r_eff <- relative_eff(llfun_logistic,
log = FALSE, # relative_eff wants likelihood not log-likelihood values
chain_id = rep(1:4, each = 1000),
data = stan_df_1,
draws = parameter_draws_1,
cores = 2)
loo_i(i = 1, llfun_logistic, r_eff = r_eff, data = stan_df_1, draws = parameter_draws_1)
```
```
$pointwise
elpd_loo mcse_elpd_loo p_loo looic influence_pareto_k
1 -0.3314552 0.0002887608 0.0003361772 0.6629103 -0.05679886
...
```
# Approximate LOO-CV using PSIS-LOO and subsampling
We can then use the `loo_subsample()` function to compute the efficient PSIS-LOO
approximation to exact LOO-CV using subsampling:
```{r, eval=FALSE}
set.seed(4711)
loo_ss_1 <-
loo_subsample(
llfun_logistic,
observations = 100, # take a subsample of size 100
cores = 2,
# these next objects were computed above
r_eff = r_eff,
draws = parameter_draws_1,
data = stan_df_1
)
print(loo_ss_1)
```
```
Computed from 4000 by 100 subsampled log-likelihood
values from 3020 total observations.
Estimate SE subsampling SE
elpd_loo -1968.5 15.6 0.3
p_loo 3.1 0.1 0.4
looic 3936.9 31.2 0.6
------
Monte Carlo SE of elpd_loo is 0.0.
MCSE and ESS estimates assume MCMC draws (r_eff in [0.9, 1.0]).
All Pareto k estimates are good (k < 0.7).
See help('pareto-k-diagnostic') for details.
```
The `loo_subsample()` function creates an object of class `psis_loo_ss`, that
inherits from `psis_loo, loo` (the classes of regular `loo` objects).
The printed output above shows the estimates
$\widehat{\mbox{elpd}}_{\rm loo}$ (expected log predictive density),
$\widehat{p}_{\rm loo}$ (effective number of parameters), and ${\rm looic} =-2\,
\widehat{\mbox{elpd}}_{\rm loo}$ (the LOO information criterion). Unlike when
using `loo()`, when using `loo_subsample()` there is an additional column
giving the "subsampling SE", which reflects the additional uncertainty due
to the subsampling used.
The line at the bottom of the printed output provides information about the
reliability of the LOO approximation (the interpretation of the $k$ parameter is
explained in `help('pareto-k-diagnostic')` and in greater detail in Vehtari,
Simpson, Gelman, Yao, and Gabry (2019)). In this case, the message tells us that all of the
estimates for $k$ are fine _for this given subsample_.
## Adding additional subsamples
If we are not satisfied with the subsample size (i.e., the accuracy) we can
simply add more samples until we are satisfied using the `update()` method.
```{r, eval=FALSE}
set.seed(4711)
loo_ss_1b <-
update(
loo_ss_1,
observations = 200, # subsample 200 instead of 100
r_eff = r_eff,
draws = parameter_draws_1,
data = stan_df_1
)
print(loo_ss_1b)
```
```
Computed from 4000 by 200 subsampled log-likelihood
values from 3020 total observations.
Estimate SE subsampling SE
elpd_loo -1968.3 15.6 0.2
p_loo 3.2 0.1 0.4
looic 3936.7 31.2 0.5
------
Monte Carlo SE of elpd_loo is 0.0.
MCSE and ESS estimates assume MCMC draws (r_eff in [0.9, 1.0]).
All Pareto k estimates are good (k < 0.7).
See help('pareto-k-diagnostic') for details.
```
## Specifying estimator and sampling method
The performance relies on two components: the estimation method and the
approximation used for the elpd. See the documentation for `loo_subsample()`
more information on which estimators and approximations are implemented. The
default implementation is using the point log predictive density evaluated at
the mean of the posterior (`loo_approximation="plpd"`) and the difference
estimator (`estimator="diff_srs"`). This combination has a focus on fast
inference. But we can easily use other estimators as well as other elpd
approximations, for example:
```{r, eval=FALSE}
set.seed(4711)
loo_ss_1c <-
loo_subsample(
x = llfun_logistic,
r_eff = r_eff,
draws = parameter_draws_1,
data = stan_df_1,
observations = 100,
estimator = "hh_pps", # use Hansen-Hurwitz
loo_approximation = "lpd", # use lpd instead of plpd
loo_approximation_draws = 100,
cores = 2
)
print(loo_ss_1c)
```
```
Computed from 4000 by 100 subsampled log-likelihood
values from 3020 total observations.
Estimate SE subsampling SE
elpd_loo -1968.9 15.4 0.5
p_loo 3.5 0.2 0.5
looic 3937.9 30.7 1.1
------
Monte Carlo SE of elpd_loo is 0.0.
MCSE and ESS estimates assume MCMC draws (r_eff in [0.9, 1.0]).
All Pareto k estimates are good (k < 0.7).
See help('pareto-k-diagnostic') for details.
```
See the documentation and references for `loo_subsample()` for details on
the implemented approximations.
# Approximate LOO-CV using PSIS-LOO with posterior approximations
Using posterior approximations, such as variational inference and Laplace
approximations, can further speed-up LOO-CV for large data. Here we demonstrate
using a Laplace approximation in Stan.
```{r, eval=FALSE}
fit_laplace <- optimizing(stan_mod, data = standata, draws = 2000,
importance_resampling = TRUE)
parameter_draws_laplace <- fit_laplace$theta_tilde # draws from approximate posterior
log_p <- fit_laplace$log_p # log density of the posterior
log_g <- fit_laplace$log_g # log density of the approximation
```
Using the posterior approximation we can then do LOO-CV by correcting for the
posterior approximation when we compute the elpd. To do this we use the
`loo_approximate_posterior()` function.
```{r, eval=FALSE}
set.seed(4711)
loo_ap_1 <-
loo_approximate_posterior(
x = llfun_logistic,
draws = parameter_draws_laplace,
data = stan_df_1,
log_p = log_p,
log_g = log_g,
cores = 2
)
print(loo_ap_1)
```
The function creates a class, `psis_loo_ap` that inherits from `psis_loo, loo`.
```
Computed from 2000 by 3020 log-likelihood matrix
Estimate SE
elpd_loo -1968.4 15.6
p_loo 3.2 0.2
looic 3936.8 31.2
------
Posterior approximation correction used.
Monte Carlo SE of elpd_loo is 0.0.
MCSE and ESS estimates assume independent draws (r_eff=1).
All Pareto k estimates are good (k < 0.7).
See help('pareto-k-diagnostic') for details.
```
## Combining the posterior approximation method with subsampling
The posterior approximation correction can also be used together with
subsampling:
```{r, eval=FALSE}
set.seed(4711)
loo_ap_ss_1 <-
loo_subsample(
x = llfun_logistic,
draws = parameter_draws_laplace,
data = stan_df_1,
log_p = log_p,
log_g = log_g,
observations = 100,
cores = 2
)
print(loo_ap_ss_1)
```
```
Computed from 2000 by 100 subsampled log-likelihood
values from 3020 total observations.
Estimate SE subsampling SE
elpd_loo -1968.2 15.6 0.4
p_loo 2.9 0.1 0.5
looic 3936.4 31.1 0.8
------
Posterior approximation correction used.
Monte Carlo SE of elpd_loo is 0.0.
MCSE and ESS estimates assume independent draws (r_eff=1).
All Pareto k estimates are good (k < 0.7).
See help('pareto-k-diagnostic') for details.
```
The object created is of class `psis_loo_ss`, which inherits from the
`psis_loo_ap` class previously described.
## Comparing models
To compare this model to an alternative model for the same data we can use the
`loo_compare()` function just as we would if using `loo()` instead of
`loo_subsample()` or `loo_approximate_posterior()`. First we'll fit a second
model to the well-switching data, using `log(arsenic)` instead of `arsenic` as a
predictor:
```{r, eval=FALSE}
standata$X[, "arsenic"] <- log(standata$X[, "arsenic"])
fit_2 <- sampling(stan_mod, data = standata)
parameter_draws_2 <- extract(fit_2)$beta
stan_df_2 <- as.data.frame(standata)
# recompute subsampling loo for first model for demonstration purposes
# compute relative efficiency (this is slow and optional but is recommended to allow
# for adjusting PSIS effective sample size based on MCMC effective sample size)
r_eff_1 <- relative_eff(
llfun_logistic,
log = FALSE, # relative_eff wants likelihood not log-likelihood values
chain_id = rep(1:4, each = 1000),
data = stan_df_1,
draws = parameter_draws_1,
cores = 2
)
set.seed(4711)
loo_ss_1 <- loo_subsample(
x = llfun_logistic,
r_eff = r_eff_1,
draws = parameter_draws_1,
data = stan_df_1,
observations = 200,
cores = 2
)
# compute subsampling loo for a second model (with log-arsenic)
r_eff_2 <- relative_eff(
llfun_logistic,
log = FALSE, # relative_eff wants likelihood not log-likelihood values
chain_id = rep(1:4, each = 1000),
data = stan_df_2,
draws = parameter_draws_2,
cores = 2
)
loo_ss_2 <- loo_subsample(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
observations = 200,
cores = 2
)
print(loo_ss_2)
```
```
Computed from 4000 by 100 subsampled log-likelihood
values from 3020 total observations.
Estimate SE subsampling SE
elpd_loo -1952.0 16.2 0.2
p_loo 2.6 0.1 0.3
looic 3903.9 32.4 0.4
------
Monte Carlo SE of elpd_loo is 0.0.
MCSE and ESS estimates assume MCMC draws (r_eff in [1.0, 1.1]).
All Pareto k estimates are good (k < 0.7).
See help('pareto-k-diagnostic') for details.
```
We can now compare the models on LOO using the `loo_compare` function:
```{r, eval=FALSE}
# Compare
comp <- loo_compare(loo_ss_1, loo_ss_2)
print(comp)
```
```
Warning: Different subsamples in 'model2' and 'model1'. Naive diff SE is used.
elpd_diff se_diff subsampling_se_diff
model2 0.0 0.0 0.0
model1 16.5 22.5 0.4
```
This new object `comp` contains the estimated difference of expected
leave-one-out prediction errors between the two models, along with the standard
error. As the warning indicates, because different subsamples were used the
comparison will not take the correlations between different observations into
account. Here we see that the naive SE is 22.5 and we cannot see any difference
in performance between the models.
To force subsampling to use the same observations for each of the models
we can simply extract the observations used in `loo_ss_1` and use them in
`loo_ss_2` by supplying the `loo_ss_1` object to the `observations` argument.
```{r, eval=FALSE}
loo_ss_2 <-
loo_subsample(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
observations = loo_ss_1,
cores = 2
)
```
We could also supply the subsampling indices using the `obs_idx()`
helper function:
```{r, eval=FALSE}
idx <- obs_idx(loo_ss_1)
loo_ss_2 <- loo_subsample(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
observations = idx,
cores = 2
)
```
```
Simple random sampling with replacement assumed.
```
This results in a message indicating that we assume these observations to have
been sampled with simple random sampling, which is true because we had used the
default `"diff_srs"` estimator for `loo_ss_1`.
We can now compare the models and estimate the difference based on the same
subsampled observations.
```{r, eval=FALSE}
comp <- loo_compare(loo_ss_1, loo_ss_2)
print(comp)
```
```
elpd_diff se_diff subsampling_se_diff
model2 0.0 0.0 0.0
model1 16.1 4.4 0.1
```
First, notice that now the `se_diff` is now around 4 (as opposed to 20 when using
different subsamples). The first column shows the difference in ELPD relative to
the model with the largest ELPD. In this case, the difference in `elpd` and its
scale relative to the approximate standard error of the difference) indicates a
preference for the second model (`model2`). Since the subsampling uncertainty is
so small in this case it can effectively be ignored. If we need larger
subsamples we can simply add samples using the `update()` method demonstrated
earlier.
It is also possible to compare a subsampled loo computation with a full loo object.
```{r, eval=FALSE}
# use loo() instead of loo_subsample() to compute full PSIS-LOO for model 2
loo_full_2 <- loo(
x = llfun_logistic,
r_eff = r_eff_2,
draws = parameter_draws_2,
data = stan_df_2,
cores = 2
)
loo_compare(loo_ss_1, loo_full_2)
```
```
Estimated elpd_diff using observations included in loo calculations for all models.
```
Because we are comparing a non-subsampled loo calculation to a subsampled
calculation we get the message that only the observations that are included in
the loo calculations for both `model1` and `model2` are included in the
computations for the comparison.
```
elpd_diff se_diff subsampling_se_diff
model2 0.0 0.0 0.0
model1 16.3 4.4 0.3
```
Here we actually see an increase in `subsampling_se_diff`, but this is due to a
technical detail not elaborated here. In general, the difference should be
better or negligible.
# References
Gelman, A., and Hill, J. (2007). *Data Analysis Using Regression and Multilevel Hierarchical Models.* Cambridge University Press.
Stan Development Team (2017). _The Stan C++ Library, Version 2.17.0._ https://mc-stan.org/
Stan Development Team (2018) _RStan: the R interface to Stan, Version 2.17.3._ https://mc-stan.org/
Magnusson, M., Riis Andersen, M., Jonasson, J. and Vehtari, A. (2020). Leave-One-Out Cross-Validation for Model Comparison in Large Data.
Proceedings of the 23rd International Conference on Artificial Intelligence and Statistics (AISTATS), in PMLR 108. [arXiv preprint arXiv:2001.00980](https://arxiv.org/abs/2001.00980).
Magnusson, M., Andersen, M., Jonasson, J. & Vehtari, A. (2019). Bayesian leave-one-out cross-validation for large data. Proceedings of the 36th International Conference on Machine Learning, in PMLR 97:4244-4253
[online](http://proceedings.mlr.press/v97/magnusson19a.html),
[arXiv preprint arXiv:1904.10679](https://arxiv.org/abs/1904.10679).
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.
[online](https://link.springer.com/article/10.1007/s11222-016-9696-4),
[arXiv preprint arXiv:1507.04544](https://arxiv.org/abs/1507.04544).
Vehtari, A., Simpson, D., Gelman, A., Yao, Y., and Gabry, J. (2024).
Pareto smoothed importance sampling. *Journal of Machine Learning Research*,
25(72):1-58. [PDF](https://jmlr.org/papers/v25/19-556.html)
|