Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $, so
that
x$y
functions the same as
foo(x, y)
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Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Ali -
Sent: Wednesday, April 27, 2005 3:11 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Defining binary indexing operators
Assume we have a function like:
foo - function(x, y)
how is it possible
, Reid
Sent: Wednesday, April 27, 2005 4:10 PM
To: 'Ali -'; r-help@stat.math.ethz.ch
Subject: RE: [R] Defining binary indexing operators
That sounds like a recipe for headaches. If you want to use x$y because
you want a certain kind of x to act like a list with components for
certain y, then you
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $, so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
It's not necessary to be that complicated, is it? AFAIK, the '$'
operator is treated specially by the parser so that its RHS is treated
as a string, not a variable name. Hence, a method for $ can just take
the indexing argument directly as given -- no need for any fancy
language tricks
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $,
so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class =
Excuse me! I misunderstood the question, and indeed, it is necessary be
that complicated when you try to make x$y behave the same as foo(x,y),
rather than foo(x,y) (doing the former would be inadvisible, as I
think someelse pointed out too.)
Tony Plate wrote:
It's not necessary to be that
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Assume we have a function like:
foo - function(x, y)
how is it possible to define a binary indexing operator, denoted by $,
so
that
x$y
functions the same as
foo(x, y)
Here is an example. Note that $ does
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i)); foo(x,
i)
}
x$y # structure(8, class = myclass)
If I got it right, in the
On 4/27/05, Ali - [EMAIL PROTECTED] wrote:
Here is an example. Note that $ does not evaluate y so you have
to do it yourself:
x - structure(3, class = myclass)
y - 5
foo - function(x,y) x+y
$.myclass - function(x, i) { i - eval.parent(parse(text=i));
foo(x,
i)
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