I've a small function to read in files with certain string using str_subset which works if I pass the argument in quotes but I want to able to do it without. I thought I could do this with curly curly but isn't working.
Working example passing with quotes:
#creating csv file
library(tidyverse)
write_csv(mtcars, "C:\\Users\\testSTACK.csv")
#reading function
read_in_fun <- function(x) {
setwd("C:\\Users")
d <- list.files() #lists all files in folder
file <- d %>%
str_subset(pattern = x)
#read in
df <- read_csv(file)
arg_name <- deparse(substitute(x))
var_name <- paste("df_new", arg_name, sep = "_")
assign(var_name, df, env = .GlobalEnv)
}
read_in_fun("STACK")
#this works, returns df called:
df_new_"STACK"
now if i try to be able to pass with no quotes using curly curly approach:
read_in_fun <- function(x) {
setwd("C:\\Users")
d <- list.files() #lists all files in folder
file <- d %>%
str_subset(pattern = {{x}})
#read in
df <- read_csv(file)
arg_name <- deparse(substitute(x))
var_name <- paste("df_new", arg_name, sep = "_")
assign(var_name, df, env = .GlobalEnv)
}
read_in_fun(STACK)
#Error in type(pattern) : object 'STACK' not found
also tried using enquo
read_in_fun <- function(x) {
x_quo <- enquo(x)
setwd("C:\\Users")
d <- list.files() #lists all files in folder
file <- d %>%
str_subset(pattern = !! as_label(x_quo)) #OR !!(x_quo)
#read in
df <- read_csv(file)
arg_name <- deparse(substitute(x))
var_name <- paste("df_new", arg_name, sep = "_")
assign(var_name, df, env = .GlobalEnv)
}
read_in_fun(STACK)
# Error during wrapup: Quosures can only be unquoted within a quasiquotation context.
My desired output is a df called df_new_STACK. Can curly curly be used in this way? Thanks
Using ensym should work.
read_in_fun <- function(x) {
x_sym <- ensym(x)
d <- list.files()
file <- d %>%
str_subset(pattern = as_label(x_sym))
#read in
df <- read_csv(file)
arg_name <- deparse(substitute(x))
var_name <- paste("df_new", arg_name, sep = "_")
assign(var_name, df, env = .GlobalEnv)
}
read_in_fun(STACK)
df_new_STACK