-----Original Message-----
From: arun [mailto:smartpink...@yahoo.com]
Sent: Friday, December 07, 2012 10:36 AM
To: Doran, Harold
Cc: R help; David L Carlson; David Winsemius
Subject: Re: [R] Vectorizing integrate()
Hi,
Using David's function:
fun <- function(u, m, s) 1/ (1 + exp(- (B[1] + B[2] *
(m + u)))) * dnorm(u, 0, s)
res<-mapply(function(i) integrate(fun,-
Inf,Inf,m=x[i],s=sem1[i])$value,1:nrow(X))
res
# [1] 0.5212356 0.6214989 0.5306124 0.5789414 0.3429795 0.6972879
0.5952949
#[8] 0.7531899 0.4740685 0.7576587
identical(res,eta)
#[1] TRUE
A.K.
----- Original Message -----
From: "Doran, Harold" <hdo...@air.org>
To: David Winsemius <dwinsem...@comcast.net>
Cc: "r-help@r-project.org" <r-help@r-project.org>
Sent: Friday, December 7, 2012 10:14 AM
Subject: Re: [R] Vectorizing integrate()
David et al
Thanks, I should have made the post more complete. I routinely use
apply functions, but often avoid mapply() as I find it so non-
intuitive. In this instance, I think the situation demands I change
that position, though.
Reproducible code for the current implementation of the function is
B <- c(0,1)
sem1 = runif(10, 1, 2)
x <- rnorm(10)
X <- cbind(1, x)
eta <- numeric(10)
for(j in 1:nrow(X)){
fun <- function(u) 1/ (1 + exp(- (B[1] + B[2] * (x[j] + u)))) *
dnorm(u, 0, sem1[j])
eta[j] <- integrate(fun, -Inf, Inf)$value
}
I can't get my head around how mapply() would work here. It accepts as
its first argument a function. But, in my case I have two functions:
the user defined integrand, fun(), an then of course calling the R
function integrate().
I was thinking maybe along these lines, but this is obviously wrong.
mapply(integrate(function(u) 1/ (1 + exp(- (B[1] + B[2] * (x + u)))) *
dnorm(u, 0, sem1), -Inf, Inf)$value, MoreArgs = list(B, x, sem1))
-----Original Message-----
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Thursday, December 06, 2012 1:59 PM
To: Doran, Harold
Cc: r-help@r-project.org
Subject: Re: [R] Vectorizing integrate()
On Dec 6, 2012, at 10:10 AM, Doran, Harold wrote:
I have written a program to solve a particular logistic regression
problem
using IRLS. In one step, I need to integrate something out of the
linear
predictor. The way I'm doing it now is within a loop and it is as you
would
expect slow to process, especially inside an iterative algorithm.
I'm hoping there is a way this can be vectorized, but I have not
found
it so far. The portion of code I'd like to vectorize is this
for(j in 1:nrow(X)){
fun <- function(u) 1/ (1 + exp(- (B[1] + B[2] * (x[j] + u)))) *
dnorm(u, 0,
sd[j])
eta[j] <- integrate(fun, -Inf, Inf)$value }
The Vectorize function is just a wrapper to mapply. If yoou are able
to get
that code to execute properly for your un-posted test cases, then why
not
use mapply?
Here X is an n x p model matrix for the fixed effects, B is a
vector with the
estimates of the fixed effects at iteration t, x is a predictor
variable in the jth
row of X, and sd is a variable corresponding to x[j].
Is there a way this can be done without looping over the rows of X?
Thanks,
Harold
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PLEASE do read the posting guide
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David Winsemius, MD
Alameda, CA, USA
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
