Reputation: 2283
I have a function that I want to wrap another function around, passing the arguments through as the ...
arguments parameters. I'm having trouble learning how to structure the underlying function call with lazyeval
Here's a decent MWE,
library(dplyr)
pythag <- function(a, b){
sqrt(a^2 + b^2)
}
pythag_wrapper <- function(data, ...){
dplyr::mutate_(data,
root = lazyeval::interp(~pythag(x), x = ...)
)
}
In this my pythag_wrapper
will do some additional data munging. and pythag
in my case has many more than two arguments. The function works just fine, and as intended.
test_data <- dplyr::data_frame(a = runif(10), b = runif(10), c = runif(10))
test_data %>%
dplyr::mutate(
root = pythag(a = b, b = c)
)
## # A tibble: 10 × 4
## a b c root
## <dbl> <dbl> <dbl> <dbl>
## 1 0.19805337 0.05567241 0.9956758 0.9972311
## 2 0.22642799 0.18871552 0.8690659 0.8893195
## 3 0.09352032 0.57328658 0.7475573 0.9420719
## 4 0.40589832 0.71270806 0.8014196 1.0724860
## 5 0.35896302 0.85889027 0.8197176 1.1872782
## 6 0.66409819 0.02206298 0.1304790 0.1323312
## 7 0.45102742 0.76048535 0.5501899 0.9386410
## 8 0.48249177 0.93670363 0.8280114 1.2502066
## 9 0.05545819 0.12281684 0.9219704 0.9301148
## 10 0.47588862 0.40196106 0.0192433 0.4024214
I've tried all sorts of combinations of lazyeval::interp
, lazy_eval::lazy_dots
, etc., but I can't understand what exactly is supposed to happen, much less how to solve my problem.
pythag_wrapper(test_data, a = "a", b = "b")
## Error: object 'x' not found
Upvotes: 2
Views: 357
Reputation: 1481
The problem in your code is on how you deal with the dots argument ...
.
Slightly changing your code and 'manually' rewriting the formula inside the wrapper it works fine:
pythag_wrapper <- function(data, ...){
# From ... argument get names and values
dots = list(...)
# 'Write' the formula: ' ~ pythag(a = val1, b = val2)'
yourformula = as.formula(
paste0(" ~ pythag(",
paste0(names(dots), " = ", unlist(dots), collapse = ", "),
")")
)
# Apply the mutate_. The setNames here is what you need to
# apply the right name to the resulting column
dplyr::mutate_(data, .dots = setNames(list(yourformula), 'root'))
}
Upvotes: 1