当有一个角色作为 ID 的变量时，为什么部署带有香根草的 tidymodel 会抛出错误？

Question

我无法部署带有香根草的 tidymodel，并且无法在模型包含一个变量并将角色作为配方中的 ID 时得到预测。看到图中出现以下错误：

{ "error": "500 - 内部服务器错误", “消息”：“错误：缺少以下必需的列：'Fake_ID'。\n” }

虚拟示例的代码如下。我是否需要从模型和配方中删除 ID 变量才能使管道工 API 工作？

#Load libraries
library(recipes)
library(parsnip)
library(workflows)
library(pins)
library(plumber)
library(stringi)



#Upload data
data(Sacramento, package = "modeldata")


#Create fake IDs for testing
Sacramento$Fake_ID <- stri_rand_strings(nrow(Sacramento), 10)


# Train model
Sacramento_recipe <- recipe(formula = price ~ type + sqft + beds + baths + zip + Fake_ID, data = Sacramento) %>% 
  update_role(Fake_ID, new_role = "ID") %>% 
  step_zv(all_predictors())

rf_spec <- rand_forest(mode = "regression") %>% set_engine("ranger")

rf_fit <-
  workflow() %>%
  add_model(rf_spec) %>%
  add_recipe(Sacramento_recipe) %>%
  fit(Sacramento)


# Create vetiver object
v <- vetiver::vetiver_model(rf_fit, "sacramento_rf")
v


# Allow for model versioning and sharing
model_board <- board_temp()
model_board %>% vetiver_pin_write(v)


# Deploying model
pr() %>%
  vetiver_api(v) %>%
  pr_run(port = 8088)

Running the example of the Plumber API

Answer 1

从今天开始，香根草寻找“霉菌”workflows::extract_mold(rf_fit) 并且只得到预测因子来创建 ptype。但是，当您从工作流中进行预测时，它确实需要所有变量，包括 non-predictors。如果您使用 non-predictors 训练了一个模型，从今天开始，您可以通过传入自定义 ptype:

来使 API 工作

library(recipes)
#> Loading required package: 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
#> 
#> Attaching package: 'recipes'
#> The following object is masked from 'package:stats':
#> 
#>     step
library(parsnip)
library(workflows)
library(pins)
library(plumber)
library(stringi)

data(Sacramento, package = "modeldata")
Sacramento$Fake_ID <- stri_rand_strings(nrow(Sacramento), 10)


Sacramento_recipe <- 
    recipe(formula = price ~ type + sqft + beds + baths + zip + Fake_ID, 
           data = Sacramento) %>% 
    update_role(Fake_ID, new_role = "ID") %>% 
    step_zv(all_predictors())

rf_spec <- rand_forest(mode = "regression") %>% set_engine("ranger")

rf_fit <-
    workflow() %>%
    add_model(rf_spec) %>%
    add_recipe(Sacramento_recipe) %>%
    fit(Sacramento)


library(vetiver)
## this is probably easiest because this model uses a simple formula
## if there is more complex preprocessing, select the variables
## from `Sacramento` via dplyr or similar
sac_ptype <- extract_recipe(rf_fit) %>% 
    bake(new_data = Sacramento, -all_outcomes()) %>% 
    vctrs::vec_ptype()

v <- vetiver_model(rf_fit, "sacramento_rf", save_ptype = sac_ptype)
v
#> 
#> ── sacramento_rf ─ <butchered_workflow> model for deployment 
#> A ranger regression modeling workflow using 6 features

pr() %>%
    vetiver_api(v)
#> # Plumber router with 2 endpoints, 4 filters, and 0 sub-routers.
#> # Use `pr_run()` on this object to start the API.
#> ├──[queryString]
#> ├──[body]
#> ├──[cookieParser]
#> ├──[sharedSecret]
#> ├──/ping (GET)
#> └──/predict (POST)

^{由 reprex package (v2.0.1)}

创建于 2022-03-10

您是否正在使用 non-predictor 变量训练生产模型？您介意 opening an issue on GitHub 多解释一下您的用例吗？

当有一个角色作为 ID 的变量时，为什么部署带有香根草的 tidymodel 会抛出错误？

Why does deploying a tidymodel with vetiver throw a error when there's a variable with role as ID?

r

tidymodels

plumber