I am asking this question again because those answers don't work for my environment (see "Session Info" below), and thus I assume, don't work for the current versions of the relevant packages.
How can I take the elements of a <dist> column output by fabletools::forecast() and place each element in a new column of a fable (convertible to a data.frame)? E.g., all the mu elements in mu, sigma elements in sigma, etc.
library(magrittr)
data(aus_production, package = "tsibbledata")
aus_production %>%
fabletools::model(ets_log = fable::ETS(log(Beer) ~ error("M") + trend("Ad") + season("A")),
ets = fable::ETS(Beer ~ error("M") + trend("Ad") + season("A"))) %>%
fabletools::forecast(h = "6 months")
#> # A fable: 4 x 4 [1Q]
#> # Key: .model [2]
#> .model Quarter Beer .mean
#> <chr> <qtr> <dist> <dbl>
#> 1 ets_log 2010 Q3 t(N(6, 0.0013)) 407.
#> 2 ets_log 2010 Q4 t(N(6.2, 0.0014)) 483.
#> 3 ets 2010 Q3 N(408, 237) 408.
#> 4 ets 2010 Q4 N(483, 340) 483.
beer_fc <- aus_production %>%
fabletools::model(ets_log = fable::ETS(log(Beer) ~ error("M") + trend("Ad") + season("A")),
ets = fable::ETS(Beer ~ error("M") + trend("Ad") + season("A"))) %>%
fabletools::forecast(h = "6 months")
str(beer_fc$Beer[1])
#> dist [1:1]
#> $ :List of 3
#> ..$ dist :List of 2
#> .. ..$ mu : num 6.01
#> .. ..$ sigma: num 0.036
#> .. ..- attr(*, "class")= chr [1:2] "dist_normal" "dist_default"
#> ..$ transform:function (Beer)
#> .. ..- attr(*, "class")= chr "transformation"
#> .. ..- attr(*, "inverse")=function (Beer)
#> ..$ inverse :function (Beer)
#> ..- attr(*, "class")= chr [1:2] "dist_transformed" "dist_default"
#> @ vars: chr "Beer"
class(beer_fc$Beer)
#> [1] "distribution" "vctrs_vctr" "list"
##### Answer 1 fails (https://stackoverflow.com/a/64960991/15723919)
beer_fc$Beer2 <- purrr::map(beer_fc$Beer, ~ .x[[1]]$x)
beer_fc %>%
tidyr::unnest(c(Beer2))
#> # A tibble: 0 × 5
#> # … with 5 variables: .model <chr>, Quarter <qtr>, Beer <dist>, .mean <dbl>,
#> # Beer2 <???>
##### Answer 2 fails (https://stackoverflow.com/a/64963665/15723919)
beer_fc %>%
dplyr::mutate(value = purrr::map(Beer, purrr::pluck, 'dist', 'x')) %>%
tidyr::unnest(value)
#> # A tibble: 0 × 6
#> # … with 6 variables: .model <chr>, Quarter <qtr>, Beer <dist>, .mean <dbl>,
#> # Beer2 <list>, value <???>
##### Change 'x' to something else for Answers 1 and 2
beer_fc$Beer2 <- purrr::map(beer_fc$Beer, ~ .x[[1]][["mu"]])
#> Error in `vec_slice()`:
#> ! Can't use character names to index an unnamed vector.
#> Backtrace:
#> ▆
#> 1. ├─purrr::map(beer_fc$Beer, ~.x[[1]][["mu"]])
#> 2. │ └─global .f(.x[[i]], ...)
#> 3. │ └─.x[[1]][["mu"]]
#> 4. │ ├─base (local) `[[.distribution`(.x[[1]], "mu")
#> 5. │ └─vctrs:::`[.vctrs_vctr`(.x[[1]], "mu")
#> 6. │ └─vctrs:::vec_index(x, i, ...)
#> 7. │ └─vctrs::vec_slice(x, i)
#> 8. └─rlang::abort(message = message)
beer_fc %>%
tidyr::unnest(c(Beer2))
#> # A tibble: 0 × 5
#> # … with 5 variables: .model <chr>, Quarter <qtr>, Beer <dist>, .mean <dbl>,
#> # Beer2 <???>
beer_fc %>%
dplyr::mutate(value = purrr::map(Beer, purrr::pluck, 'dist', 'mu')) %>%
tidyr::unnest(value)
#> # A tibble: 0 × 6
#> # … with 6 variables: .model <chr>, Quarter <qtr>, Beer <dist>, .mean <dbl>,
#> # Beer2 <list>, value <???>
Created on 2022-11-14 with reprex v2.0.2
My ideal output looks like unnesting the elements of the dist column, however it's critical to extract the mu and sigma elements (or whatever the distribution parameters might be) in a rowwise fashion:
#> # A fable: 12 x 8 [1Q]
#> # Key: .model [1]
#> .model Quarter Beer .mean mu sigma transform inverse
#> <chr> <qtr> <dist> <dbl> <dbl> <dbl> <chr> <chr>
#> 1 ets_log 2010 Q3 t(N(6, 0.0013)) 407. 6.01 0.036 log exp
#> 2 ets_log 2010 Q4 t(N(6.2, 0.0014)) 483. ... ... ... ...
#> 3 ets 2010 Q3 N(408, 237) 408. 408 237 NA NA
#> 4 ets 2010 Q4 N(483, 340) 483. ... ... ... ...
sessioninfo::session_info()
#> ─ Session info ───────────────────────────────────────────────────────────────
#> setting value
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#> collate en_US.UTF-8
#> ctype en_US.UTF-8
#> tz Etc/UTC
#> date 2022-11-14
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#>
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#> ──────────────────────────────────────────────────────────────────────────────
You can obtain the parameters from a distribution using the parameters() function. You can obtain the shape of the distribution using the family() function.
For example, you might have a Normally distributed forecast:
library(distributional)
fc <- dist_normal(10, 3)
Obtain the parameters from the distribution
parameters(fc)
#> mu sigma
#> 1 10 3
Obtain the shape of the distribution
family(fc)
#> [1] "normal"
For your example…
library(fable)
#> Loading required package: fabletools
fc <- tsibbledata::aus_production %>%
model(ets = ETS(log(Beer) ~ error("M") + trend("Ad") + season("A"))) %>%
forecast(h = "3 years")
You could extract the parameters and family
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
fc %>%
mutate(parameters(Beer), family(Beer))
#> # A fable: 12 x 8 [1Q]
#> # Key: .model [1]
#> .model Quarter Beer .mean dist trans…¹ inverse famil…²
#> <chr> <qtr> <dist> <dbl> <dist> <list> <list> <chr>
#> 1 ets 2010 Q3 t(N(6, 0.0013)) 407. N(6, 0.0013) <fn> <fn> transf…
#> 2 ets 2010 Q4 t(N(6.2, 0.0014)) 483. N(6.2, 0.0014) <fn> <fn> transf…
#> 3 ets 2011 Q1 t(N(6, 0.0014)) 419. N(6, 0.0014) <fn> <fn> transf…
#> 4 ets 2011 Q2 t(N(6, 0.0015)) 384. N(6, 0.0015) <fn> <fn> transf…
#> 5 ets 2011 Q3 t(N(6, 0.0019)) 405. N(6, 0.0019) <fn> <fn> transf…
#> 6 ets 2011 Q4 t(N(6.2, 0.0022)) 481. N(6.2, 0.0022) <fn> <fn> transf…
#> 7 ets 2012 Q1 t(N(6, 0.0023)) 417. N(6, 0.0023) <fn> <fn> transf…
#> 8 ets 2012 Q2 t(N(5.9, 0.0025)) 383. N(5.9, 0.0025) <fn> <fn> transf…
#> 9 ets 2012 Q3 t(N(6, 0.0032)) 403. N(6, 0.0032) <fn> <fn> transf…
#> 10 ets 2012 Q4 t(N(6.2, 0.0036)) 479. N(6.2, 0.0036) <fn> <fn> transf…
#> 11 ets 2013 Q1 t(N(6, 0.0039)) 416. N(6, 0.0039) <fn> <fn> transf…
#> 12 ets 2013 Q2 t(N(5.9, 0.0043)) 382. N(5.9, 0.0043) <fn> <fn> transf…
#> # … with abbreviated variable names ¹transform, ²`family(Beer)`
Which produces the columns ‘dist’, ‘transform’ and ‘inverse’ - the parameters of the transformed distribution
However you’re probably more interested in the parameters of the Normal, which you can obtain from the parameters() of dist.
fc %>%
mutate(parameters(Beer), family(Beer)) %>%
mutate(parameters(dist), family(dist))
#> # A fable: 12 x 11 [1Q]
#> # Key: .model [1]
#> .model Quarter Beer .mean dist trans…¹ inverse famil…²
#> <chr> <qtr> <dist> <dbl> <dist> <list> <list> <chr>
#> 1 ets 2010 Q3 t(N(6, 0.0013)) 407. N(6, 0.0013) <fn> <fn> transf…
#> 2 ets 2010 Q4 t(N(6.2, 0.0014)) 483. N(6.2, 0.0014) <fn> <fn> transf…
#> 3 ets 2011 Q1 t(N(6, 0.0014)) 419. N(6, 0.0014) <fn> <fn> transf…
#> 4 ets 2011 Q2 t(N(6, 0.0015)) 384. N(6, 0.0015) <fn> <fn> transf…
#> 5 ets 2011 Q3 t(N(6, 0.0019)) 405. N(6, 0.0019) <fn> <fn> transf…
#> 6 ets 2011 Q4 t(N(6.2, 0.0022)) 481. N(6.2, 0.0022) <fn> <fn> transf…
#> 7 ets 2012 Q1 t(N(6, 0.0023)) 417. N(6, 0.0023) <fn> <fn> transf…
#> 8 ets 2012 Q2 t(N(5.9, 0.0025)) 383. N(5.9, 0.0025) <fn> <fn> transf…
#> 9 ets 2012 Q3 t(N(6, 0.0032)) 403. N(6, 0.0032) <fn> <fn> transf…
#> 10 ets 2012 Q4 t(N(6.2, 0.0036)) 479. N(6.2, 0.0036) <fn> <fn> transf…
#> 11 ets 2013 Q1 t(N(6, 0.0039)) 416. N(6, 0.0039) <fn> <fn> transf…
#> 12 ets 2013 Q2 t(N(5.9, 0.0043)) 382. N(5.9, 0.0043) <fn> <fn> transf…
#> # … with 3 more variables: mu <dbl>, sigma <dbl>, `family(dist)` <chr>, and
#> # abbreviated variable names ¹transform, ²`family(Beer)`
Or directly to the underlying (untransformed) distribution
fc %>%
mutate(parameters(parameters(Beer)$dist))
#> # A fable: 12 x 6 [1Q]
#> # Key: .model [1]
#> .model Quarter Beer .mean mu sigma
#> <chr> <qtr> <dist> <dbl> <dbl> <dbl>
#> 1 ets 2010 Q3 t(N(6, 0.0013)) 407. 6.01 0.0360
#> 2 ets 2010 Q4 t(N(6.2, 0.0014)) 483. 6.18 0.0377
#> 3 ets 2011 Q1 t(N(6, 0.0014)) 419. 6.04 0.0380
#> 4 ets 2011 Q2 t(N(6, 0.0015)) 384. 5.95 0.0390
#> 5 ets 2011 Q3 t(N(6, 0.0019)) 405. 6.00 0.0437
#> 6 ets 2011 Q4 t(N(6.2, 0.0022)) 481. 6.17 0.0467
#> 7 ets 2012 Q1 t(N(6, 0.0023)) 417. 6.03 0.0482
#> 8 ets 2012 Q2 t(N(5.9, 0.0025)) 383. 5.95 0.0504
#> 9 ets 2012 Q3 t(N(6, 0.0032)) 403. 6.00 0.0562
#> 10 ets 2012 Q4 t(N(6.2, 0.0036)) 479. 6.17 0.0600
#> 11 ets 2013 Q1 t(N(6, 0.0039)) 416. 6.03 0.0624
#> 12 ets 2013 Q2 t(N(5.9, 0.0043)) 382. 5.94 0.0653
Created on 2022-11-15 by the reprex package (v2.0.1)