On 29/03/14 01:34, peter dalgaard wrote:
On 28 Mar 2014, at 02:37 , Rolf Turner wrote:
So there you are. Feel enlightened?
Somewhat, actually, but not to such an extent as to have reached nirvana.
"Promises" blow me away.
Here's the most useful part of the post: to get what you want,
On 28 Mar 2014, at 02:37 , Rolf Turner wrote:
>>
>> So there you are. Feel enlightened?
>
> Somewhat, actually, but not to such an extent as to have reached nirvana.
> "Promises" blow me away.
>
>>
>> Here's the most useful part of the post: to get what you want, use
>>
>> do.call(plot,
On 28/03/2014, 2:04 AM, Philippe Grosjean wrote:
On 28 Mar 2014, at 00:57, Duncan Murdoch wrote:
On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally
On 28 Mar 2014, at 00:57, Duncan Murdoch wrote:
> On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
>> You get what you wanted from
>>
>> do.call(plot,list(x=quote(x),y=quote(y)))
>>
>> By the time do.call() gets the arguments it doesn't know how y was
>> originally computed, just what values it has
On 28/03/14 12:49, Duncan Murdoch wrote:
On 27/03/2014, 7:17 PM, Rolf Turner wrote:
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x <- 1:10
y <- (x-5.5)^2
do.call(plot)
I get the expected plot but
On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally computed, just what values it has.
This works, but it doesn't make sense to me. The arguments end
On 27/03/2014, 7:17 PM, Rolf Turner wrote:
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x <- 1:10
y <- (x-5.5)^2
do.call(plot)
I get the expected plot but with the y-values (sur
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally computed, just what values it has.
-thomas
On Thu, Mar 27, 2014 at 6:17 PM, Rolf Turner wrote:
>
>
> I was under the impression that
>
>
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