在R中使用lapply从函数调用中的向量返回实际函数参数

请考虑以下几点.

我想使用lapply()来将存储在字符向量中的几个函数参数应用于其他函数.一个最小的可重复示例可以是将两个或多个"族"应用于glm()函数.请注意，该示例对于应用此类族可能毫无意义，仅用于说明目的.

以下内容摘自?glm()中的示例

counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
data.frame(treatment, outcome, counts) # showing data

我们现在可以使用"高斯"或"泊松"族运行GLM

glm(counts ~ outcome + treatment, family = "gaussian")
glm(counts ~ outcome + treatment, family = "poisson")

这也可以通过创建具有以下族名称的字符向量来"自动"完成:

families <- c("poisson", "gaussian")

在lapply()函数中使用这个.

但一旦运行，返回的函数调用不再返回族名称，而是返回匿名函数参数x.

lapply(families, function(x) glm(counts ~ outcome + treatment, family = x))
#> [[1]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = x)
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   3.045e+00   -4.543e-01   -2.930e-01   -3.242e-16   -2.148e-16  
#> 
#> Degrees of Freedom: 8 Total (i.e. Null);  4 Residual
#> Null Deviance:       10.58 
#> Residual Deviance: 5.129     AIC: 56.76
#> 
#> [[2]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = x)
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   2.100e+01   -7.667e+00   -5.333e+00    2.221e-15    2.971e-15  
#> 
#> Degrees of Freedom: 8 Total (i.e. Null);  4 Residual
#> Null Deviance:       176 
#> Residual Deviance: 83.33     AIC: 57.57

Question:

Desired outcome:

#> [[1]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = "gaussian")
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   3.045e+00   -4.543e-01   -2.930e-01   -3.242e-16   -2.148e-16  
#> 
#> Degrees of Freedom: 8 Total (i.e. Null);  4 Residual
#> Null Deviance:       10.58 
#> Residual Deviance: 5.129     AIC: 56.76
#> 
#> [[2]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = "poisson")
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   2.100e+01   -7.667e+00   -5.333e+00    2.221e-15    2.971e-15  
#> 
#> Degrees of Freedom: 8 Total (i.e. Null);  4 Residual
#> Null Deviance:       176 
#> Residual Deviance: 83.33     AIC: 57.57

我try 了这里建议的eval(bquote(x)):R: Passing named function arguments from vector，但这不起作用.请参阅:

lapply(families, function(x) glm(counts ~ outcome + treatment, family = eval(bquote(x))))
#> [[1]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = eval(bquote(x)))
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   3.045e+00   -4.543e-01   -2.930e-01   -3.242e-16   -2.148e-16  
#> 
#> Degrees of Freedom: 8 Total (i.e. Null);  4 Residual
#> Null Deviance:       10.58 
#> Residual Deviance: 5.129     AIC: 56.76
#> 
#> [[2]]
#> 
#> Call:  glm(formula = counts ~ outcome + treatment, family = eval(bquote(x)))
#> 
#> Coefficients:
#> (Intercept)     outcome2     outcome3   treatment2   treatment3  
#>   2.100e+01   -7.667e+00   -5.333e+00    2.221e-15    2.971e-15  
#> 
#> Degrees of Freedom: 8 Total (i.e. Null);  4 Residual
#> Null Deviance:       176 
#> Residual Deviance: 83.33     AIC: 57.57

^{由reprex package(v2.0.1)于2022-07-22创建}

非常感谢.