В Mon, 29 Jul 2024 09:52:22 +0530
Christofer Bogaso пишет:
> LL = function(b0, b1)
help(optim) documents that the function to be optimised takes a single
argument, a vector containing the parameters. Here's how your LL
function can be adapted to this interface:
LL <- function(par) {
b0 <- par[
Hi,
I am trying to fit a GLM on below data. While R does provide direct
estimation, I wanted to go with manual calculation as below
dat = structure(list(PurchasedProb = c(0.37212389963679, 0.572853363351896,
0.908207789994776, 0.201681931037456, 0.898389684967697, 0.944675268605351,
0.66079779248
Ben,
Thank you for the incredibly helpful suggestions and links. I've been
exploring each over the past few days, and for anyone else's future
reference, here's what I've found.
(1) I was able to use SANN to specify how to choose new candidate
solutions, but I wasn't able to easily use SANN for a
Lucas Merrill Brown gmail.com> writes:
>
> I've been programming maximum likelihood estimation models using the
> function "optim." My current research requires modeling a particular
> parameter as a categorical variable (what R calls a "factor"), not as a
> continuous parameter.
>
> (The resea
I've been programming maximum likelihood estimation models using the
function "optim." My current research requires modeling a particular
parameter as a categorical variable (what R calls a "factor"), not as a
continuous parameter.
(The research question is, at what level of X does a subject in ou
Great! Thank you for your help!
-Charlotte
>>
>> On Mon, Apr 26, 2010 at 1:12 AM, Tal Galili wrote:
>>>
>>> mm...
>>> I also noticed the function you wrote didn't use parenthesis, mixed b and c
>>> and used different names for K.
>>> Your code is a great exercise in debugging (no offense inte
mm...
I also noticed the function you wrote didn't use parenthesis, mixed b and c
and used different names for K.
Your code is a great exercise in debugging (no offense intended :) )
Try using:
bird<-bird.density[0] # I assume this exists
eqn<- function(K1, b1 = 1.22, c1 = .55) {
Hi Charlotte ,
I can't reproduce your code, but skimming through it -
It would appear that:
1) in
eqn1<- function(K1, bird)
you didn't define "bird" (you did define it before the function, so I'd
suggest just removing it from the function call like this:
eqn1<- function(K1)
2) you didn't "return"
Hello!
I'm a college undergrad desperately trying to finish up my thesis. I
have a dataset on the distribution of a grassland bird from the
Breeding Bird Survey. I have a very straightforward and simple version
of the logistic growth model to describe changes in this bird's
abundance over time. Th
Duncan Murdoch wrote:
On 01/01/2010 11:45 PM, Erin Hodgess wrote:
Dear R People:
I know that you can use "optim" for a function with several parameters.
Is there an equivalent for 2 functions, please? Or should I put
together a finite difference type of matrix, etc., please?
What is the g
On 01/01/2010 11:45 PM, Erin Hodgess wrote:
Dear R People:
I know that you can use "optim" for a function with several parameters.
Is there an equivalent for 2 functions, please? Or should I put
together a finite difference type of matrix, etc., please?
What is the goal? optim optimizes a s
Dear R People:
I know that you can use "optim" for a function with several parameters.
Is there an equivalent for 2 functions, please? Or should I put
together a finite difference type of matrix, etc., please?
Thanks,
Happy New Year,
Erin
--
Erin Hodgess
Associate Professor
Department of Com
I know "optim" should do a minimisation, therefor I used as the
optimisation function
opt.power <- function(val, x, y) {
a <- val[1];
b <- val[2];
sum(y - b/(2*pi*a^2*gamma(2/b))*exp(-(x/a)^b));
}
I call: (with xm and ym the data from the table)
a1 <- c(0.2, 100)
opt <- optim(a1, opt.powe
Hello,
I am trying to fit a exponential power distribution
y = b/(2*pi*a^2*gamma(2/b))*exp(-(x/a)^b)
to a bunch of data for x and y I have in a table.
> data
x y
1 2527
2 7559
3125 219
...
25912925 1
26012975
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