Reputation: 733
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?
Upvotes: 4
Views: 464
Reputation: 1693
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
Upvotes: 0
Reputation: 887881
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
Upvotes: 6