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

Question

I'm unable to deploy a tidymodel with vetiver and get a prediction when the model includes a variable with role as ID in the recipe. See the following error in the image:

{ "error": "500 - Internal server error", "message": "Error: The following required columns are missing: 'Fake_ID'.\n" }

The code for the dummy example is below. Do I need to remove the ID-variable from both the model and recipe to make the Plumber API work?

#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

Julia Silge · Accepted Answer · 2022-03-11T14:46:11.640

As of today, vetiver looks for the "mold" workflows::extract_mold(rf_fit) and only get the predictors out to create the ptype. But then when you predict from a workflow, it does require all the variables, including non-predictors. If you have trained a model with non-predictors, as of today you can make the API work by passing in a custom ptype:

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)

^{Created on 2022-03-10 by the reprex package (v2.0.1)}

Are you training models for production with non-predictor variables? Would you mind opening an issue on GitHub to explain your use case a little more?

Thanks Julia for such a quick answer. I see that the solution implies that the ID-variable will show up as a predictor in the API, in the sense that it will need to be populated to be able to get a prediction. And that the accepted values are only those IDs used in the training data. In such case, I don't think it's a good idea to train the production model with non-predictor variables so I will exclude the ID-variable and retrain the model. Thanks again for your help! — StuckityStuck, Mar 11 '22 at 08:42
Sounds like a good plan! If you have other issues as you get started, please post another question or open an issue. — Julia Silge, Mar 11 '22 at 14:44
We've made some changes in hardhat so that this now works (perhaps as you expected) out of the box. You'll need to install hardhat from GitHub for this as of today. — Julia Silge, Apr 21 '22 at 22:12

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

1 Answers1