运行 tidymodels "Error in `dplyr::select()`: Can't subset columns that don't exist." 中的 `predict()` 错误
Error on running `predict()` in tidymodels "Error in `dplyr::select()`: Can't subset columns that don't exist."
问题
在 tidymodels 框架中 运行 predict 时出现错误。
该错误似乎与在配方中选择变量有关（参见下面的代码）。
我试过的
有一些相关的 SO 帖子，例如 , , or ，但它们似乎处理不同的问题（例如操纵配方中的结果）。
但是，我想了解为什么我的代码首先会抛出错误。
抛出错误的代码
library(tidyverse)
library(tidymodels)
data("mtcars")
d_train <- mtcars %>% slice(1:20)
d_test <- mtcars %>% slice(21:nrow(mtcars))
preds_chosen <- c("hp", "disp", "am")
rec1 <-
recipe( ~ ., data = d_train) %>%
step_select(all_of(preds_chosen), mpg) %>%
update_role(all_of(preds_chosen), new_role = "predictor") %>%
update_role(mpg, new_role = "outcome")
model_lm <- linear_reg()
wf1 <-
workflow() %>%
add_model(model_lm) %>%
add_recipe(rec1)
lm_fit1 <-
wf1 %>%
fit(d_train)
preds <-
lm_fit1 %>%
predict(d_test)
#> Error in `dplyr::select()`:
#> ! Can't subset columns that don't exist.
#> ✖ Column `mpg` doesn't exist.
可能的解决方案
如果我按以下方式更改配方，整个代码运行时不会出错：
rec2 <- recipe(mpg ~ hp + disp + am, data = d_train)
rec3 <-
recipe(mpg ~ ., data = d_train) %>%
update_role(all_predictors(), new_role = "id") %>%
update_role(all_of(preds_chosen), new_role = "predictor") %>%
update_role(mpg, new_role = "outcome")
会话信息
sessionInfo()
#> R version 4.1.3 (2022-03-10)
#> Platform: x86_64-apple-darwin17.0 (64-bit)
#> Running under: macOS Big Sur/Monterey 10.16
#>
#> Matrix products: default
#> BLAS: /Library/Frameworks/R.framework/Versions/4.1/Resources/lib/libRblas.0.dylib
#> LAPACK: /Library/Frameworks/R.framework/Versions/4.1/Resources/lib/libRlapack.dylib
#>
#> locale:
#> [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
#>
#> attached base packages:
#> [1] stats graphics grDevices utils datasets methods base
#>
#> loaded via a namespace (and not attached):
#> [1] rstudioapi_0.13 knitr_1.39 magrittr_2.0.3 rlang_1.0.2
#> [5] fastmap_1.1.0 fansi_1.0.3 stringr_1.4.0 styler_1.5.1
#> [9] highr_0.9 tools_4.1.3 xfun_0.30 utf8_1.2.2
#> [13] cli_3.3.0 withr_2.5.0 htmltools_0.5.2 ellipsis_0.3.2
#> [17] yaml_2.3.5 digest_0.6.29 tibble_3.1.7 lifecycle_1.0.1
#> [21] crayon_1.5.1 purrr_0.3.4 vctrs_0.4.1 fs_1.5.2
#> [25] glue_1.6.2 evaluate_0.15 rmarkdown_2.14 reprex_2.0.1
#> [29] stringi_1.7.6 compiler_4.1.3 pillar_1.7.0 backports_1.4.1
#> [33] pkgconfig_2.0.3
Created on 2022-05-21 by the reprex package (v2.0.1)
```
我们可以在skip_select中使用skip = TRUE。根据?skip_select
skip - A logical. Should the step be skipped when the recipe is baked by bake()? While all operations are baked when prep() is run, some operations may not be able to be conducted on new data (e.g. processing the outcome variable(s)). Care should be taken when using skip = TRUE as it may affect the computations for subsequent operations.
rec1 <-
recipe( ~ ., data = d_train) %>%
step_select(all_of(preds_chosen), mpg, skip = TRUE) %>%
update_role(all_of(preds_chosen), new_role = "predictor") %>%
update_role(mpg, new_role = "outcome")
然后使用 OP 的代码
> preds <-
+ lm_fit1 %>%
+ predict(d_test)
> preds
# A tibble: 12 × 1
.pred
<dbl>
1 22.6
2 17.2
3 17.4
4 12.1
5 14.9
6 28.2
7 26.3
8 25.6
9 14.6
10 21.8
11 11.7
12 25.4
关于如何使用 skip = TRUE 的答案可能适用于您的情况，但如果您使用的是非常简单的预处理并且除了指定角色之外并不真正需要配方，您可能需要 look into using add_variables():
library(tidyverse)
library(tidymodels)
data("mtcars")
d_train <- mtcars %>% slice(1:20)
d_test <- mtcars %>% slice(21:nrow(mtcars))
preds_chosen <- c("hp", "disp", "am")
wf1 <-
workflow() %>%
add_model(linear_reg()) %>%
add_variables(outcomes = mpg, predictors = !! preds_chosen)
lm_fit1 <- fit(wf1, d_train)
predict(lm_fit1, d_test)
#> # A tibble: 12 × 1
#> .pred
#> <dbl>
#> 1 22.6
#> 2 17.2
#> 3 17.4
#> 4 12.1
#> 5 14.9
#> 6 28.2
#> 7 26.3
#> 8 25.6
#> 9 14.6
#> 10 21.8
#> 11 11.7
#> 12 25.4
由 reprex package (v2.0.1)
创建于 2022-05-22
问题
在 tidymodels 框架中 运行 predict 时出现错误。
该错误似乎与在配方中选择变量有关（参见下面的代码）。
我试过的
有一些相关的 SO 帖子，例如
但是，我想了解为什么我的代码首先会抛出错误。
抛出错误的代码
library(tidyverse)
library(tidymodels)
data("mtcars")
d_train <- mtcars %>% slice(1:20)
d_test <- mtcars %>% slice(21:nrow(mtcars))
preds_chosen <- c("hp", "disp", "am")
rec1 <-
recipe( ~ ., data = d_train) %>%
step_select(all_of(preds_chosen), mpg) %>%
update_role(all_of(preds_chosen), new_role = "predictor") %>%
update_role(mpg, new_role = "outcome")
model_lm <- linear_reg()
wf1 <-
workflow() %>%
add_model(model_lm) %>%
add_recipe(rec1)
lm_fit1 <-
wf1 %>%
fit(d_train)
preds <-
lm_fit1 %>%
predict(d_test)
#> Error in `dplyr::select()`:
#> ! Can't subset columns that don't exist.
#> ✖ Column `mpg` doesn't exist.
可能的解决方案
如果我按以下方式更改配方，整个代码运行时不会出错：
rec2 <- recipe(mpg ~ hp + disp + am, data = d_train)
rec3 <-
recipe(mpg ~ ., data = d_train) %>%
update_role(all_predictors(), new_role = "id") %>%
update_role(all_of(preds_chosen), new_role = "predictor") %>%
update_role(mpg, new_role = "outcome")
会话信息
sessionInfo()
#> R version 4.1.3 (2022-03-10)
#> Platform: x86_64-apple-darwin17.0 (64-bit)
#> Running under: macOS Big Sur/Monterey 10.16
#>
#> Matrix products: default
#> BLAS: /Library/Frameworks/R.framework/Versions/4.1/Resources/lib/libRblas.0.dylib
#> LAPACK: /Library/Frameworks/R.framework/Versions/4.1/Resources/lib/libRlapack.dylib
#>
#> locale:
#> [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
#>
#> attached base packages:
#> [1] stats graphics grDevices utils datasets methods base
#>
#> loaded via a namespace (and not attached):
#> [1] rstudioapi_0.13 knitr_1.39 magrittr_2.0.3 rlang_1.0.2
#> [5] fastmap_1.1.0 fansi_1.0.3 stringr_1.4.0 styler_1.5.1
#> [9] highr_0.9 tools_4.1.3 xfun_0.30 utf8_1.2.2
#> [13] cli_3.3.0 withr_2.5.0 htmltools_0.5.2 ellipsis_0.3.2
#> [17] yaml_2.3.5 digest_0.6.29 tibble_3.1.7 lifecycle_1.0.1
#> [21] crayon_1.5.1 purrr_0.3.4 vctrs_0.4.1 fs_1.5.2
#> [25] glue_1.6.2 evaluate_0.15 rmarkdown_2.14 reprex_2.0.1
#> [29] stringi_1.7.6 compiler_4.1.3 pillar_1.7.0 backports_1.4.1
#> [33] pkgconfig_2.0.3
Created on 2022-05-21 by the reprex package (v2.0.1)
```
我们可以在skip_select中使用skip = TRUE。根据?skip_select
skip - A logical. Should the step be skipped when the recipe is baked by bake()? While all operations are baked when prep() is run, some operations may not be able to be conducted on new data (e.g. processing the outcome variable(s)). Care should be taken when using skip = TRUE as it may affect the computations for subsequent operations.
rec1 <-
recipe( ~ ., data = d_train) %>%
step_select(all_of(preds_chosen), mpg, skip = TRUE) %>%
update_role(all_of(preds_chosen), new_role = "predictor") %>%
update_role(mpg, new_role = "outcome")
然后使用 OP 的代码
> preds <-
+ lm_fit1 %>%
+ predict(d_test)
> preds
# A tibble: 12 × 1
.pred
<dbl>
1 22.6
2 17.2
3 17.4
4 12.1
5 14.9
6 28.2
7 26.3
8 25.6
9 14.6
10 21.8
11 11.7
12 25.4
关于如何使用 skip = TRUE 的答案可能适用于您的情况，但如果您使用的是非常简单的预处理并且除了指定角色之外并不真正需要配方，您可能需要 look into using add_variables():
library(tidyverse)
library(tidymodels)
data("mtcars")
d_train <- mtcars %>% slice(1:20)
d_test <- mtcars %>% slice(21:nrow(mtcars))
preds_chosen <- c("hp", "disp", "am")
wf1 <-
workflow() %>%
add_model(linear_reg()) %>%
add_variables(outcomes = mpg, predictors = !! preds_chosen)
lm_fit1 <- fit(wf1, d_train)
predict(lm_fit1, d_test)
#> # A tibble: 12 × 1
#> .pred
#> <dbl>
#> 1 22.6
#> 2 17.2
#> 3 17.4
#> 4 12.1
#> 5 14.9
#> 6 28.2
#> 7 26.3
#> 8 25.6
#> 9 14.6
#> 10 21.8
#> 11 11.7
#> 12 25.4
由 reprex package (v2.0.1)
创建于 2022-05-22