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.
__
R-help@r-project.org mailing
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 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 a
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
Try this:
deriv(expression(x^2+y^3, x^5+y^6), c("x","y"))
On Thu, Oct 29, 2009 at 3:31 PM, 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 t
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
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