Shortcut to parse() and eval() evaluate R expression in a
character string, and turns it into actual R code. This function is targeted
for interaction with external files (where expression is stored in text
format) ; for tidy elements where code expression is generated using
dplyr::mutate(), combined with paste0() ; in for while, map, etc.
loops where character string expression can be indexed or iteratively
generated and evaluated ; objects to be created (using assign, <- or <<- obj)
where the name of the R object is stored in a string. Some issues may occur
when parceval is used in a different environment, such as in a function.
Prefer eval(parse(text = ...) instead.
parceval(...)String character to be parsed and evaluated
Any output generated by the evaluation of the string character.
{
##### Example 1 -------------------------------------------------------------
# Simple assignation will assign 'b' in parceval environment (which is
# associated to a function and different from .GlobalEnv, by definition).
# Double assignation will put 'b' in .GlobalEnv.
# (similar to assign(x = "b",value = 1,envir = .GlobalEnv))
a <- 1
parceval("print(a)")
##### Example 2 -------------------------------------------------------------
# use rowwise to directly use parceval in a tibble, or use a for loop.
library(dplyr)
library(tidyr)
tibble(cars) %>%
mutate(
to_eval = paste0(speed,"/",dist)) %>%
rowwise() %>%
mutate(
eval = parceval(to_eval))
##### Example 3 -------------------------------------------------------------
# parceval can be parcevaled itself!
code_R <-
'as_tibble(cars) %>%
mutate(
to_eval = paste0(speed,"/",dist)) %>%
rowwise() %>%
mutate(
eval = parceval(to_eval))'
cat(code_R)
parceval(code_R)
}
#> Error in eval(parse(text = str_squish(...) %>% str_remove_all("\\\r"))): object 'a' not found