automatically name dataframe based on file name

Question

If I observe a convention of saving my file paths as variables with the common prefix "file_", it seems I could create a wrapper function for read_rds() that would name my read files based on whatever text came after "file_" in the file path's name.

I run into trouble when I evaluate the name I want the read file to take.

library(here)
library(readr)
library(stringr)

file_survey <- here("my_survey_2019.rds")

my_read_rds <- function(file){
  name <- deparse(substitute(file))
  name <- stringr::str_remove(name, "^file_")
  eval(name) <- readr::read_rds(file) # Does not work
}

my_read_rds(file_survey) # would ideally create a dataframe named `survey`

This type of behavior isn't encouraged in R. Functions shouldn't have "side effects" or create variables outside their scope. The "normal" way to do this is survey <- read_rds(file). This makes functions much easier to deal with and to be called from inside other functions. — MrFlick
– MrFlick, Commented Apr 26, 2019 at 17:01
@MrFlick "Side effect" is a functional programming terminology, but it's not the only way to program in R. For example, data.table does not follow a functional programming paradigm, yet it's one of the most popular libraries. People should program the way that they want to, if it makes sense to them or if they have a system. — thc
– thc, Commented Apr 26, 2019 at 17:13

thc · Accepted Answer · 2019-04-26 17:16:05Z

1

You can use assign.

my_read_rds <- function(file){
  name <- deparse(substitute(file))
  name <- stringr::str_remove(name, "^file_")
  assign(name, readr::read_rds(file), envir=globalenv())
}

edited Apr 26, 2019 at 17:16

answered Apr 26, 2019 at 16:58

thc

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