如何使用 dplyr 和 ggplot2 将列名作为函数参数传递？

Question

我正在尝试编写一个会吐出模型诊断图的函数。

to_plot <- function(df, model, response_variable, indep_variable) {
  resp_plot <- 
    df %>%
    mutate(model_resp = predict.glm(model, df, type = 'response')) %>%
    group_by(indep_variable) %>%
    summarize(actual_response = mean(response_variable),
              predicted_response = mean(model_resp)) %>%
    ggplot(aes(indep_variable)) + 
    geom_line(aes(x = indep_variable, y = actual_response, colour = "actual")) + 
    geom_line(aes(x = indep_variable, y = predicted_response, colour = "predicted")) +
    ylab(label = 'Response')

}

当我在数据集上运行它时，dplyr 会抛出一个我不明白的错误：

fit <- glm(data = mtcars, mpg ~ wt + qsec + am, family = gaussian(link = 'identity')
to_plot(mtcars, fit, mpg, wt)

 Error in grouped_df_impl(data, unname(vars), drop) : 
  Column `indep_variable` is unknown 

基于一些粗略的调试，我发现错误发生在 group_by 步骤中，因此它可能与我在函数中调用列的方式有关。谢谢！

Answer 1

这段代码似乎可以修复它。正如上面的评论者所提到的，传入函数的变量必须包装在“enquo”函数中，然后用 !! 解包。注意 aes() 函数在处理字符串时变为 aes_()。

library(tidyverse)

to_plot <- function(df, model, response_variable, indep_variable) {
  response_variable <- enquo(response_variable)
  indep_variable <- enquo(indep_variable)

  resp_plot <- 
    df %>%
    mutate(model_resp = predict.glm(model, df, type = 'response')) %>%
    group_by(!!indep_variable) %>%
    summarize(actual_response = mean(!!response_variable),
              predicted_response = mean(model_resp)) %>%
    ggplot(aes_(indep_variable)) + 
    geom_line(aes_(x = indep_variable, y = quote(actual_response)), colour = "blue") + 
    geom_line(aes_(x = indep_variable, y = quote(predicted_response)), colour = "red") +
    ylab(label = 'Response')

  return(resp_plot)
}

fit <- glm(data = mtcars, mpg ~ wt + qsec + am, family = gaussian(link = 'identity'))
to_plot(mtcars, fit, mpg, wt)