R help - Oct 2009 - deriv() to take vector of expressions as 1st arg?

The deriv() function takes an 'expression' as its first argument). I was
wondering if the this function can take an array or a vector of
expressions as its first argument. Aside, I saw how to give a vector
argument to the second argument.

like to have something like:
deriv(c(~x^2+y^3, ~x^5+y^6), c("x","y"))

the documentation for this function talks about being able to "returns
the function values with a gradient attribute containing the gradient
matrix[!!!]". and I indeed want a matrix of expressions to be returned.

In effect, I want to get the Jacobian.

Thank you, if you have some hints on this.

Tom

Try this:

deriv(expression(x^2+y^3, x^5+y^6), c("x","y"))


On Thu, Oct 29, 2009 at 3:31 PM,  <t121 at gmx.net>
wrote:
> The deriv() function takes an 'expression' as its first argument).
I was
> wondering if the this function can take an array or a vector of
> expressions as its first argument. Aside, I saw how to give a vector
> argument to the second argument.
>
> like to have something like:
> deriv(c(~x^2+y^3, ~x^5+y^6), c("x","y"))
>
> the documentation for this function talks about being able to "returns
> the function values with a gradient attribute containing the gradient
> matrix[!!!]". and I indeed want a matrix of expressions to be
returned.
>
> In effect, I want to get the Jacobian.
>
> Thank you, if you have some hints on this.
>
> Tom
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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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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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Thanks a lot for the hints. The sapply method may work for me.

But how can I extract just the ".grad" expression from the return
value
of the deriv function? (and secondly store in a matrix of expressions?)

Thanks again.