difference between enquo() and toString() within a function

Question

I am still a bit confused with the use of enquo and toString. In the example below I basically just try to filter a data frame and sum the rows in the end. I don't really understand why is enquo and toString doing the same for the first thing I want to do (filter --> option 1 and 2 gives the same result) but not for the second thing I wanna do (sum --> option 1 works but option 2 gives me an error). Is it just because I use it within a dplyr pipe?

library(dplyr)
library(tidyverse)


### define dataframe
dataframe_test <- data.frame(
  column_test = c(100,99,99,90,89,50),
  month_test = c("2020-09-01", "2020-09-01","2020-09-01", "2020-09-01","2020-10-01","2020-10-01")
)



test_function <- function(df, df_col_indicator, df_col_month, char_month) {
  
  
  ### define variables for enquo, ensym, toString
  df_col_indicator_enquo <- enquo(df_col_indicator)
  df_col_indicator_ensym <- ensym(df_col_indicator)
  df_col_indicator_toString <- toString(df_col_indicator)
  
  df_col_month_ensym <- ensym(df_col_month)

  
  dataframe2 <- df %>%
    filter(!!df_col_month_ensym == char_month) %>% # filter for month
    slice_max(!!df_col_indicator_ensym, n = 3) %>% # slice top 3 observations
  
    ## two options for filter
    # option 1
    filter(!!df_col_indicator_ensym == df[2, df_col_indicator_toString]) # filter for observations with same observation as second row
    # option 2
    #filter(!!df_col_indicator_ensym == df[2, !!df_col_indicator_enquo])
  
  
  ## two options for sum
  # option 1
  bb <- sum(dataframe2[ , df_col_indicator_toString]) # sum up observations
  
  # option 2
  #bb <- sum(dataframe2[ , !!df_col_indicator_enquo])
  
  return(bb)
  
}


test_function(df = dataframe_test, df_col_indicator = "column_test", df_col_month = "month_test" , char_month = "2020-09-01")

EDIT:

Thank you all for your answers. Hehe, ok I have to admitt that the example is a bit stupid, but I tried to keep it as simple as possible here. My initial problem is actually this one (see below). I basically try to select top 5 numbers of a column. There are three different outcomes. 1) If more than 5 are ==100, then I wanna randomly store 5 oberservation in list(indicator) the other observations in list(asterisk). 2) If not all observations are ==100 but there are ties (5th, 6th place have the same value), I wanna randomly pick those with ties and again put some in list(indicator)the other observations in list(asterisk). 3) If thre are no ties, just pick top 5 observations. My main problem now is if I want to run my function over a loop (with all the columns) at the very bottom. Somehow I always just get the first row as an outcome... I think I somehow don't understand how to propperly set variable names for the function within a loop...?

library(dplyr)
library(tidyverse)

remove(list = ls())


dataframe_test <- data.frame(
  county_name = c("a", "b","c", "d","e", "f", "g", "h"),
  column_test1 = c(100,100,100,100,100,100,50,50),
  column_test2 = c(40,90,50,40,40,100,13,14),
  column_test3 = c(100,90,50,40,30,40,100,50),
  month = c("2020-09-01", "2020-09-01" ,"2020-09-01" ,"2020-09-01" ,"2020-09-01" ,"2020-09-01" ,"2020-08-01","2020-08-01"))


choose_top_5 <- function(df, df_col_indicator, df_col_month, char_month, numb_top, df_col_county) {

  ### enquo / ensym / deparse
  df_col_indicator_enquo <- enquo(df_col_indicator)
  df_col_indicator_ensym <- ensym(df_col_indicator)

  df_col_month_ensym <- ensym(df_col_month)
  df_col_month_enquo = enquo(df_col_month)


  ### filter month and top 5 observations
  df_top <- df %>%
    filter(!!df_col_month_ensym == char_month) %>%
    slice_max(!!df_col_indicator_ensym, n = numb_top) %>%
    select(!!df_col_county, !!df_col_month_ensym, !!df_col_indicator_ensym)


  ### if there are more than "numb_top" values and all equals to 100 --> randomly pick "numb_top"
  if (nrow(df_top) > numb_top &
      sum(df_top[ , df_col_indicator  ]) == 100*nrow(df_top)  ) {

    ## randomly pick "numb_top" out of all
    random_shuffle <- df_top[sample(nrow(df_top)),]
    indicator <- random_shuffle[1:numb_top,]
    asterisk <- random_shuffle[(numb_top+1):nrow(random_shuffle),]

    ## return "numb_top" and put names of others in asterisk
    return_list <- list(indicator, asterisk)


    ### if there are more than "numb_top" values but not all 100 (e.g. 100, 100, 100, 99, 99, 99)
    ## --> pick randomly 99 values
  } else if (nrow(df_top) > numb_top) {

    ### filter for all observations that have the same value as "numb_top"
    df_treshold <- df_top %>%
      filter(!!df_col_indicator_ensym == df_top[numb_top, df_col_indicator])

    ## randomly shuffle the observations
    random_shuffle <- df_treshold[sample(nrow(df_treshold)),]

    ## combine observations again an pick "numb_top"
    combine <- rbind(df_top[1:(nrow(df_top)-nrow(df_treshold)), ], random_shuffle)
    indicator <- combine[1:numb_top,]
    asterisk <- combine[(numb_top+1):nrow(combine),]

    ## return "numb_top" and put names of others in asterisk
    return_list <- list(indicator, asterisk)

    ### if there are not more than "numb_top" values
  } else {


    indicator <- df_top
    asterisk <- NA

    ## return "numb_top", asterisk is NA
    return_list <- list(indicator, asterisk)
  }

  return(return_list)


}



### function for 1 column
a=choose_top_5(df = dataframe_test, df_col_indicator = "column_test3",
             df_col_month = "month", char_month = "2020-09-01", numb_top = 5,
             df_col_county = "county_name")
a



### function over all columns and store in list

all_indicators <- c("column_test1","column_test2","column_test3")

my_list <- list()

for (i in all_indicators) {

  my_list[[i]] <- choose_top_5(df = dataframe_test, df_col_indicator = i,
                                        df_col_month = "month", char_month = "2020-09-01", numb_top = 5,
                                        df_col_county = "county_name")
}

my_list

Answer 1

To add to Allans answer above:

Normally, you cannot use the forcing operator !! inside functions that do not support quasiquotation. However, as Lionel Henry pointed out here and here an upcoming version of {rlang} is likely to contain a function called blast() for quasiquotation with immediate evaluation. Below I use your example. In the last row you can see how !! can be used inside base R functions which themselves do not support quasiquoation.

library(tidyverse)
library(rlang)

### define dataframe
dataframe_test <- data.frame(
  column_test = c(100,99,99,90,89,50),
  month_test = c("2020-09-01", "2020-09-01","2020-09-01", "2020-09-01","2020-10-01","2020-10-01")
)

blast <- function(expr, env = caller_env()) {
  eval_bare(enexpr(expr), env)
}

test_function <- function(df, df_col_indicator, df_col_month, char_month) {
  
  df_col_indicator_enquo <- enquo(df_col_indicator)
  df_col_indicator_ensym <- ensym(df_col_indicator)
  df_col_month_enquo     <- enquo(df_col_month)
  
  temp <- df %>%
    filter(!!df_col_month_enquo == char_month) %>% 
    slice_max(!!df_col_indicator_enquo, n = 3) %>% 
    filter(!!df_col_indicator_enquo == df %>% 
             select(!!df_col_indicator_enquo) %>% 
             pluck(1, 2)) 
  
  # with `blast()` we can use the forcing operator !! inside sum(`$`...)
  blast(sum(`$`(temp, !! df_col_indicator_ensym)))
}

test_function(df = dataframe_test, column_test, month_test , "2020-09-01")
#> [1] 198

^{Created on 2020-11-05 by the reprex package (v0.3.0)}

Answer 2

It's worth pointing out that toString isn't really doing anything here. If you pass a character to toString with no other arguments, it remains unchanged:

test_toString <- function(x) identical(x, toString(x))

test_toString("hello")
#> [1] TRUE

So df_col_indicator_toString could be taken out of your function and replaced with df_col_indicator.

It's also worth pointing out that when you enquo a variable that is a string, then unquote it with the double bang (!!) operator, you are left with the original string, so there is no point using enquo in your function if you are passing a string.

That means that we can simplify your function by doing:

test_function <- function(df, df_col_indicator, df_col_month, char_month) {

  df_col_indicator_ensym <- ensym(df_col_indicator)
  df_col_month_ensym     <- ensym(df_col_month)

  dataframe2 <- df %>%
    filter(!!df_col_month_ensym == char_month) %>% 
    slice_max(!!df_col_indicator_ensym, n = 3) %>% 
    filter(!!df_col_indicator_ensym == df[2, df_col_indicator]) 

  sum(dataframe2[ , df_col_indicator])
}

test_function(df = dataframe_test, "column_test", "month_test" , "2020-09-01")
#> [1] 198

The use of enquo and double-bang (together known as quasiquotation) inside user-defined functions is more commonly used to be able to pass columns without needing quotation marks around column names. For example, if your change your function to:

test_function <- function(df, df_col_indicator, df_col_month, char_month) {

  df_col_indicator_enquo <- enquo(df_col_indicator)
  df_col_month_enquo     <- enquo(df_col_month)

  df %>%
    filter(!!df_col_month_enquo == char_month) %>% 
    slice_max(!!df_col_indicator_enquo, n = 3) %>% 
    filter(!!df_col_indicator_enquo == df %>% 
                                        select(!!df_col_indicator_enquo) %>% 
                                        pluck(1, 2)) %>% 
    summarize(total = sum(!!df_col_indicator_enquo)) %>%
    pluck(1, 1)
}

Then you can do:

test_function(df = dataframe_test, column_test, month_test , "2020-09-01")
#> [1] 198

To answer your broader question about why you are getting the error, then the answer is that you can only use this syntax inside a quasiquotation context. Don't confuse this with the pipe; there are certain tidyverse functions where you can use quasiquotation, but in base R functions you can't, which is why method 2 doesn't work. Essentially, the function in which you use the !! operator has to be written specifically to handle quasiquotation; base R isn't written that way.

As a final point, your function logic seems a bit unusual, and you should check that this is really what you are trying to do. You are filtering the month, selecting the top three entries, but then filtering that according to whether the top three entries are equal to the value in the second row of your original data frame, before summing. This seems like a strange thing to want to do.