很抱歉今天所有与 purrr 相关的问题,仍在试图弄清楚如何有效地利用它。
因此,在 SO 的帮助下,我设法根据来自 data.frame 的输入值运行随机森林护林员模型。这是使用purrr::pmap
. 但是,我不明白返回值是如何从被调用函数生成的。考虑这个例子:
library(ranger)
data(iris)
Input_list <- list(iris1 = iris, iris2 = iris) # let's assume these are different input tables
# the data.frame with the values for the function
hyper_grid <- expand.grid(
Input_table = names(Input_list),
mtry = c(1,2),
Classification = TRUE,
Target = "Species")
> hyper_grid
Input_table mtry Classification Target
1 iris1 1 TRUE Species
2 iris2 1 TRUE Species
3 iris1 2 TRUE Species
4 iris2 2 TRUE Species
# the function to be called for each row of the `hyper_grid`df
fit_and_extract_metrics <- function(Target, Input_table, Classification, mtry,...) {
RF_train <- ranger(
dependent.variable.name = Target,
mtry = mtry,
data = Input_list[[Input_table]], # referring to the named object in the list
classification = Classification) # otherwise regression is performed
RF_train$confusion.matrix
}
# the pmap call using a row of hyper_grid and the function in parallel
purrr::pmap(hyper_grid, fit_and_extract_metrics)
它应该返回 3*3 混淆矩阵的 4 倍,因为 中有 3 个级别iris$Species
,而是返回巨大的混淆矩阵。有人可以向我解释发生了什么吗?
第一行:
> purrr::pmap(hyper_grid, fit_and_extract_metrics)
[[1]]
predicted
true 4.4 4.7 4.8 4.9 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 6 6.1 6.2 6.3 6.4
4.3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4.4 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4.5 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4.6 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4.7 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4.8 0 0 1 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4.9 0 0 1 2 2 0 0 0 0 0 0 0 0 0 1 0 0 0 0
5 0 0 0 1 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5.1 0 0 0 0 0 8 0 0 0 1 0 0 0 0 0 0 0 0 0