在函数中使用“marg” - 如何正确引用？

Posted 2023-02-14

【中文标题】在函数中使用“marg” - 如何正确引用？

【英文标题】：Using "marg" in a function - how to enquote correctly?

【发布时间】：2022-01-03 22:43:01

【问题描述】：

想象一下我有一个类似的回归：

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，我有许多具有不同变量的模型来计算边距，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 没有正确引用。我玩过enquo，但没有运气。怎么走？

【参考方案1】：

使用传递的对象命名list可以使用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
...

---

或者可能需要来自dplyr/purrr的lst

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
...

【讨论】：

像魅力一样工作