当使用 expand.grid 和 purrr::pmap 时，R ranger confusion.matrix 比预期的要大

Question

抱歉今天所有与 purrr 相关的问题，仍在努力弄清楚如何有效地使用它。

因此，在 SO 的帮助下，我设法根据来自 data.frame 的输入值获得了随机森林管理员模型运行。这是使用 purrr::pmap 完成的。但是，我不明白 return 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)

它应该是 return 4 倍的 3*3 混淆矩阵，因为 iris$Species 中有 3 个级别，而不是 return 的巨大混淆矩阵。谁能给我解释一下这是怎么回事？

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

Answer 1

这里的问题是因为传递给函数的参数是级别，而不是字符。这触发了 ranger 函数。要解决这个问题，您需要做的就是在 expand.grid:

中设置 stringsAsFactors = FALSE

hyper_grid <- expand.grid(
    Input_table = names(Input_list),
    mtry = c(1,2),
    Classification = TRUE,
    Target = "Species", stringsAsFactors = FALSE)

您将获得：

[[1]]
            predicted
true         setosa versicolor virginica
  setosa         50          0         0
  versicolor      0         46         4
  virginica       0          4        46

[[2]]
            predicted
true         setosa versicolor virginica
  setosa         50          0         0
  versicolor      0         46         4
  virginica       0          5        45

[[3]]
            predicted
true         setosa versicolor virginica
  setosa         50          0         0
  versicolor      0         47         3
  virginica       0          3        47

[[4]]
            predicted
true         setosa versicolor virginica
  setosa         50          0         0
  versicolor      0         47         3
  virginica       0          3        47

当使用 expand.grid 和 purrr::pmap 时，R ranger confusion.matrix 比预期的要大

R ranger confusion.matrix is larger than supposed when using expand.grid and purrr::pmap

r

purrr

confusion-matrix

r-ranger