在函数中使用 "marg" - 如何正确引用？

Question

假设我有一个像这样的回归：

regression <- glm(mpg ~ am * hp, data = mtcars)

我可以使用 modmarg 包轻松计算边际效应：

library(modmarg)
margins <- marg(
    regression, var_interest = "am",
    at = list("hp" = 52:110),
    type = "effects")

但是，我有许多模型具有不同的变量来计算边距，所以我想将其放入一个函数中：

plotting_marg <- function(reg, cov) {
  margins <- marg(
    reg, var_interest = "truth",
    at = list(cov = 52:110),
    type = "effects") 
}
plotting_marg(regression, "hp", -3, 3)

Error in .marg(mod = mod, var_interest = var_interest, data = data, weights = weights,  : 
  var_interest %in% names(data) is not TRUE

这不起作用，大概是因为 cov 没有被正确引用。我玩过 enquo，但运气不好。去这里的路是什么？

Answer 1

用传递的对象命名 list 可以用 setNames

完成

plotting_marg <- function(reg, cov) {
  marg(
    reg, var_interest = "am",
    at = setNames(list(52:110), cov),
    type = "effects") 
}

-测试

> plotting_marg(regression, "hp")
$`hp = 52`
   Label   Margin Standard.Error Test.Stat    P.Value Lower CI (95%) Upper CI (95%)
1 am = 0 0.000000       0.000000       NaN        NaN       0.000000       0.000000
2 am = 1 5.238604       1.918006  2.731276 0.01079466       1.309746       9.167462

$`hp = 53`
   Label   Margin Standard.Error Test.Stat    P.Value Lower CI (95%) Upper CI (95%)
1 am = 0 0.000000       0.000000       NaN        NaN       0.000000       0.000000
2 am = 1 5.239007       1.904537  2.750803 0.01030278       1.337739       9.140274

$`hp = 54`
   Label   Margin Standard.Error Test.Stat     P.Value Lower CI (95%) Upper CI (95%)
1 am = 0 0.000000       0.000000       NaN         NaN       0.000000       0.000000
2 am = 1 5.239409       1.891116  2.770539 0.009827057       1.365635       9.113184
...

或者可能需要 lst 来自 dplyr/purrr

plotting_marg <- function(reg, cov) {
  marg(
    reg, var_interest = "am",
    at = dplyr::lst(!!cov := 52:110),
    type = "effects") 
}

-测试

> plotting_marg(regression, "hp")
$`hp = 52`
   Label   Margin Standard.Error Test.Stat    P.Value Lower CI (95%) Upper CI (95%)
1 am = 0 0.000000       0.000000       NaN        NaN       0.000000       0.000000
2 am = 1 5.238604       1.918006  2.731276 0.01079466       1.309746       9.167462

$`hp = 53`
   Label   Margin Standard.Error Test.Stat    P.Value Lower CI (95%) Upper CI (95%)
1 am = 0 0.000000       0.000000       NaN        NaN       0.000000       0.000000
2 am = 1 5.239007       1.904537  2.750803 0.01030278       1.337739       9.140274

$`hp = 54`
   Label   Margin Standard.Error Test.Stat     P.Value Lower CI (95%) Upper CI (95%)
1 am = 0 0.000000       0.000000       NaN         NaN       0.000000       0.000000
2 am = 1 5.239409       1.891116  2.770539 0.009827057       1.365635       9.113184
...

在函数中使用 "marg" - 如何正确引用？

Using "marg" in a function - how to enquote correctly?

r

regression