Using rlang Package to Parse Quoted Argument










3














I'm hoping to string-split a single argument into two arguments and use each in different sections of a function.



Is it possible to do this using quasiquotation (!!) or other rlang functions?



Thanks!



Data:



person <- tibble(id = 1, age = 20)
friends <- tibble(id = c(2, 3, 4, 5), age = c(48, 29, 20, 48))


(unfunctional) Function:



different_age_friends <- function(condition, person = person, friends = friends ) 

person <- person
friends <- friends

condition <- str_split(condition, " ~ ", simplify = T)
condition_statement <- condition[1]
filter_statement <- condition[2]

if(!!condition_statement)
different_age_friends <- friends %>%
filter(!!filter_statement)


return(return_same_age_friends)



Call:



different_age_friends(condition = "age == 20 ~ age == 48")


Desired Output



id age
2 48
5 48









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  • 1




    This doesn't fix the issue, but you've got condition = TRUE where you want condition == TRUE, or just condition, to do a comparison
    – camille
    Nov 13 '18 at 2:32










  • Also, I'm unclear on the difference between the condition and the filter. I don't get what's substantially different between these two types of statements. Are you trying to test the condition against person, and then if true, use the second piece to filter friends?
    – camille
    Nov 13 '18 at 2:36










  • yes exactly, thanks for clarifying!
    – CFB
    Nov 13 '18 at 4:19
















3














I'm hoping to string-split a single argument into two arguments and use each in different sections of a function.



Is it possible to do this using quasiquotation (!!) or other rlang functions?



Thanks!



Data:



person <- tibble(id = 1, age = 20)
friends <- tibble(id = c(2, 3, 4, 5), age = c(48, 29, 20, 48))


(unfunctional) Function:



different_age_friends <- function(condition, person = person, friends = friends ) 

person <- person
friends <- friends

condition <- str_split(condition, " ~ ", simplify = T)
condition_statement <- condition[1]
filter_statement <- condition[2]

if(!!condition_statement)
different_age_friends <- friends %>%
filter(!!filter_statement)


return(return_same_age_friends)



Call:



different_age_friends(condition = "age == 20 ~ age == 48")


Desired Output



id age
2 48
5 48









share|improve this question



















  • 1




    This doesn't fix the issue, but you've got condition = TRUE where you want condition == TRUE, or just condition, to do a comparison
    – camille
    Nov 13 '18 at 2:32










  • Also, I'm unclear on the difference between the condition and the filter. I don't get what's substantially different between these two types of statements. Are you trying to test the condition against person, and then if true, use the second piece to filter friends?
    – camille
    Nov 13 '18 at 2:36










  • yes exactly, thanks for clarifying!
    – CFB
    Nov 13 '18 at 4:19














3












3








3







I'm hoping to string-split a single argument into two arguments and use each in different sections of a function.



Is it possible to do this using quasiquotation (!!) or other rlang functions?



Thanks!



Data:



person <- tibble(id = 1, age = 20)
friends <- tibble(id = c(2, 3, 4, 5), age = c(48, 29, 20, 48))


(unfunctional) Function:



different_age_friends <- function(condition, person = person, friends = friends ) 

person <- person
friends <- friends

condition <- str_split(condition, " ~ ", simplify = T)
condition_statement <- condition[1]
filter_statement <- condition[2]

if(!!condition_statement)
different_age_friends <- friends %>%
filter(!!filter_statement)


return(return_same_age_friends)



Call:



different_age_friends(condition = "age == 20 ~ age == 48")


Desired Output



id age
2 48
5 48









share|improve this question















I'm hoping to string-split a single argument into two arguments and use each in different sections of a function.



Is it possible to do this using quasiquotation (!!) or other rlang functions?



Thanks!



Data:



person <- tibble(id = 1, age = 20)
friends <- tibble(id = c(2, 3, 4, 5), age = c(48, 29, 20, 48))


(unfunctional) Function:



different_age_friends <- function(condition, person = person, friends = friends ) 

person <- person
friends <- friends

condition <- str_split(condition, " ~ ", simplify = T)
condition_statement <- condition[1]
filter_statement <- condition[2]

if(!!condition_statement)
different_age_friends <- friends %>%
filter(!!filter_statement)


return(return_same_age_friends)



Call:



different_age_friends(condition = "age == 20 ~ age == 48")


Desired Output



id age
2 48
5 48






r tidyverse rlang






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 '18 at 4:17

























asked Nov 12 '18 at 23:45









CFB

544




544







  • 1




    This doesn't fix the issue, but you've got condition = TRUE where you want condition == TRUE, or just condition, to do a comparison
    – camille
    Nov 13 '18 at 2:32










  • Also, I'm unclear on the difference between the condition and the filter. I don't get what's substantially different between these two types of statements. Are you trying to test the condition against person, and then if true, use the second piece to filter friends?
    – camille
    Nov 13 '18 at 2:36










  • yes exactly, thanks for clarifying!
    – CFB
    Nov 13 '18 at 4:19













  • 1




    This doesn't fix the issue, but you've got condition = TRUE where you want condition == TRUE, or just condition, to do a comparison
    – camille
    Nov 13 '18 at 2:32










  • Also, I'm unclear on the difference between the condition and the filter. I don't get what's substantially different between these two types of statements. Are you trying to test the condition against person, and then if true, use the second piece to filter friends?
    – camille
    Nov 13 '18 at 2:36










  • yes exactly, thanks for clarifying!
    – CFB
    Nov 13 '18 at 4:19








1




1




This doesn't fix the issue, but you've got condition = TRUE where you want condition == TRUE, or just condition, to do a comparison
– camille
Nov 13 '18 at 2:32




This doesn't fix the issue, but you've got condition = TRUE where you want condition == TRUE, or just condition, to do a comparison
– camille
Nov 13 '18 at 2:32












Also, I'm unclear on the difference between the condition and the filter. I don't get what's substantially different between these two types of statements. Are you trying to test the condition against person, and then if true, use the second piece to filter friends?
– camille
Nov 13 '18 at 2:36




Also, I'm unclear on the difference between the condition and the filter. I don't get what's substantially different between these two types of statements. Are you trying to test the condition against person, and then if true, use the second piece to filter friends?
– camille
Nov 13 '18 at 2:36












yes exactly, thanks for clarifying!
– CFB
Nov 13 '18 at 4:19





yes exactly, thanks for clarifying!
– CFB
Nov 13 '18 at 4:19













1 Answer
1






active

oldest

votes


















2














Use rlang::parse_expr to convert strings to expressions and eval to evaluate them. eval() allows you to provide context for the expression in its second argument, where we supply the person data frame to it. In case of filter, the context is already understood to be the dataframe on the left-hand side of the %>% pipe.



Another difference in how we handle the two expressions is that filter() has an additional internal layer of quasiquoation. Since you already have an expression, you don't need it to be quoted again, so you would use !! to unquote it.



different_age_friends <- function(condition, p = person, f = friends) 

stmts <- str_split(condition, " ~ ")[[1]] %>% map( rlang::parse_expr )

if( eval(stmts[[1]], p) ) # Effectively: eval(age == 20, person)
f %>% filter(!!stmts[[2]]) # Effectively: friends %>% filter(age == 48)
else
f


different_age_friends(condition = "age == 20 ~ age == 48")
# # A tibble: 2 x 2
# id age
# <dbl> <dbl>
# 1 2 48
# 2 5 48


Minor note:



  • You didn't provide a value for different_age_friends when the condition is false. I made an assumption that in this case, the entire friends list is to be returned.





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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Use rlang::parse_expr to convert strings to expressions and eval to evaluate them. eval() allows you to provide context for the expression in its second argument, where we supply the person data frame to it. In case of filter, the context is already understood to be the dataframe on the left-hand side of the %>% pipe.



    Another difference in how we handle the two expressions is that filter() has an additional internal layer of quasiquoation. Since you already have an expression, you don't need it to be quoted again, so you would use !! to unquote it.



    different_age_friends <- function(condition, p = person, f = friends) 

    stmts <- str_split(condition, " ~ ")[[1]] %>% map( rlang::parse_expr )

    if( eval(stmts[[1]], p) ) # Effectively: eval(age == 20, person)
    f %>% filter(!!stmts[[2]]) # Effectively: friends %>% filter(age == 48)
    else
    f


    different_age_friends(condition = "age == 20 ~ age == 48")
    # # A tibble: 2 x 2
    # id age
    # <dbl> <dbl>
    # 1 2 48
    # 2 5 48


    Minor note:



    • You didn't provide a value for different_age_friends when the condition is false. I made an assumption that in this case, the entire friends list is to be returned.





    share|improve this answer



























      2














      Use rlang::parse_expr to convert strings to expressions and eval to evaluate them. eval() allows you to provide context for the expression in its second argument, where we supply the person data frame to it. In case of filter, the context is already understood to be the dataframe on the left-hand side of the %>% pipe.



      Another difference in how we handle the two expressions is that filter() has an additional internal layer of quasiquoation. Since you already have an expression, you don't need it to be quoted again, so you would use !! to unquote it.



      different_age_friends <- function(condition, p = person, f = friends) 

      stmts <- str_split(condition, " ~ ")[[1]] %>% map( rlang::parse_expr )

      if( eval(stmts[[1]], p) ) # Effectively: eval(age == 20, person)
      f %>% filter(!!stmts[[2]]) # Effectively: friends %>% filter(age == 48)
      else
      f


      different_age_friends(condition = "age == 20 ~ age == 48")
      # # A tibble: 2 x 2
      # id age
      # <dbl> <dbl>
      # 1 2 48
      # 2 5 48


      Minor note:



      • You didn't provide a value for different_age_friends when the condition is false. I made an assumption that in this case, the entire friends list is to be returned.





      share|improve this answer

























        2












        2








        2






        Use rlang::parse_expr to convert strings to expressions and eval to evaluate them. eval() allows you to provide context for the expression in its second argument, where we supply the person data frame to it. In case of filter, the context is already understood to be the dataframe on the left-hand side of the %>% pipe.



        Another difference in how we handle the two expressions is that filter() has an additional internal layer of quasiquoation. Since you already have an expression, you don't need it to be quoted again, so you would use !! to unquote it.



        different_age_friends <- function(condition, p = person, f = friends) 

        stmts <- str_split(condition, " ~ ")[[1]] %>% map( rlang::parse_expr )

        if( eval(stmts[[1]], p) ) # Effectively: eval(age == 20, person)
        f %>% filter(!!stmts[[2]]) # Effectively: friends %>% filter(age == 48)
        else
        f


        different_age_friends(condition = "age == 20 ~ age == 48")
        # # A tibble: 2 x 2
        # id age
        # <dbl> <dbl>
        # 1 2 48
        # 2 5 48


        Minor note:



        • You didn't provide a value for different_age_friends when the condition is false. I made an assumption that in this case, the entire friends list is to be returned.





        share|improve this answer














        Use rlang::parse_expr to convert strings to expressions and eval to evaluate them. eval() allows you to provide context for the expression in its second argument, where we supply the person data frame to it. In case of filter, the context is already understood to be the dataframe on the left-hand side of the %>% pipe.



        Another difference in how we handle the two expressions is that filter() has an additional internal layer of quasiquoation. Since you already have an expression, you don't need it to be quoted again, so you would use !! to unquote it.



        different_age_friends <- function(condition, p = person, f = friends) 

        stmts <- str_split(condition, " ~ ")[[1]] %>% map( rlang::parse_expr )

        if( eval(stmts[[1]], p) ) # Effectively: eval(age == 20, person)
        f %>% filter(!!stmts[[2]]) # Effectively: friends %>% filter(age == 48)
        else
        f


        different_age_friends(condition = "age == 20 ~ age == 48")
        # # A tibble: 2 x 2
        # id age
        # <dbl> <dbl>
        # 1 2 48
        # 2 5 48


        Minor note:



        • You didn't provide a value for different_age_friends when the condition is false. I made an assumption that in this case, the entire friends list is to be returned.






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 13 '18 at 4:30

























        answered Nov 13 '18 at 4:02









        Artem Sokolov

        4,70221936




        4,70221936



























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