Re: [R] Create a vector from another vector

From: Robin Hankin <r.hankin_at_noc.soton.ac.uk>
Date: Thu 31 Aug 2006 - 01:10:26 EST

OK,

With this extra detail, a
third solution follows, which may be closer in spirit to your application.
It may or may not be faster than the other two, depending on the exact parameters used:

library(partitions)
1+blockparts(n=15,y=c(3,4,2,5,6)-1,include.fewer=T)

(720 distinct solutions)

rksh

On 30 Aug 2006, at 15:49, Doran, Harold wrote:

> Hi Duncan
>
> Here is a bit more detail, this is a bit tough to explain, sorry
> for not
> being clear. Ordering is not important because the vector I am
> creating
> is used as a sufficient statistic in an optimization routine to get
> some
> MLEs. So, any combination of the vector that sums to X is OK. But, the
> condition that x2[i] <= x[i] must be maintained. So, the example below
> would not work because x2[1] > x[1] as you note below.
>
>> I don't think it's really clear what you mean by "ordering is
>> not important". Would
>>
>> x2 <- c(6,5,2,4,2)
>> be acceptable (a re-ordering of your first two examples),
>> even though x2[1] > x1[1]?
>
> To be concrete, the following is the optimization function. This is a
> psychometric problem where the goal is to get the MLE for a test taker
> conditional on their response pattern (i.e., number of points on the
> test) and the item parameters.
>
> pcm.max3 <- function(score, d){
> pcm <- function(theta, d, score)
> exp(sum(theta-d[1:score]))/sum(exp(cumsum(theta-d)))
> opt <- function(theta) -sum(log(mapply(pcm, d, theta = theta,
> score=
> score )))
> start_val <- log(sum(score-1)/(length(score-1)/sum(score-1)))
> out <- optim(start_val, opt, method = "BFGS", hessian = TRUE)
> cat('theta is about', round(out$par, 2), ', se',
> 1/sqrt(out$hes),'\n')
> }
>
> Suppose we have a three item test. I store the item parameters in a
> list
> as
>
> items <- list(c(0,.5,1), c(0,1), c(0, -1, .5, 1))
>
> We can get the total possible number correct as
>
> (x <- sapply(items, length))
> [1] 3 2 4
>
> But, you cannot actually get the MLE for this because the
> likelihood is
> unbounded in this case.
>
> So, let's say the student scored in the following categories for each
> item:
>
> x2 <- c(3,1,4)
>
> By x2[i] <= x[i], I mean that there are 3 possible categories for
> item 1
> above. So, a student can only score in categories 1,2 or 3. He cannot
> score in category 4. This is why the condition that x2[i] <= x[i] is
> critical.
>
> But, because total score is a sufficient statistic, (i.e.,
> "ordering is
> not important") we could either vector in the function pcm.
>
> x3 <- c(3,2,3)
>
> Using the function
>
> pcm.max3(x2, items)
> pcm.max3(x3, items)
>
> Gives the same MLE.
>
> But, the vector
>
> X_bad <- c(4,1,3)
>
> Would not work. You can see that the elements of this vector actually
> serve as indices denoting which category a test taker scored in for
> each
> item in the list "items"
>
> I hope this is helpful and appreciate your time.
>
> Harold
>
>
>>>
>>> ______________________________________________
>>> R-help@stat.math.ethz.ch 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.
>>
>>
>
> ______________________________________________
> R-help@stat.math.ethz.ch 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.

--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
  tel  023-8059-7743

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Received on Thu Aug 31 11:32:04 2006

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