Getting filter within a function to work with tidy evaluation

Question

I'm trying to use dplyr to filter based on a dynamic variable.

I've figured out that to get filter to work, I need to enclose the variable name in parentheses. However, if I program this into a fuction, it does not work properly.

df_ex <- data.frame(a = 1:10, b = 11:20)

param <- quo(a)

# returns df_ex with column a, only, as expected
df_ex %>%
dplyr::select(!!param)

# returns expected df
df_ex %>%
dplyr::filter((!!param)==5)

# Now for the function
testfun <- function(test_df, filt_var){
   filt_var_mod <- quo(filt_var)

   test_df %>%
    dplyr::filter((!!filt_var_mod)==5)
}

# returns empty df, not as expected
testfun(df_ex, "a")

I would like to learn to find the answers to these questions types of questions about dplyr for myself, so please feel free to refer me to the relevant part of the programming vignette

Answer 1

If your function accepts column name as character, then there is no need to quote it, on the other hand you need to convert it to a symbol and evaluate them in the filter function immediately with UQ or !! in the nse syntax:

testfun <- function(test_df, filt_var){
    test_df %>%
        dplyr::filter((!!rlang::sym(filt_var)) == 5)
}

testfun(df_ex, "a")
#  a  b
#1 5 15

---

If you want to type the column names without quotes, then you need enquo, which

takes a symbol referring to a function argument, quotes the R code that was supplied to this argument, captures the environment where the function was called (and thus where the R code was typed), and bundles them in a quosure.

testfun <- function(test_df, filt_var){
    filt_var_mod <- enquo(filt_var)
    test_df %>%
        dplyr::filter((!!filt_var_mod) == 5)
}

testfun(df_ex, a)
#  a  b
#1 5 15

Answer 2

Technically you don't need rlang or tidyeval or tibbles or dplyr for this kind of problem, base R leaves practically no sacred cows with what you can do using quote, eval, parse, and the other NSE tools that are baked in from the bottom up.

Edit: Much more elegant solution proposed by @thelatemail

df_ex <- data.frame(a = 1:10, b = 11:20)

testfun <- function(test_df, filt_var) {
  test_df[test_df[,filt_var] == 5,]
}    

testfun(df_ex, "a")

Returns

  a  b
5 5 15

Just for fun, a data.table option could work as well:

library(data.table)

df_ex <- data.frame(a = 1:10, b = 11:20)

testfun <- function(test_df, filt_var) {
  setDT(test_df,key = filt_var)[.(5)]
}

testfun(df_ex, "a")

Returns:

   a  b
1: 5 15

Answer 3

Sometimes the scoped versions verbs can be used in place of tideval when making simple functions.

For example, if you want to pass the name as a string in your function, filter_at is an option.

Below is what using filter_at looks like with your variable as a string. You pass the variables you want to filter on and then give the predicate function within either all_vars or any_vars. With filtering with a single variable it doesn't matter which of those you use.

filter_at(df_ex, "a", all_vars(. == 5) )

 a  b
1 5 15

filter_at can easily be used in a function.

testfun = function(test_df, filt_var){

    test_df %>%
        dplyr::filter_at(filt_var, all_vars(. == 5) )
}

testfun(df_ex, "a")

  a  b
1 5 15

Answer 4

Base NSE appear to work, too:

testfun2 <- function(test_df, filt_var){
  filt_var_mod <- substitute(filt_var)
  test_df %>% 
    dplyr::filter(eval(filt_var_mod) == 5)
}

testfun2(df_ex, a)


  a  b
1 5 15