tidymodel 配方和 `step_lag()`:使用 `predict()` 时出错

tidymodel recipe and `step_lag()`: Error when using `predict()`

这可能是一个用法误解,但我希望下面的玩具示例能够工作。我想在我的配方中有一个滞后的预测器,但是一旦我将它包含在配方中,并尝试使用具有配方的工作流来预测相同的数据,它就无法识别列 foo 并且无法计算它的滞后。

现在,如果我:

,我可以让它工作:
  1. 从已拟合的工作流中拉出拟合。
  2. 独立准备和烘焙我想要拟合的数据。

我在失败的工作流适合后编写代码,它成功了。根据文档,我应该能够将工作流放入预测槽中:https://www.tidymodels.org/start/recipes/#predict-workflow

我可能从根本上误解了工作流应该如何运作。我有我认为的解决方法,但我不明白为什么失败的语句没有按照解决方法的方式工作。我希望失败的工作流构造能够像我的解决方法一样在幕后工作。

简而言之,如果 work_df 是一个数据框,the_rec 是一个基于 work_df 的食谱,rf_mod 是一个模型,并且您创建工作流 rf_workflow,那么我是否应该期望 predict() 函数在下面的两个 predict() 调用中工作相同?

## Workflow
rf_workflow <-
    workflow() %>%
    add_model(rf_mod) %>%
    add_recipe(the_rec)

## fit
rf_workflow_fit <-
    rf_workflow %>%
    fit(data = work_df)

## Predict with workflow.  I expect since a workflow has a fit model and
## a recipe as part of it, it should know how to do the following:
predict(rf_workflow_fit, work_df)
#> Error: Problem with `mutate()` input `lag_1_foo`.
#> x object 'foo' not found
#> i Input `lag_1_foo` is `dplyr::lag(x = foo, n = 1L, default = NA)`.


## Predict by explicitly prepping and baking the data, and pulling out the
## fit from the workflow:
predict(
    rf_workflow_fit %>%
        pull_workflow_fit(),
    bake(prep(the_rec), work_df))
#> # A tibble: 995 x 1
#>     .pred
#>     <dbl>
#>  1  2.24 
#>  2  0.595
#>  3  0.262

下面是完整的 reprex 示例。

library(tidymodels)
#> -- Attaching packages -------------------------------------------------------------------------------------- tidymodels 0.1.1 --
#> v broom     0.7.1      v recipes   0.1.13
#> v dials     0.0.9      v rsample   0.0.8 
#> v dplyr     1.0.2      v tibble    3.0.3 
#> v ggplot2   3.3.2      v tidyr     1.1.2 
#> v infer     0.5.3      v tune      0.1.1 
#> v modeldata 0.0.2      v workflows 0.2.1 
#> v parsnip   0.1.3      v yardstick 0.0.7 
#> v purrr     0.3.4
#> -- Conflicts ----------------------------------------------------------------------------------------- tidymodels_conflicts() --
#> x purrr::discard() masks scales::discard()
#> x dplyr::filter()  masks stats::filter()
#> x dplyr::lag()     masks stats::lag()
#> x recipes::step()  masks stats::step()
library(dplyr)

set.seed(123)

### Create autocorrelated timeseries: https://stafoo.stackexchange.com/a/29242/17203
work_df <-
    tibble(
        foo = stats::filter(rnorm(1000), filter=rep(1,5), circular=TRUE) %>%
            as.numeric()
    )
# plot(work_df$foo)
work_df
#> # A tibble: 1,000 x 1
#>         foo
#>       <dbl>
#>  1 -0.00375
#>  2  0.589  
#>  3  0.968  
#>  4  3.24   
#>  5  3.93   
#>  6  1.11   
#>  7  0.353  
#>  8 -0.222  
#>  9 -0.713  
#> 10 -0.814  
#> # ... with 990 more rows

## Recipe
the_rec <-
    recipe(foo ~ ., data = work_df) %>%
    step_lag(foo, lag=1:5) %>%
    step_naomit(all_predictors())

the_rec %>% prep() %>% juice()
#> # A tibble: 995 x 6
#>       foo lag_1_foo lag_2_foo lag_3_foo lag_4_foo lag_5_foo
#>     <dbl>     <dbl>     <dbl>     <dbl>     <dbl>     <dbl>
#>  1  1.11      3.93      3.24      0.968     0.589  -0.00375
#>  2  0.353     1.11      3.93      3.24      0.968   0.589  
#>  3 -0.222     0.353     1.11      3.93      3.24    0.968  
#>  4 -0.713    -0.222     0.353     1.11      3.93    3.24   
#>  5 -0.814    -0.713    -0.222     0.353     1.11    3.93   
#>  6  0.852    -0.814    -0.713    -0.222     0.353   1.11   
#>  7  1.65      0.852    -0.814    -0.713    -0.222   0.353  
#>  8  1.54      1.65      0.852    -0.814    -0.713  -0.222  
#>  9  2.10      1.54      1.65      0.852    -0.814  -0.713  
#> 10  2.24      2.10      1.54      1.65      0.852  -0.814  
#> # ... with 985 more rows

## Model
rf_mod <-
    rand_forest(
        mtry = 4,
        trees = 1000,
        min_n = 13) %>%
    set_mode("regression") %>%
    set_engine("ranger")

## Workflow
rf_workflow <-
    workflow() %>%
    add_model(rf_mod) %>%
    add_recipe(the_rec)

## fit
rf_workflow_fit <-
    rf_workflow %>%
    fit(data = work_df)

## Predict
predict(rf_workflow_fit, work_df)
#> Error: Problem with `mutate()` input `lag_1_foo`.
#> x object 'foo' not found
#> i Input `lag_1_foo` is `dplyr::lag(x = foo, n = 1L, default = NA)`.


## Perhaps I just need to pull off the fit and work with that?... Nope.
predict(
    rf_workflow_fit %>%
        pull_workflow_fit(),
    work_df)
#> Error: Can't subset columns that don't exist.
#> x Columns `lag_1_foo`, `lag_2_foo`, `lag_3_foo`, `lag_4_foo`, and `lag_5_foo` don't exist.

## Maybe I need to bake it first... and that works.
## But doesn't that defeat the purpose of a workflow?
predict(
    rf_workflow_fit %>%
        pull_workflow_fit(),
    bake(prep(the_rec), work_df))
#> # A tibble: 995 x 1
#>     .pred
#>     <dbl>
#>  1  2.24 
#>  2  0.595
#>  3  0.262
#>  4 -0.977
#>  5 -1.24 
#>  6 -0.140
#>  7  1.36 
#>  8  1.30 
#>  9  1.78 
#> 10  2.42 
#> # ... with 985 more rows

## Session info
sessioninfo::session_info()
#> - Session info ---------------------------------------------------------------
#>  setting  value                       
#>  version  R version 3.6.3 (2020-02-29)
#>  os       Windows 10 x64              
#>  system   x86_64, mingw32             
#>  ui       RTerm                       
#>  language (EN)                        
#>  collate  English_United States.1252  
#>  ctype    English_United States.1252  
#>  tz       America/Chicago             
#>  date     2020-10-13                  
#> 
#> - Packages -------------------------------------------------------------------
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#>  backports     1.1.10     2020-09-15 [1] CRAN (R 3.6.3)
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#> [1] C:/Users/IRINZN/Documents/R/R-3.6.3/library

reprex package (v0.3.0)

于 2020-10-13 创建

您遇到错误的原因是您根据 结果 创建了一个预测变量。当需要预测新数据时,结果不可用;我们正在 预测 新数据的结果,而不是假设它已经存在。

这是 tidymodels 框架的一个相当强大的假设,用于建模或预处理,以防止信息泄漏。您可以阅读 about this a bit more here.

您可能已经了解这些资源,但如果您正在使用时间序列模型,我建议您查看这些资源: