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This is a very basic question, but I am a bit confused with optim. I want to get the MLEs using optim which could replace the newton-raphson code I have below which also gives the MLEs. The function takes as input a vector x denoting whether a respondent answered an item correctly (x=1) or not (x=0). It also takes as input a vector b_vector, and these are parameters of test items (Rasch estimates

Hello R, I am looking for non-parametric simulation in IRT. Is there any IRT package that does non-parametric simulation? helen L [[alternative HTML version deleted]]

Dear List: Is any one aware of a package that would extend the D() function and allow for one to maximize a likelihood function symbolically? Something akin to Solve[x==0, parameter] function in Mathematica? Clearly R has the capacity to _compute_ MLEs given a set of data. But, I'm looking for a package that would allow for me to define the likelihood function, find the 1st order partial

Martin In terms of context of the actual problem, sapply is called millions of times because the work involves scoring individual students who took a test. A score for student A is generated and then student B and such and there are millions of students. The psychometric process of scoring students is complex and our code makes use of sapply many times for each student. The toy example used

It sounds as thought you are interested in Hoyt's Anova which is a form of generalizability theory. This is usually estimated using by getting the variance components from ANOVA. > -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Karl Knoblick > Sent: Tuesday, May 16, 2006 6:10 AM > To: r-help at

Could your code use vapply instead of sapply? vapply forces you to declare the type and dimensions of FUN's output and stops if any call to FUN does not match the declaration. It can use much less memory and time than sapply because it fills in the output array as it goes instead of calling lapply() and seeing how it could be simplified. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue,

Quite possibly, and I?ll look into that. Aside from the work I was doing, however, I wonder if there is a way such that sapply could avoid the overhead of having to call the identical function to determine the conditional path. From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:14 PM To: Doran, Harold <HDoran at air.org> Cc: Martin Morgan <martin.morgan

Dear List: When I click on the link to download a reference manual for a package on cran, I get an error message that the file is damaged and could not be repaired. I randomly chose various packages and the same error message appears. Are the links actually broken? I have also restarted my machine and closed and re-opened acrobat. I am using Windows XP, Acrobat Professional 6.0.0.5, and

FYI, in R devel (to become 3.5.0), there's isFALSE() which will cut some corners compared to identical(): > microbenchmark::microbenchmark(identical(FALSE, FALSE), isFALSE(FALSE)) Unit: nanoseconds expr min lq mean median uq max neval identical(FALSE, FALSE) 984 1138 1694.13 1218.0 1337.5 13584 100 isFALSE(FALSE) 713 761 1133.53 809.5 871.5

While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function if (!identical(simplify, FALSE) && length(answer)) This seems superfluous to me, in particular this part: !identical(simplify, FALSE) The preceding

You?re right, it sure does. My suggestion causes it to fail when simplify = ?array? From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:11 PM To: Doran, Harold <HDoran at air.org> Cc: r-help at r-project.org Subject: Re: [R] Possible Improvement to sapply Wouldn't that change how simplify='array' is handled? > str(sapply(1:3,

I have some matrices stored as elements in a list that I am working with. On example is provided below as TP[[18]] > TP[[18]] level2 level1 1 2 3 4 1 79 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 Now, using prop.table on this gives > prop.table(TP[[18]],1) level2 level1 1 2 3 4 1 1 0 0 0 2 3

Dear List I apologize for the multiple postings. After being in the weeds on this problem for a while I think my original post may have been a little cryptic. I think I can be clearer. Essentially, I need the following a <- c(2,3) b <- c(4,5,6) (2*4) + (2*5) + (2*6) + (3*4) + (3*5) +(3*6) But I do not know of a built in function that would do this. Any suggestions? -----Original

Dear R: Am I having trouble downloading the LME4 library. I am using Windows and am using ver 1.7 I have tried the following: 1) Install package from CRAN, but LME4 is not listed 2) Downloaded LME4 from http://cran.us.r-project.org/, however, I cannot open the file when I try install from local drive. I get the following error: Error in file(file, "r") : unable to open connection

Hi all: I'm having difficulty with something I believe is very simple, but I'm stuck. I have a large data frame that took days to clean and prepare. All I now need to do is concatenate three variables into a single column. For example, I have tenn$up, tenn$down, and tenn$stable which all have values of 1 or 0. I simply want to put all three columns together to create a pattern (e.g.,

Just out of curiosity, has anyone ever written a function that sends an email to you when an R process has finished? For instance, I often work with very large data sets and certain tasks (e.g., merging records, lmer runs) can take a long time and I find myself constantly looking over to see if R is done. But, it would just be neat if there was a way R could send me an email alerting me that a

I need to wrap a loop inside a function and am having a small bit of difficulty getting the results I need. Below is a replicable example. # define functions pcm <- function(theta,d,score){ exp(rowSums(outer(theta,d[1:score],'-')))/ apply(exp(apply(outer(theta,d, '-'), 1, cumsum)), 2, sum) } foo <- function(theta,items, score){ like.mat <-

Dear list Suppose I have the following vector: x <- c(3,4,2,5,6) Obviously, this sums to 20. Now, I want to have a second vector, call it x2, that sums to x where 5 <= x <= 20, but there are constraints. 1) The new vector must be same length as x 2) No element of the new vector can be 0 3) Element x2[i] of the new vector cannot be larger than element x[i] of the original vector 4)

>>>>> Henrik Bengtsson <henrik.bengtsson at gmail.com> >>>>> on Tue, 13 Mar 2018 10:12:55 -0700 writes: > FYI, in R devel (to become 3.5.0), there's isFALSE() which will cut > some corners compared to identical(): > > microbenchmark::microbenchmark(identical(FALSE, FALSE), isFALSE(FALSE)) > Unit: nanoseconds > expr

Wouldn't that change how simplify='array' is handled? > str(sapply(1:3, function(x)diag(x,5,2), simplify="array")) int [1:5, 1:2, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=TRUE)) int [1:10, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=FALSE)) List of 3 $ : int [1:5, 1:2] 1 0 0 0 0 0 1 0 0