在R中用lapply调用glm时，如何保留向量中的family参数名称

方法1：用bquote()构造并执行调用 #

通过bquote()将x替换为实际的family字符串，再用eval()执行，确保模型调用记录真实参数：

lapply(families, function(x) {
  eval(bquote(glm(counts ~ outcome + treatment, family = .(x))))
})

方法2：手动修改模型的call属性 #

先运行模型，再修改返回对象的call部分，替换x为实际family名称：

lapply(families, function(x) {
  mod <- glm(counts ~ outcome + treatment, family = x)
  # 将call中的family参数替换为带引号的实际名称
  mod$call$family <- str2lang(sprintf('"%s"', x))
  mod
})

方法3：tidyverse风格的inject()实现 #

如果使用tidyverse工具，可结合purrr::map()和rlang::inject()注入真实参数：

library(purrr)
library(rlang)

map(families, function(x) {
  inject(glm(counts ~ outcome + treatment, family = !!x))
})

验证结果 #

以上任意方法运行后，返回的模型调用都会显示真实的family名称，例如：

#> [[1]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = "poisson")
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   3.045e+00   -4.543e-01   -2.930e-01   -3.242e-16   -2.148e-16  
#> ...
#> 
#> [[2]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = "gaussian")
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   2.100e+01   -7.667e+00   -5.333e+00    2.221e-15    2.971e-15  
#> ...

内容的提问来源于stack exchange，提问作者Frederick