I need to select variables dynamically on multiple expressions. Consider the following example:
library(tidyverse)
set.seed(42)
df <- tibble(
grey_dog = runif(n = 69),
white_bear = runif(n = 69),
blue_oyster = runif(n = 69),
white_lobster = runif(n = 69),
green_dog = runif(n = 69)
)
df %>%
dplyr::select(
(contains("dog") & contains("green")) |
(contains("white") & contains("bear"))
)
Instead of explicitly selecting, I have vectors containing the information I want to base my selection on:
x <- c(green = "dog", white = "bear")
So I was hoping to concatenate a string that can be used as a tidy-select:
s <- paste0("(", paste0("contains(", names(x),") & contains(", x, ")"), ")", collapse = " | ")
dplyr::select(df, s)
This fails with:
Error: Can't subset columns that don't exist.
x Column `(contains(green) & contains(dog)) | (contains(white) & contains(bear))` doesn't exist.
Run `rlang::last_error()` to see where the error occurred.
Any ideas on how to accomplish this?
You can't just pass in a string. A string is not the same as an expression. One way is to use purrr and rlang to build the expression and then inject that into the select
library(purrr)
library(rlang)
query <- map2(
map(x, ~expr(contains(!!.x))),
map(names(x), ~expr(contains(!!.x))),
~expr((!!.x & !!.y))) %>%
reduce(~expr(!!.x | !!.y))
dplyr::select(df, !!query)
Though if you really wanted to build the code as a string, then you would just need to parse that string into an expression first using rlang::parse_expr. You just need to add some quotes to your string so it exactly matches the code you used before
s <- paste0("(", paste0("contains(\"", names(x),"\") & contains(\"", x, "\")"), ")", collapse = " | ")
dplyr::select(df, !!rlang::parse_expr(s))