當有一個角色為 ID 的變量時，為什么使用香根草部署 tidymodel 會引發錯誤？

Question

當模型在配方中包含一個角色為 ID 的變量時，我無法使用香根草部署 tidymodel 並獲得預測。 在圖像中看到以下錯誤：

{“錯誤”：“500 - 內部服務器錯誤”，“消息”：“錯誤：缺少以下必需列：'Fake_ID'。\n”}

虛擬示例的代碼如下。 我是否需要從模型和配方中刪除 ID 變量才能使 Plumber 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)

運行 Plumber API 的示例

Answer 1

到今天為止，香根草尋找“模具” workflows::extract_mold(rf_fit)並且只取出預測器來創建 ptype。 但是，當您從工作流進行預測時，它確實需要所有變量，包括非預測變量。 如果您已經用非預測器訓練了一個模型，那么從今天開始，您可以通過傳入自定義ptype來使 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 包於 2022-03-10 創建 (v2.0.1)}

您是否正在使用非預測變量訓練生產模型？ 您介意在 GitHub 上打開一個問題來進一步解釋您的用例嗎？