I'm trying to better understand tidy evaluation and the use of rlang, but there's a specific use case I can't figure out. Let's say I want to run a linear mixed model and save the output of one predictor that interests me. Running this code outside of a function would look like this:
library(tidyverse)
library(lme4)
library(lmerTest)
library(broom.mixed)
tibble %>%
lmer(y ~ x*z + (1|ID), data = .) %>%
tidy() %>%
filter(term == "x:z_2") %>%
select(term, estimate, p.value) %>%
mutate(var = "y") %>%
relocate(var)
I want to wrap this operation in a function where I can specify the dependent variable and the term by which I filter. I’ve managed to call the linear mixed model in this way:
between_group_lmm <- function(df, yvar) {
ysym <- rlang::ensym(yvar)
# Construct and inject the formula dynamically into lmer
rlang::inject(
lmer(!!ysym ~ time * group + (1 | sno), data = df)
) %>%
tidy()
}
However, I don’t understand how to use tidy evaluation for the function’s arguments that I want to pass as strings. Additionally, I’d prefer to specify the filtering value (e.g., "x:z_2" in my example) without using quotation marks, so that both the dependent variable and filtering term can be called consistently without parentheses.
desired outcome
between_group_lmm(df, y, x:z_2)
How can I use tidy evaluation with rlang to handle both the dependent variable (for both use cases) and filtering term as function arguments?
You can use deparse(substitute(yvar)) to capture the expression as a string:
between_group_lmm <- function(df, yvar, filter_var) {
ysym <- rlang::ensym(yvar)
filter_var <- deparse(substitute(filter_var))
yvar <- deparse(substitute(yvar))
rlang::inject(lmer(!!ysym ~ time * group + (1 | sno), data = df)) %>%
tidy() %>%
filter(term == filter_var) %>%
select(term, estimate, p.value) %>%
mutate(var = yvar) %>%
relocate(var)
}
Or, if you want to stick to rlang syntax, you can use enquo and as_label:
between_group_lmm <- function(df, yvar, filter_var) {
ysym <- rlang::ensym(yvar)
yvar <- rlang::enquo(yvar) %>% rlang::as_label()
filter_var <- rlang::enquo(filter_var) %>% rlang::as_label()
rlang::inject(lmer(!!ysym ~ time * group + (1 | sno), data = df)) %>%
tidy() %>%
filter(term == filter_var) %>%
select(term, estimate, p.value) %>%
mutate(var = yvar) %>%
relocate(var)
}
Both of these versions do the same thing, so that if your data is something like this:
set.seed(1)
tib <- tibble(sno = rep(1:10, 10),
time = rnorm(100, 10, 2),
group = factor(sample(10, 100, TRUE)),
yvar = time * as.numeric(group)/180 + sno + rnorm(100))
Then you can do
between_group_lmm(tib, outcome, time:group6)
#> # A tibble: 1 x 4
#> var term estimate p.value
#> <chr> <chr> <dbl> <dbl>
#> 1 y time:group6 0.0891 0.864