Let's say I want to sum over all columns in a tibble to create a new column called "total". I could do:
library(tibble)
library(dplyr)
set.seed(42)
N <- 10
Df <- tibble(p_1 = rnorm(N),
p_2 = rnorm(N),
q_1 = rnorm(N),
q_2 = rnorm(N))
# Works fine
Df %>% mutate(total = apply(., 1, sum))
I could make a helper function like so,
myfun <- function(Df){
apply(Df, 1, sum)
}
# Works fine
Df %>% mutate(total = myfun(.))
But let's say this myfun was usually going to be used in this way, i.e. within a dplyr verb function, then the "." referencing the data frame is a but superfluous, and it would be nice if the myfun function could replace this with a default value. I'd like something like this:
myfun2 <- function(Df=.){
apply(Df, 1, sum)
}
which does not work.
Df %>% mutate(total = myfun2())
Error in mutate_impl(.data, dots) :
Evaluation error: object '.' not found.
Because I am not even sure how the "." works, I don't think I can formulate the question better, but basically, I want to know if there a way of saying, in effect, if the Df is not defined in myfun2, get the data-frame that is normally referenced by "."?
One option would be to quote the function and then evaluate with !!
library(tidyverse)
myfun <- function() {
quote(reduce(., `+`))
}
r1 <- Df %>%
mutate(total = !! myfun())
r1
# A tibble: 10 x 5
# p_1 p_2 q_1 q_2 total
# <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 1.37 1.30 -0.307 0.455 2.82
# 2 -0.565 2.29 -1.78 0.705 0.645
# 3 0.363 -1.39 -0.172 1.04 -0.163
# 4 0.633 -0.279 1.21 -0.609 0.960
# 5 0.404 -0.133 1.90 0.505 2.67
# 6 -0.106 0.636 -0.430 -1.72 -1.62
# 7 1.51 -0.284 -0.257 -0.784 0.186
# 8 -0.0947 -2.66 -1.76 -0.851 -5.37
# 9 2.02 -2.44 0.460 -2.41 -2.38
#10 -0.0627 1.32 -0.640 0.0361 0.654
Note that the reduce was used to be more in align with tidyverse, but the OP's function can also be quoted and get the same result
myfun2 <- function() {
quote(apply(., 1, sum ))
}
r2 <- Df %>%
mutate(total = !! myfun2())
all.equal(r2$total, r1$total)
#[1] TRUE