Programmatic regression modelling with tidyeval

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I am trying to get my head around programming using tidyeval.

I want to write a function to run logistic regression models for selected outcome variables:

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)

But get:

Error in !enquo_outcome : invalid argument type

(note real scenario involves more complex function).

Is this possible?

tidyeval rlang dplyr r
2 Answers

We need to create a formula for glm to pick it up. One option is 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

If we also need to use the tidyverse functions

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

Or as @lionel mentioned in the comments get_expr can be used

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

Or a more compact approach suggested by @lionel which avoids the enquo/quo_name/sym conversion instead directly takes the argument in enexpr

 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
user_37322710

Base NSE appears to work, too:

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
user_44342630
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