I am trying to implement advice I am finding in the web but I am halfway where I want to go.
Here is a reproducible example:
library(tidyverse)
library(dplyr)
library(rlang)
data(mtcars)
filter_expr = "am == 1"
mutate_expr = "gear_carb = gear*carb"
select_expr = "mpg , cyl"
mtcars %>% filter_(filter_expr) %>% mutate_(mutate_expr) %>% select_(select_expr)
The filter expression works fine.
The mutate expression works as well but the new variable has the name gear_carb = gear*carb instead of the intended gear_carb.
Finally, the select expression returns an exception.
The filter () function in dplyr (and other similar functions from the package) use something called non-standard evaluation (NSE). In NSE, names are treated as string literals.
Convert a String to an Expression in R Programming – parse () Function parse () function in R Language is used to convert an object of character class to an object of expression class. Syntax: parse (text = character)
Last Updated : 24 Jun, 2020. parse () function in R Language is used to convert an object of character class to an object of expression class. Syntax: parse (text = character) Parameters: character: Object of character class. Example 1: x <- "sin (pi / 2)" class(x)
If we don’t want to pass a string but a name instead, the tidyverse has recently introduced a { {}} (#curly-curly’) operator for tidy evaluation. See now we can pass the variable name climbing to the group_by function using the { {}} operator.
As mentioned in the comments, the underscore versions of dplyr verbs are now deprecated. The correct approach is to use quasiquotation.
To address your issue with select
, you simply need to modify select_expr
to contain multiple expressions:
## I renamed your variables to *_str because they are, well, strings.
filter_str <- "am == 1"
mutate_str <- "gear_carb = gear*carb"
select_str <- "mpg; cyl" # Note the ;
Use rlang::parse_expr
to convert these strings to unevaluated expressions:
## Notice the plural parse_exprs, which parses a list of expressions
filter_expr <- rlang::parse_expr( filter_str )
mutate_expr <- rlang::parse_expr( mutate_str )
select_expr <- rlang::parse_exprs( select_str )
Given the unevaluated expressions, we can now pass them to the dplyr
verbs. Writing filter( filter_expr )
won't work because filter
will look for a column named filter_expr
in your data frame. Instead, we want to access the expression stored inside filter_expr
. To do this we use the !!
operator to let dplyr
verbs know that the argument should be expanded to its contents (which is the unevaluated expressions we are interested in):
mtcars %>% filter( !!filter_expr )
# mpg cyl disp hp drat wt qsec vs am gear carb
# 1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
# 2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
# 3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
# 4 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
mtcars %>% mutate( !!mutate_expr )
# mpg cyl disp hp drat wt qsec vs am gear carb gear_carb = gear * carb
# 1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 16
# 2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 16
# 3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 4
# 4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 3
In case of select
, we have multiple expressions, which is handled by !!!
instead:
mtcars %>% select( !!!select_expr )
# mpg cyl
# Mazda RX4 21.0 6
# Mazda RX4 Wag 21.0 6
# Datsun 710 22.8 4
P.S. It's also worth mentioning that select
works directly with string vectors, without having to rlang::parse_expr()
them first:
mtcars %>% select( c("mpg", "cyl") )
# mpg cyl
# Mazda RX4 21.0 6
# Mazda RX4 Wag 21.0 6
# Datsun 710 22.8 4
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With