How can I use dplyr/magrittr's pipe inside functions in R? - r

Question

I'm trying to write a function which takes as argument a dataframe and the name of the function. When I try to write the function with the standard R syntax, I can get the good result using eval and substitute as recommanded by #hadley in http://adv-r.had.co.nz/Computing-on-the-language.html
> df <- data.frame(y = 1:10)
> f <- function(data, x) {
+ out <- mean(eval(expr = substitute(x), envir = data))
+ return(out)
+ }
> f(data = df, x = y)
[1] 5.5
Now, when I try to write the same function using the %>% operator, it doesn't work :
> df <- data.frame(y = 1:10)
> f <- function(data, x) {
+ data %>%
+ eval(expr = substitute(x), envir = .) %>%
+ mean()
+ }
> f(data = df, x = y)
Show Traceback
Rerun with Debug
Error in eval(expr, envir, enclos) : objet 'y' introuvable
>
How can I using the combine the piping operator with the use of eval and substitute ? It's seems really tricky for me.

Answer 1

A workaround would be
f <- function(data, x) {
v <- substitute(x)
data %>%
eval(expr = v, envir = .) %>%
mean()
}
The problem is that the pipe functions (%>%) are creating another level of closure which interferes with the evaluation of substitute(x). You can see the difference with this example
df <- data.frame(y = 1:10)
f1 <- function(data, x) {
print(environment())
eval(expr = environment(), envir = data)
}
f2 <- function(data, x) {
print(environment())
data %>%
eval(expr = environment(), envir = .)
}
f1(data = df, x = y)
# <environment: 0x0000000006388638>
# <environment: 0x0000000006388638>
f2(data = df, x = y)
# <environment: 0x000000000638a4a8>
# <environment: 0x0000000005f91ae0>
Notice how the environments differ in the matrittr version. You want to take care of substitute stuff as soon as possible when mucking about with non-standard evaluation.
I hope your use case is a bit more complex than your example, because it seems like
mean(df$y)
would be a much easier bit of code to read.

Answer 2

I've been trying to understand my problem.
First, I've written what I want with the summarise() function :
> library(dplyr)
> df <- data.frame(y = 1:10)
> summarise_(.data = df, mean = ~mean(y))
mean
1 5.5
Then I try to program my own function. I've found a solution which seems to work with the lazyeval package in this post. I use the lazy() and the interp() functions to write what I want.
The first possibility is here :
> library(lazyeval)
> f <- function(data, col) {
+ col <- lazy(col)
+ inter <- interp(~mean(x), x = col)
+ summarise_(.data = data, mean = inter)
+ }
> f(data = df, col = y)
mean
1 5.5
I can also use pipes :
> f <- function(data, col) {
+ col <- lazy(col)
+ inter <- interp(~mean(x), x = col)
+ data %>%
+ summarise_(.data = ., mean = inter)
+ }
>
> f(data = df, col = y)
mean
1 5.5

Answer 3

I would not use eval and substitute.
What follows is a simplified version of this great post suited to your question.
df <- data.frame(y = 1:10)
f <- function(data, x) {
x <- enquo(x)
df %>% summarise(mean = mean(!!x))
}
f(data = df, x = y)
There are two things happening here:
Tranforming the column name with enquo()
Prefixing the column with !!
Please see refer to the link for a more complicated example.