R help - Apr 2010 - Extracting formulae from expression() / deriv()

I am attempting to extract the derivative/ gradient from this expression 

df1p <- deriv(f1, "P")
> df1p
expression({
    .value <- s - c - a * P
    .grad <- array(0, c(length(.value), 1L), list(NULL, c("P")))
    .grad[, "P"] <- -a
    attr(.value, "gradient") <- .grad
    .value
})

So in this case I want to extract the "-a".

I have read the expression help file, which tells me that I should use [ and
[[ to extract information, but the best result I've had so far is 
> df1p[attr(f1,"gradient")]
expression()

I'm sure I'm missing something really basic, but I've now lost
several hours
trying to do this and I'm getting pretty desperate.
-- 
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Sent from the R help mailing list archive at Nabble.com.

Try this:

attributes(eval(df1p))$gradient[,'P']


On Tue, Apr 6, 2010 at 9:09 AM, nickymcp <nickymcpherson at gmail.com>
wrote:
>
> I am attempting to extract the derivative/ gradient from this expression
>
> df1p <- deriv(f1, "P")
>> df1p
> expression({
> ? ?.value <- s - c - a * P
> ? ?.grad <- array(0, c(length(.value), 1L), list(NULL,
c("P")))
> ? ?.grad[, "P"] <- -a
> ? ?attr(.value, "gradient") <- .grad
> ? ?.value
> })
>
> So in this case I want to extract the "-a".
>
> I have read the expression help file, which tells me that I should use [
and
> [[ to extract information, but the best result I've had so far is
>
>> df1p[attr(f1,"gradient")]
> expression()
>
> I'm sure I'm missing something really basic, but I've now lost
several hours
> trying to do this and I'm getting pretty desperate.
> --
> View this message in context:
http://n4.nabble.com/Extracting-formulae-from-expression-deriv-tp1752738p1752738.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> 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.
>


-- 
Henrique Dallazuanna
Curitiba-Paran?-Brasil
25? 25' 40" S 49? 16' 22" O

On Tue, Apr 6, 2010 at 9:09 AM, nickymcp <nickymcpherson at gmail.com
<https://stat.ethz.ch/mailman/listinfo/r-help>>
wrote:
>*
*>* I am attempting to extract the derivative/ gradient from this expression
*>*
*>* df1p <- deriv(f1, "P")
*>>* df1p
*>* expression({
*>*    .value <- s - c - a * P
*>*    .grad <- array(0, c(length(.value), 1L), list(NULL,
c("P")))
*>*    .grad[, "P"] <- -a
*>*    attr(.value, "gradient") <- .grad
*>*    .value
*>* })
*>*
*>* So in this case I want to extract the "-a".
*>*
*>* I have read the expression help file, which tells me that I should use [
and
*>* [[ to extract information, but the best result I've had so far is
*>*
*>>* df1p[attr(f1,"gradient")]
*>* expression()
*>*
*>* I'm sure I'm missing something really basic, but I've now
lost several hours
*>* trying to do this and I'm getting pretty desperate.

Try this **ugly* but *straightforward approach:
> x = df1p[[1]][[4]][[3]]
> x
-a
> eval(x,list(a=1))
-1
> eval(x,list(a=5))
[1] -5

Best,
Dannemora




*

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