未嵌套<dist>来自 `fatabletools::forecast()` 的列 output</dist>
[英]Unnest <dist> column from `fabletools::forecast()` output
问题描述
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.
我再次问这个问题是因为这些答案不适用于我的环境（请参阅下面的“会话信息”），因此我假设，不适用于相关包的当前版本。
Question问题
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 )?如何通过fabletools::forecast()获取<dist>列 output 的元素并将每个元素放入fable的新列中（可转换为data.frame ）？ Eg, all the mu elements in mu , sigma elements in sigma , etc.例如， mu中的所有mu元素、 sigma中的sigma元素等。
Old Answers Fail旧答案失败
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创建于 2022-11-14，使用reprex v2.0.2
Structure of Ideal Output理想结构 Output
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:我理想的 output 看起来像是取消嵌套dist列的元素，但是以行方式提取 mu 和 sigma 元素（或任何可能的分布参数）至关重要：
#> # 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 #> version R version 4.2.1 (2022-06-23) #> os Ubuntu 20.04.5 LTS #> system x86_64, linux-gnu #> ui X11 #> language (EN) #> collate en_US.UTF-8 #> ctype en_US.UTF-8 #> tz Etc/UTC #> date 2022-11-14 #> pandoc 2.19.2 @ /usr/lib/rstudio-server/bin/quarto/bin/tools/ (via rmarkdown) #> #> ─ Packages ─────────────────────────────────────────────────────────────────── #> package * version date (UTC) lib source #> anytime 0.3.9 2020-08-27 [1] RSPM (R 4.2.0) #> assertthat 0.2.1 2019-03-21 [1] RSPM (R 4.2.0) #> cli 3.4.0 2022-09-08 [1] RSPM (R 4.2.0) #> colorspace 2.0-3 2022-02-21 [1] RSPM (R 4.2.0) #> DBI 1.1.3 2022-06-18 [1] RSPM (R 4.2.0) #> digest 0.6.29 2021-12-01 [1] RSPM (R 4.2.0) #> distributional 0.3.1 2022-09-02 [1] RSPM (R 4.2.0) #> dplyr 1.0.10 2022-09-01 [1] RSPM (R 4.2.0) #> ellipsis 0.3.2 2021-04-29 [1] RSPM (R 4.2.0) #> evaluate 0.16 2022-08-09 [1] RSPM (R 4.2.0) #> fable 0.3.2 2022-09-01 [1] RSPM (R 4.2.0) #> fabletools 0.3.2 2021-11-29 [1] RSPM (R 4.2.0) #> fansi 1.0.3 2022-03-24 [1] RSPM (R 4.2.0) #> farver 2.1.1 2022-07-06 [1] RSPM (R 4.2.0) #> fastmap 1.1.0 2021-01-25 [1] RSPM (R 4.2.0) #> fs 1.5.2 2021-12-08 [1] RSPM (R 4.2.0) #> generics 0.1.3 2022-07-05 [1] RSPM (R 4.2.0) #> ggplot2 3.3.6 2022-05-03 [1] RSPM (R 4.2.0) #> glue 1.6.2 2022-02-24 [1] RSPM (R 4.2.0) #> gtable 0.3.1 2022-09-01 [1] RSPM (R 4.2.0) #> highr 0.9 2021-04-16 [1] RSPM (R 4.2.0) #> htmltools 0.5.3 2022-07-18 [1] RSPM (R 4.2.0) #> knitr 1.40 2022-08-24 [1] RSPM (R 4.2.0) #> lifecycle 1.0.2 2022-09-09 [1] RSPM (R 4.2.0) #> lubridate 1.8.0 2021-10-07 [1] RSPM (R 4.2.0) #> magrittr * 2.0.3 2022-03-30 [1] RSPM (R 4.2.0) #> munsell 0.5.0 2018-06-12 [1] RSPM (R 4.2.0) #> numDeriv 2016.8-1.1 2019-06-06 [1] RSPM (R 4.2.0) #> pillar 1.8.1 2022-08-19 [1] RSPM (R 4.2.0) #> pkgconfig 2.0.3 2019-09-22 [1] RSPM (R 4.2.0) #> progressr 0.11.0 2022-09-02 [1] RSPM (R 4.2.0) #> purrr 0.3.4 2020-04-17 [1] RSPM (R 4.2.0) #> R6 2.5.1 2021-08-19 [1] RSPM (R 4.2.0) #> Rcpp 1.0.9 2022-07-08 [1] RSPM (R 4.2.0) #> reprex 2.0.2 2022-08-17 [1] RSPM (R 4.2.0) #> rlang 1.0.5 2022-08-31 [1] RSPM (R 4.2.0) #> rmarkdown 2.16 2022-08-24 [1] RSPM (R 4.2.0) #> rstudioapi 0.14 2022-08-22 [1] RSPM (R 4.2.0) #> scales 1.2.1 2022-08-20 [1] RSPM (R 4.2.0) #> sessioninfo 1.2.2 2021-12-06 [1] RSPM (R 4.2.0) #> stringi 1.7.8 2022-07-11 [1] RSPM (R 4.2.0) #> stringr 1.4.1 2022-08-20 [1] RSPM (R 4.2.0) #> tibble 3.1.8 2022-07-22 [1] RSPM (R 4.2.0) #> tidyr 1.2.1 2022-09-08 [1] RSPM (R 4.2.0) #> tidyselect 1.1.2 2022-02-21 [1] RSPM (R 4.2.0) #> tsibble 1.1.3 2022-10-09 [1] RSPM (R 4.2.0) #> utf8 1.2.2 2021-07-24 [1] RSPM (R 4.2.0) #> vctrs 0.4.1 2022-04-13 [1] RSPM (R 4.2.0) #> withr 2.5.0 2022-03-03 [1] RSPM (R 4.2.0) #> xfun 0.33 2022-09-12 [1] RSPM (R 4.2.0) #> yaml 2.3.5 2022-02-21 [1] RSPM (R 4.2.0) #> #> [1] /usr/local/lib/R/site-library #> [2] /usr/local/lib/R/library #> #> ──────────────────────────────────────────────────────────────────────────────1 个解决方案
解决方案1
1 已采纳 2022-11-15 00:06:34
You can obtain the parameters from a distribution using the parameters() function. You can obtain the shape of the distribution using the family() function.您可以使用parameters() function 从分布中获取参数。您可以使用family() function 获取分布的形状。
For example, you might have a Normally distributed forecast:例如，您可能有一个正态分布的预测：
library(distributional)
fc <- dist_normal(10, 3)
Obtain the parameters from the distribution从分布中获取参数
parameters(fc)
#> mu sigma
#> 1 10 3
Obtain the shape of the distribution获得分布的形状
family(fc)
#> [1] "normal"
For your example…对于你的例子......
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您可以提取参数和系列
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 .这会产生列 'dist'、'transform' 和 'inverse' - 转换分布的参数但是您可能对 Normal 的参数更感兴趣，您可以从dist的parameters()获得这些参数。
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或者直接到底层（未转换的）分布
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
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