Help for numericDeriv function

Fri, May 18, 2012 11:48 AM

You could use parse(text=a)[[1]]:
   > numericDeriv(parse(text=a)[[1]], c(t(l),recursive=TRUE))
  [1] 3
  attr(,"gradient")
       [,1] [,2] [,3] [,4] [,5] [,6]
  [1,]    1    0    3    0    0    0

or construct your expression directly:
  > lnames <- array(lapply(l, as.name), dim=dim(l))
  > aa <- call("+", call("*", lnames[[1,1]], lnames[[2,1]]), call("*", lnames[[1,2]], lnames[[1,2]]))
  > numericDeriv(aa, c(t(l),recursive=TRUE))
  [1] 3
  attr(,"gradient")
       [,1] [,2] [,3] [,4] [,5] [,6]
  [1,]    1    0    3    0    0    0

The results are identical.  The latter method is a bit more work to set up but
works no matter what the names in the expression look like.  The former
fails if the names contain spaces or other things that are not allowed in
R names.

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
Of Cengiz Zopluoglu
Sent: Friday, May 18, 2012 11:33 AM
To: r-help at r-project.org
Subject: Re: [R] Help for numericDeriv function

I missed a couple line of codes in the previous e-mail. Here is whole code
again:

load <- matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)

l <- matrix(nrow=nrow(load),ncol=ncol(load))
for(i in 1:nrow(load)) {
for(j in 1:ncol(load)) { l[i,j]=paste("l",i,j,sep="")}}

for(i in 1:nrow(load)){
for(j in 1:ncol(load)){ assign(paste("l",i,j,sep=""),load[i,j]) }}

numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))

a <- paste(
paste(l[1,1],l[2,1],sep="*"),
paste(l[1,2],l[1,2],sep="*"),
sep="+"
)

numericDeriv(???a???, c(t(l),recursive=TRUE))

Thanks

On Fri, May 18, 2012 at 1:25 PM, Cengiz Zopluo??lu <cen.zop at gmail.com> wrote:

Hi,

I am stuck on something for a couple days, I am almost about to give up.
This looks simple, but I can't figure out. I hope I can get some help here.

I am trying to do some symbolic and numerical derivations. Let me explain
the problem. Let's say, I have a matrix as follows:

load <- matrix(c(3,0,1,4,1,3),nrow=3,ncol=2,byrow=TRUE)

load

      [,1] [,2]
[1,]    3    0
[2,]    1    4
[3,]    1    3

I will create objects for each element of the matrix, and assign the
matrix elements to these objects. These objects will be also an element of
another matrix.

l <- matrix(nrow=nrow(load),ncol=ncol(load))
for(i in 1:nrow(load)) {
for(j in 1:ncol(load)) { l[i,j]=paste("l",i,j,sep="")}}

     [,1]  [,2]
[1,] "l11" "l12"
[2,] "l21" "l22"
[3,] "l31" "l32"

l11

[1] 3

l12

[1] 0

l21

[1] 1

l22

[1] 4

l31

[1] 1

l32

[1] 3

Let's say, I need to take the derivative of  (l11*l21+l12*l22) with
respect to l11,l12,l21,l22,l31,l32.

numericDeriv(quote(l11*l21+l12*l22), c(t(l),recursive=TRUE))

[1] 3
attr(,"gradient")
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    4    3    0    0    0

Everything is fine up to this point. I need what I have. But, I don't want
to manipulate the elements in "quote()" manually within the numericDeriv
function. For instance, I want to call these elements from the matrix "l"
as below:

a <- paste(

+ paste(l[1,1],l[2,1],sep="*"),
+ paste(l[1,2],l[1,2],sep="*"),
+ sep="+"
+ )

[1] "l11*l21+l12*l12"

Now, "a" is a character. Do you have any idea about how to use the
character value "l11*l21+l12*l12" assigned to object "a" within the
numericDeriv function. So, I can get the same result as above without
manipulating the expression in quote() manually.

Thanks




--

Cengiz Zopluoglu


--

Cengiz Zopluoglu

	[[alternative HTML version deleted]]

Help for numericDeriv function

Thread (3 messages)