如何将多个列表列表作为参数传递给 R 中的 mapply 函数

Question

给定三个列表列表（avg_errors、year_months、dfs）如下：

avg_errors <- list(0.7, 0.92, 2.35)

输出：

[[1]]
[1] 0.7

[[2]]
[1] 0.92

[[3]]
[1] 2.35

和：

year_months <- list(list('2021-12'), list('2021-11'), list('2021-10'))

和数据dfs:

dfs <- list(structure(list(id = c("M01", "M02", "S01"), `2021-10(actual)` = c(8.9, 
15.7, 5.3), `2021-11(actual)` = c(7.3, 14.8, 3.1), `2021-12(actual)` = c(6.1, 
14.2, 3.5), `2021-12(pred)` = c(6.113631596, 14.16243166, 3.288372517
), `2021-12(error)` = c(0.719846116, 0.154463985, 1.225992602
), mean_col = c(7.43333333333333, 14.9, 3.96666666666667), act_direction = c(-1.2, 
-0.600000000000001, 0.4), pred_direction = c(-1.186368404, -0.637568340000001, 
0.188372517)), row.names = c(NA, -3L), class = "data.frame"), 
    structure(list(id = c("M01", "M02", "S01"), `2021-09(actual)` = c(10.3, 
    17.3, 6.4), `2021-10(actual)` = c(8.9, 15.7, 5.3), `2021-11(actual)` = c(7.3, 
    14.8, 3.1), `2021-11(pred)` = c(8.352097939, 13.97318204, 
    3.164682627), `2021-11(error)` = c(1.998109138, 0.414373304, 
    0.342072615), mean_col = c(8.83333333333333, 15.9333333333333, 
    4.93333333333333), act_direction = c(-1.6, -0.899999999999999, 
    -2.2), pred_direction = c(-0.547902061, -1.72681796, -2.135317373
    )), row.names = c(NA, -3L), class = "data.frame"), structure(list(
        id = c("M01", "M02", "S01"), `2021-08(actual)` = c(12.6, 
        19.2, 8.3), `2021-09(actual)` = c(10.3, 17.3, 6.4), `2021-10(actual)` = c(8.9, 
        15.7, 5.3), `2021-10(pred)` = c(9.619846116, 15.54553601, 
        6.525992602), `2021-10(error)` = c(0.945567783, 4.883250027, 
        1.215819585), mean_col = c(10.6, 17.4, 6.66666666666667
        ), act_direction = c(-1.4, -1.6, -1.1), pred_direction = c(-0.680153884000001, 
        -1.75446399, 0.125992601999999)), row.names = c(NA, -3L
    ), class = "data.frame"))

如何将这些变量作为参数传递给函数 Plotting()，如下所示：

library(gt)
library(tidyverse)

Plotting <- function(data, year_months, errors){
  for (current_month in year_months){
    for (avg_error in errors){
      p <- data %>% 
        gt() %>% 
        tab_header(
          title = gt::html(glue("<span style='color:red'>data for {current_month}</span>")),
          subtitle = md(paste0("The average error for this month is: <em>", avg_error, '</em>.'))
        ) 
      return (p)
  }
  }
}

mapply(Plotting, dfs, year_months, errors)

当前代码给我：Error in (function (data, errors) : argument no use (dots[[3]][[1]]).

如何调用函数并传递参数？感谢您的提前帮助。

Answer 1

我不确定这是否是您想要的结果，但我的猜测是，您不需要 Plotting 函数中的 for-loop，因为您已经在循环用 mapply.

覆盖你的输入

library(gt)
library(tidyverse)
library(glue)


Plotting <- function(data, year_months, errors){
      p <- data %>% 
        gt() %>% 
        tab_header(
          title = gt::html(glue("<span style='color:red'>data for {year_months}</span>")),
          subtitle = md(paste0("The average error for this month is: <em>", errors, '</em>.'))
        ) 
      print(p)
}

mapply(Plotting, dfs, year_months, avg_errors)

^{由 reprex package (v0.3.0)}

创建于 2022-01-23

如何将多个列表列表作为参数传递给 R 中的 mapply 函数

How to pass multiple list of lists as arguments to mapply function in R

functional-programming

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function

environment-variables