使用 tidyeval 进行程序化回归建模

Question

我正在尝试使用 tidyeval 进行编程。

我想为所选结果变量的运行逻辑回归模型编写一个函数：

library(tidyverse)
set.seed(1234)

df <- tibble(id = 1:1000,
             group = sample(c("Group 1", "Group 2", "Group 3"), 1000, replace = TRUE),
             died = sample(c(0,1), 1000, replace = TRUE))

myfunc <- function(data, outcome){

enquo_var <- enquo(outcome)

fit <- tidy(glm(!!enquo_var ~ group, data=data, 
                family = binomial(link = "logit")), 
                exponentiate = TRUE, conf.int=TRUE)

fit
}


myfunc(df, died)

但是得到：

Error in !enquo_outcome : invalid argument type

（注意真实场景涉及更复杂的功能）。

这可能吗？

Answer 1

我们需要为 glm 创建一个公式来提取它。一种选择是 paste

myfunc <- function(data, outcome){
  enquo_var <- enquo(outcome)
   fit <- tidy(glm(paste(quo_name(enquo_var), "group", sep="~"), data=data, 
                family = binomial(link = "logit")), 
                exponentiate = TRUE, conf.int=TRUE)

fit
}

myfunc(df, died)
#         term  estimate std.error  statistic    p.value  conf.low conf.high
#1  (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185  1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512  1.253959
#3 groupGroup 3 1.3692735 0.1557241  2.0181864 0.04357185 1.0095739  1.859403

如果我们还需要使用tidyverse函数

myfunc <- function(data, outcome){

  quo_var <- quo_name(enquo(outcome))

   fit <- tidy(glm(rlang::expr(!! rlang::sym(quo_var) ~ group), data=data, 
            family = binomial(link = "logit")), 
            exponentiate = TRUE, conf.int=TRUE)

 fit
}

myfunc(df, died)
#           term  estimate std.error  statistic    p.value  conf.low conf.high
#1  (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185  1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512  1.253959
#3 groupGroup 3 1.3692735 0.1557241  2.0181864 0.04357185 1.0095739  1.859403

或者@lionel在评论中提到的get_expr可以使用

myfunc <- function(data, outcome){

  quo_var <- enquo(outcome)

   fit <- tidy(glm(rlang::expr(!! rlang::get_expr(quo_var) ~ group), data=data, 
            family = binomial(link = "logit")), 
            exponentiate = TRUE, conf.int=TRUE)

 fit
}

myfunc(df, died)
#         term  estimate std.error  statistic    p.value  conf.low conf.high
#1  (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185  1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512  1.253959
#3 groupGroup 3 1.3692735 0.1557241  2.0181864 0.04357185 1.0095739  1.859403

或者@lionel 建议的一种更紧凑的方法，它避免了 enquo/quo_name/sym 转换，而是直接采用 enexpr

中的参数

 myfunc <- function(data, outcome){



   fit <- tidy(glm(rlang::expr(!! rlang::enexpr(outcome) ~ group), data=data, 
            family = binomial(link = "logit")), 
            exponentiate = TRUE, conf.int=TRUE)

 fit
}

myfunc(df, died)
#         term  estimate std.error  statistic    p.value  conf.low conf.high
#1  (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185  1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512  1.253959
#3 groupGroup 3 1.3692735 0.1557241  2.0181864 0.04357185 1.0095739  1.859403

Answer 2

Base NSE 似乎也有效：

library(broom)
myfunc <- function(data, outcome){
  outcome_subst <- substitute(outcome)
  fit <- tidy(glm(paste(as.name(outcome_subst), "group", sep="~"), data=data, 
                  family = binomial(link = "logit")), 
              exponentiate = TRUE, conf.int=TRUE)

  fit
}


myfunc(df, died)



         term  estimate std.error statistic    p.value  conf.low conf.high
1  (Intercept) 0.8238636 0.1121528 -1.727556 0.08406792 0.6606245  1.025838
2 groupGroup 2 1.2587484 0.1571734  1.464102 0.14316606 0.9253116  1.713937
3 groupGroup 3 1.2490778 0.1550546  1.434369 0.15146698 0.9220209  1.693699

使用 tidyeval 进行程序化回归建模

Programmatic regression modelling with tidyeval

r

dplyr

rlang

tidyeval