[R] deriv() to take vector of expressions as 1st arg?
Gabor Grothendieck
ggrothendieck at gmail.com
Fri Oct 30 00:29:44 CET 2009
OK. Try this:
sapply(expression(x^2+y^3, x^5+y^6), deriv, c("x", "y"))
On Thu, Oct 29, 2009 at 7:11 PM, Rolf Turner <r.turner at auckland.ac.nz> wrote:
>
> On 30/10/2009, at 11:35 AM, Gabor Grothendieck wrote:
>
>> Try this:
>>
>> deriv(expression(x^2+y^3, x^5+y^6), c("x","y"))
>
> Did *you* try it Gabor? I did just now and it returns only
> the gradient of the first component:
>
> > deriv(expression(x^2+y^3, x^5+y^6), c("x","y"))
> expression({
> .value <- x^2 + y^3
> .grad <- array(0, c(length(.value), 2L), list(NULL, c("x",
> "y")))
> .grad[, "x"] <- 2 * x
> .grad[, "y"] <- 3 * y^2
> attr(.value, "gradient") <- .grad
> .value
> })
>
> cheers,
>
> Rolf Turner
>
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