How to pass a tibble to a function?

[eliza]

Hello,

I cannot figure out how to pass tibbles to a function. I receive the error:

no applicable method for 'extract_' applied to an object of class "character"

I also think there is a better formulation of my regex with a quantifier (e.g. {1,3} for [[:alpha:]]+, but I couldn't figure out how to make it work.

Thanks so much for any help.

#################################
#  data source:
# https://rsf.org/en/ranking/2020
# Reporters without borders accessed rankings
# from 2013-2020 by changing XXXX to year
# https://rsf.org/en/ranking/XXXX
# accessed on May 3, 2020
# sample data from 2 yrs below
#################################



library(tidyverse)
rankings_2013  <- tibble::tribble(~header2013,
                                  "0Vietnam71.78",
                                  "1Finland6.38",
                                  "2Netherlands6.48",
                                  "3Norway6.52",
                                  "4Luxembourg6.68")

rankings_2014  <- tibble::tribble(~header2014,
                                  "1Finland6.40",
                                  "2Netherlands6.46",
                                  "3Norway6.52",
                                  "4Luxembourg6.70",
                                  "5Andorra6.82",
                                  "6Liechtenstein7.02")


fix_df  <- function(df, col, year){
  df %>%
    extract({{col}}, into = c("rank", "country", "score"),
            regex = "(\\d+)([[:alpha:]]+.?[[:alpha:]]+.?[[:alpha:]]+.?[[:alpha:]]+)(\\d+\\.*\\d*)",
            remove = TRUE,
            convert = TRUE) %>%
    mutate(year = as.integer({{year}}))
}


fix_df(rankings_2013, header2013,  2012)
#> # A tibble: 5 x 4
#>    rank country     score  year
#>   <int> <chr>       <dbl> <int>
#> 1     0 Vietnam     71.8   2012
#> 2     1 Finland      6.38  2012
#> 3     2 Netherlands  6.48  2012
#> 4     3 Norway       6.52  2012
#> 5     4 Luxembourg   6.68  2012

fix_df(rankings_2014, header2014,  2013)
#> # A tibble: 6 x 4
#>    rank country       score  year
#>   <int> <chr>         <dbl> <int>
#> 1     1 Finland        6.4   2013
#> 2     2 Netherlands    6.46  2013
#> 3     3 Norway         6.52  2013
#> 4     4 Luxembourg     6.7   2013
#> 5     5 Andorra        6.82  2013
#> 6     6 Liechtenstein  7.02  2013

df_in  <- ls(pattern = "rankings")
df_in
#> [1] "rankings_2013" "rankings_2014"

year  <- 2012:2013
year_in   <- paste(year, sep = ",")
year_in
#> [1] "2012" "2013"

col_in  <- paste0("header", year + 1)
col_in
#> [1] "header2013" "header2014"

# doesn't work
# How can I pass the first argument of df_in to the function?
fix_df(df_in[1], col_in[1],  year_in[1])
#> Warning: `extract_()` is deprecated as of tidyr 1.0.0.
#> Please use `extract()` instead.
#> This warning is displayed once every 8 hours.
#> Call `lifecycle::last_warnings()` to see where this warning was generated.
#> Error in UseMethod("extract_"): no applicable method for 'extract_' applied to an object of class "character"

# does work
fix_df(rankings_2013, col_in[1],  year_in[1])
#> # A tibble: 5 x 4
#>    rank country     score  year
#>   <int> <chr>       <dbl> <int>
#> 1     0 Vietnam     71.8   2012
#> 2     1 Finland      6.38  2012
#> 3     2 Netherlands  6.48  2012
#> 4     3 Norway       6.52  2012
#> 5     4 Luxembourg   6.68  2012

# end goal
# l  <- list(df_in, col_in, year_in)
# pmap_dfr(l, fix_df)

^{Created on 2020-05-07 by the reprex package (v0.3.0)}

[nirgrahamuk]

df_in[1] is not a tibble. It's a character string of the name of a tibble. You can use the get() function to get the tibble with that name though.