尝试使用交叉折叠重采样，并从ranger包中拟合一个随机林。不重新采样的拟合有效，但一旦我尝试重新采样拟合，它就会失败，并出现以下错误。
考虑以下df
df<-structure(list(a = c(1379405931, 732812609, 18614430, 1961678341,
2362202769, 55687714, 72044715, 236503454, 61988734, 2524712675,
98081131, 1366513385, 48203585, 697397991, 28132854), b = structure(c(1L,
6L, 2L, 5L, 7L, 8L, 8L, 1L, 3L, 4L, 3L, 5L, 7L, 2L, 2L), .Label = c("CA",
"IA", "IL", "LA", "MA", "MN", "TX", "WI"), class = "factor"),
c = structure(c(2L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L,
2L, 2L, 2L, 1L), .Label = c("R", "U"), class = "factor"),
d = structure(c(3L, 3L, 1L, 3L, 3L, 1L, 1L, 3L, 1L, 3L, 1L,
3L, 2L, 3L, 1L), .Label = c("CAH", "LTCH", "STH"), class = "factor"),
e = structure(c(3L, 2L, 3L, 3L, 1L, 3L, 3L, 3L, 2L, 4L, 2L,
2L, 3L, 3L, 3L), .Label = c("cancer", "general long term",
"psychiatric", "rehabilitation"), class = "factor")), row.names = c(NA,
-15L), class = c("tbl_df", "tbl", "data.frame"))
以下简单的配合工作没有问题
library(tidymodels)
library(ranger)
rf_spec <- rand_forest(mode = 'regression') %>%
set_engine('ranger')
rf_spec %>%
fit(a ~. , data = df)
但只要我想通过运行交叉验证
rf_folds <- vfold_cv(df, strata = c)
fit_resamples(a ~ . ,
rf_spec,
rf_folds)
跟随错误
model：解析中出错。formula(formula，data，env=parent.frame(((：错误：公式接口中的列名非法。修复列名称或在ranger中使用替代接口。
上面的注释是正确的，这里的问题来源是因子列中的空格。重新采样的函数和普通旧拟合的函数目前处理这一问题的方式不同，我们正在积极研究如何为用户解决这一问题。谢谢你的耐心！
同时，我建议设置一个简单的workflow()和一个recipe()，它们将一起为您处理所有必要的伪变量munging。
library(tidymodels)
rf_spec <- rand_forest(mode = "regression") %>%
set_engine("ranger")
rf_wf <- workflow() %>%
add_model(rf_spec) %>%
add_recipe(recipe(a ~ ., data = df))
fit(rf_wf, data = df)
#> ══ Workflow [trained] ═══════════════════════════════════════════════════════════════════════════════════════════
#> Preprocessor: Recipe
#> Model: rand_forest()
#>
#> ── Preprocessor ─────────────────────────────────────────────────────────────────────────────────────────────────
#> 0 Recipe Steps
#>
#> ── Model ────────────────────────────────────────────────────────────────────────────────────────────────────────
#> Ranger result
#>
#> Call:
#> ranger::ranger(formula = formula, data = data, num.threads = 1, verbose = FALSE, seed = sample.int(10^5, 1))
#>
#> Type: Regression
#> Number of trees: 500
#> Sample size: 15
#> Number of independent variables: 4
#> Mtry: 2
#> Target node size: 5
#> Variable importance mode: none
#> Splitrule: variance
#> OOB prediction error (MSE): 4.7042e+17
#> R squared (OOB): 0.4341146
rf_folds <- vfold_cv(df, strata = c)
fit_resamples(rf_wf,
rf_folds)
#> # 10-fold cross-validation using stratification
#> # A tibble: 9 x 4
#> splits id .metrics .notes
#> <list> <chr> <list> <list>
#> 1 <split [13/2]> Fold1 <tibble [2 × 3]> <tibble [0 × 1]>
#> 2 <split [13/2]> Fold2 <tibble [2 × 3]> <tibble [0 × 1]>
#> 3 <split [13/2]> Fold3 <tibble [2 × 3]> <tibble [0 × 1]>
#> 4 <split [13/2]> Fold4 <tibble [2 × 3]> <tibble [0 × 1]>
#> 5 <split [13/2]> Fold5 <tibble [2 × 3]> <tibble [0 × 1]>
#> 6 <split [13/2]> Fold6 <tibble [2 × 3]> <tibble [0 × 1]>
#> 7 <split [14/1]> Fold7 <tibble [2 × 3]> <tibble [0 × 1]>
#> 8 <split [14/1]> Fold8 <tibble [2 × 3]> <tibble [0 × 1]>
#> 9 <split [14/1]> Fold9 <tibble [2 × 3]> <tibble [0 × 1]>
由reprex包(v0.3.0(于2020-03-20创建
Julia和我打赌，这样她就得到了因果报应。我得到了同样的答案(我做她做的事，但速度较慢(：
这是一个类型的错误，我们一直在努力使其不出错。这很复杂。让我解释一下。
ranger是少数几个公式方法不创建伪变量的R包之一(因为它不需要它们(。
tune中的基础结构使用workflows包来处理公式，然后将结果数据移交给ranger默认情况下，工作流确实会创建伪变量，并且由于某些因子级别不是有效的R列名(例如"general long term"(，ranger()会引发错误。
(我知道你没有使用工作流，但这就是幕后发生的事情(。
我们正在做最好的事情，因为大多数用户不知道许多基于树的模型包不会产生伪变量。更复杂的是，parsnip(还(不使用工作流，也没有给你一个错误。
目前的解决方案
使用简单的配方而不是配方：
library(tidymodels)
#> ── Attaching packages ─────────────────────────────────────────────────────────────────────────────────── tidymodels 0.1.0 ──
#> ✓ broom 0.5.4 ✓ recipes 0.1.10
#> ✓ dials 0.0.4 ✓ rsample 0.0.5
#> ✓ dplyr 0.8.5 ✓ tibble 2.1.3
#> ✓ ggplot2 3.3.0 ✓ tune 0.0.1
#> ✓ infer 0.5.1 ✓ workflows 0.1.0
#> ✓ parsnip 0.0.5 ✓ yardstick 0.0.5
#> ✓ purrr 0.3.3
#> ── Conflicts ────────────────────────────────────────────────────────────────────────────────────── tidymodels_conflicts() ──
#> x purrr::discard() masks scales::discard()
#> x dplyr::filter() masks stats::filter()
#> x dplyr::lag() masks stats::lag()
#> x ggplot2::margin() masks dials::margin()
#> x recipes::step() masks stats::step()
df<-structure(list(a = c(1379405931, 732812609, 18614430, 1961678341,
2362202769, 55687714, 72044715, 236503454, 61988734, 2524712675,
98081131, 1366513385, 48203585, 697397991, 28132854), b = structure(c(1L,
6L, 2L, 5L, 7L, 8L, 8L, 1L, 3L, 4L, 3L, 5L, 7L, 2L, 2L), .Label = c("CA",
"IA", "IL", "LA", "MA", "MN", "TX", "WI"), class = "factor"),
c = structure(c(2L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L,
2L, 2L, 2L, 1L), .Label = c("R", "U"), class = "factor"),
d = structure(c(3L, 3L, 1L, 3L, 3L, 1L, 1L, 3L, 1L, 3L, 1L,
3L, 2L, 3L, 1L), .Label = c("CAH", "LTCH", "STH"), class = "factor"),
e = structure(c(3L, 2L, 3L, 3L, 1L, 3L, 3L, 3L, 2L, 4L, 2L,
2L, 3L, 3L, 3L), .Label = c("cancer", "general long term",
"psychiatric", "rehabilitation"), class = "factor")), row.names = c(NA,
-15L), class = c("tbl_df", "tbl", "data.frame"))
library(tidymodels)
library(ranger)
rf_spec <- rand_forest(mode = 'regression') %>%
set_engine('ranger')
rf_folds <- vfold_cv(df, strata = c)
fit_resamples(recipe(a ~ ., data = df), rf_spec, rf_folds)
#> # 10-fold cross-validation using stratification
#> # A tibble: 9 x 4
#> splits id .metrics .notes
#> * <list> <chr> <list> <list>
#> 1 <split [13/2]> Fold1 <tibble [2 × 3]> <tibble [0 × 1]>
#> 2 <split [13/2]> Fold2 <tibble [2 × 3]> <tibble [0 × 1]>
#> 3 <split [13/2]> Fold3 <tibble [2 × 3]> <tibble [0 × 1]>
#> 4 <split [13/2]> Fold4 <tibble [2 × 3]> <tibble [0 × 1]>
#> 5 <split [13/2]> Fold5 <tibble [2 × 3]> <tibble [0 × 1]>
#> 6 <split [13/2]> Fold6 <tibble [2 × 3]> <tibble [0 × 1]>
#> 7 <split [14/1]> Fold7 <tibble [2 × 3]> <tibble [0 × 1]>
#> 8 <split [14/1]> Fold8 <tibble [2 × 3]> <tibble [0 × 1]>
#> 9 <split [14/1]> Fold9 <tibble [2 × 3]> <tibble [0 × 1]>
# FYI `tune` 0.0.2 will require a different argument order:
# rf_spec %>% fit_resamples(recipe(a ~ ., data = df), rf_folds)
由reprex包(v0.3.0(于2020-03-20创建