如何在R中的函数内使用dplyr / magrittr的管道？

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