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r - Tidymodels 包 : Visualising a random forest model using ggplot() to show the most important predictors

转载 作者:行者123 更新时间:2023-12-04 15:12:41 24 4
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概览

我正在按照教程(见下文)从袋装树、随机森林、提升树和一般线性模型中找到最适合的模型。

教程(参见下面的示例)

https://bcullen.rbind.io/post/2020-06-02-tidymodels-decision-tree-learning-in-r/

问题

在这种情况下,我想进一步探索数据,并为我在随机森林模型中的数据可视化最重要的预测变量(见下图)。

我的数据框称为FID随机森林模型中的预测变量涉及:

  1. 年份(数字)
  2. 月份(因子)
  3. 天数(数字)

因变量是频率(数字)

当我尝试运行绘图以可视化最重要的预测变量时,我不断收到此错误消息:-

Error: Problem with `mutate()` input `oob_rmse`.
x non-numeric argument to mathematical function
ℹ Input `oob_rmse` is `map_dbl(fit, ~sqrt(.x$prediction.error))`.
Run `rlang::last_error()` to see where the error occurred.
Called from: signal_abort(cnd)

如果有人对如何修复错误消息有任何建议,我将不胜感激。

提前致谢

如何从教程中的 R 代码生成绘图的示例

enter image description here enter image description here

可视化模型

enter image description here

绘制以显示教程中 R 代码中最重要的预测变量

enter image description here

我的 R 代码

##Open libraries
library(tidymodels)
library(parsnip)
library(forcats)
library(ranger)
library(baguette)

###########################################################
#split this single dataset into two: a training set and a testing set
data_split <- initial_split(FID)
# Create data frames for the two sets:
train_data <- training(data_split)
test_data <- testing(data_split)

# resample the data with 10-fold cross-validation (10-fold by default)
cv <- vfold_cv(train_data, v=3)
###########################################################

##Produce the recipe

rec <- recipe(Frequency ~ ., data = FID) %>%
step_nzv(all_predictors(), freq_cut = 0, unique_cut = 0) %>% # remove variables with zero variances
step_novel(all_nominal()) %>% # prepares test data to handle previously unseen factor levels
step_medianimpute(all_numeric(), -all_outcomes(), -has_role("id vars")) %>% # replaces missing numeric observations with the median
step_dummy(all_nominal(), -has_role("id vars")) # dummy codes categorical variables

###################################################################################


###################################################
##Random forests
###################################################

mod_rf <-rand_forest(trees = 1e3) %>%
set_engine("ranger",
num.threads = parallel::detectCores(),
importance = "permutation",
verbose = TRUE) %>%
set_mode("regression")

##Create Workflow

wflow_rf <- workflow() %>%
add_model(mod_rf) %>%
add_recipe(rec)

##Fit the model

plan(multisession)

fit_rf<-fit_resamples(
wflow_rf,
cv,
metrics = metric_set(rmse, rsq),
control = control_resamples(save_pred = TRUE,
extract = function(x) extract_model(x)))


# extract roots
rf_tree_roots <- function(x){
map_chr(1:1000,
~ranger::treeInfo(x, tree = .)[1, "splitvarName"])
}

rf_roots <- function(x){
x %>%
dplyr::select(.extracts) %>%
unnest(cols = c(.extracts)) %>%
dplyr::mutate(fit = map(.extracts,
~.x$fit$fit$fit),
oob_rmse = map_dbl(fit,
~sqrt(.x$prediction.error)),
roots = map(fit,
~rf_tree_roots(.))
) %>%
dplyr::select(roots) %>%
unnest(cols = c(roots))
}

##Open a plotting window
dev.new()

# plot
rf_roots(fit_rf) %>%
group_by(roots) %>%
count() %>%
dplyr::arrange(desc(n)) %>%
dplyr::filter(n > 75) %>%
ggplot(aes(fct_reorder(roots, n), n)) +
geom_col() +
coord_flip() +
labs(x = "root", y = "count")

##Error message

Error: Problem with `mutate()` input `oob_rmse`.
x non-numeric argument to mathematical function
ℹ Input `oob_rmse` is `map_dbl(fit, ~sqrt(.x$prediction.error))`.
Run `rlang::last_error()` to see where the error occurred.
Called from: signal_abort(cnd)

数据框 - FID

  structure(list(Year = c(2015, 2015, 2015, 2015, 2015, 2015, 2015, 
2015, 2015, 2015, 2015, 2015, 2016, 2016, 2016, 2016, 2016, 2016,
2016, 2016, 2016, 2016, 2016, 2016, 2017, 2017, 2017, 2017, 2017,
2017, 2017, 2017, 2017, 2017, 2017, 2017), Month = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L,
5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 10L, 11L, 12L), .Label = c("January", "February", "March",
"April", "May", "June", "July", "August", "September", "October",
"November", "December"), class = "factor"), Frequency = c(36,
28, 39, 46, 5, 0, 0, 22, 10, 15, 8, 33, 33, 29, 31, 23, 8, 9,
7, 40, 41, 41, 30, 30, 44, 37, 41, 42, 20, 0, 7, 27, 35, 27,
43, 38), Days = c(31, 28, 31, 30, 6, 0, 0, 29, 15,
29, 29, 31, 31, 29, 30, 30, 7, 0, 7, 30, 30, 31, 30, 27, 31,
28, 30, 30, 21, 0, 7, 26, 29, 27, 29, 29)), row.names = c(NA,
-36L), class = "data.frame")

最佳答案

它不会正确提取错误,如果您查看包含所有模型的小标题:

fit_rf$.extracts
[[1]]
# A tibble: 1 x 1
.extracts
<list>
1 <ranger>

它嵌入在列表或列表中,但没有名称:

names(fit_rf$.extracts[[1]][[1]])
NULL

因此这部分将失败:

map(fit_rf$.extracts,~.x$fit$fit$fit)

如果您查看第一个 unnest 之后的结构,这已经是合适的:

fit_rf %>% dplyr::select(.extracts) %>% unnest(cols = c(.extracts)) 
# A tibble: 3 x 1
.extracts
<list>
1 <ranger>
2 <ranger>
3 <ranger>

所以我们可以这样做:

rf_roots <- function(x){
x %>%
select(.extracts) %>%
unnest(cols = c(.extracts)) %>%
mutate(oob_rmse = map_dbl(.extracts,
~sqrt(.x$prediction.error)),
roots = map(.extracts,
~rf_tree_roots(.))
) %>%
dplyr::select(roots) %>%
unnest(cols = c(roots))
}

这现在可以工作了:

rf_roots(fit_rf)
# A tibble: 3,000 x 1
roots
<chr>
1 Month_August
2 Year
3 Month_July
4 Month_September
5 Month_December
6 Month_March
7 Month_July
8 Month_September
9 Month_December
10 Days

附加组件:如果目标是获取每个模型中每棵树的根变量,可以简单地执行:

root_vars = unnest(fit_rf,.extracts) %>% 
pull(.extracts) %>%
map(rf_tree_roots)

或以 R 为基数:

lapply(fit_rf$.extracts,function(i)rf_tree_roots(i[[1]][[1]]))

你可以很容易地取消列出它来制作条形图。

关于r - Tidymodels 包 : Visualising a random forest model using ggplot() to show the most important predictors,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64939433/

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