Question

I have been defined a bind function ,

b <- function(f,...) function(x) f(x, ...)

so I can do things like (this is a fake example)

d = data.frame(x=c(1,2,1), y=c(10, 20, 5))
ddply(d, ~x, b(transform, y=sum(y)))

instead of

ddply(d, ~x, function (df) { transform(df, y=sum(y)) }

Now, I’m trying to define an operator

'%b%' <- function(x,...) b(x,...)

and try

ddply(d, ~x, transform %b% (z=y*10)))

It doesn’t work. What is the difference ?

When I do

> b(transform, y=y/sum(y))(d)
x         y c.1..2..1.
1 1 0.2857143          1
2 2 0.5714286          2
3 1 0.1428571          1

That works, but

> transform %b% (y=y/sum(y))(d)
Error in transform %b% (y = y/sum(y))(d) : object 'y' not found

I understand, there is a ‘capture’ difference, what can I do ?

Answer 1

To make your example work, you need to match the function argument to an existing function, using match.fun():

`%b%` <- function(x,...) match.fun(b)(x,...)
ddply(d, ~x, transform %b% (z=y*10))

  x  y
1 1 10
2 1  5
3 2 20

It will also be safer if you use match.fun() in your first definition of f:

b <- function(f, ...) function(x) match.fun(f)(x, ...)
ddply(d, ~x, b(transform, y=sum(y)))
  x  y
1 1 10
2 1  5
3 2 20

---

Having answered your question, I now have to point out that I don’t understand why you want to do this, since the plyr functions like ddply as well as the base R apply functions already do this.

So, I would write your original example simply like this:

ddply(d, ~x, transform, y=sum(y))
  x  y
1 1 15
2 1 15
3 2 20

ddply(d, ~x, transform, y=y/sum(y))
  x         y
1 1 0.6666667
2 1 0.3333333
3 2 1.0000000

---

Edit:

I had another look at your question. You simply made a syntax error in your function definition. This works perfectly fine with an infix operator:

'%b%' <- function(f,...) function(x) f(x, ...)
ddply(d, ~x, transform %b% (z=y*10))
  x  y
1 1 10
2 1  5
3 2 20