On 28 Mar 2014, at 00:57, Duncan Murdoch murdoch.dun...@gmail.com 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
On 28/03/2014, 2:04 AM, Philippe Grosjean wrote:
On 28 Mar 2014, at 00:57, Duncan Murdoch murdoch.dun...@gmail.com 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
On 28 Mar 2014, at 02:37 , Rolf Turner r.tur...@auckland.ac.nz 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
On 29/03/14 01:34, peter dalgaard wrote:
On 28 Mar 2014, at 02:37 , Rolf Turner r.tur...@auckland.ac.nz 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:
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,list(x=x,y=y))
I get the expected plot but with the y-values (surrounded by c()) being
printed
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 r.tur...@auckland.ac.nzwrote:
I was under the
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
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 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
Dear all,
I am trying to use do.call, but I don't think I totally understand this
function.
Here is an simple example.
B - matrix(c(.5,.1,.2,.3),2,2)
B
[,1] [,2]
[1,] 0.5 0.2
[2,] 0.1 0.3
x - c(.1,.2)
X - cbind(1,x)
X
x
[1,] 1
On 08/08/2011 4:34 AM, Kathie wrote:
Dear all,
I am trying to use do.call, but I don't think I totally understand this
function.
Here is an simple example.
B- matrix(c(.5,.1,.2,.3),2,2)
B
[,1] [,2]
[1,] 0.5 0.2
[2,] 0.1 0.3
x-
On 08/08/2011 8:21 AM, Duncan Murdoch wrote:
On 08/08/2011 4:34 AM, Kathie wrote:
Dear all,
I am trying to use do.call, but I don't think I totally understand this
function.
Here is an simple example.
B- matrix(c(.5,.1,.2,.3),2,2)
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