similar to: Concatenating variables

I am working with a longitudinal data set in the long format. This data set has three observations per grade level per year. Here are the first 10 rows of the data frame: >tenn.dat[1:10,] year schid type grade gain se new cohort 6 2001 100005 5 4 33.1 3.5 4 3 7 2002 100005 5 4 33.9 3.9 4 2 8 2003 100005 5 4 32.3 4.2 4 1 10 2001 100005

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

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

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

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

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

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,

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,

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

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

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 <-

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

>>>>> 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

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

This is just posed out of curiosity, (not as a criticism per se). But what is the functional role of the argument na.rm inside the mean() function? If there are missing values, mean() will always return an NA as in the example below. But, is there ever a purpose in computing a mean only to receive NA as a result? In 10 years of using R, I have always used mean() in order to get a result, which is

Given the following data, I want a scatterplot with the data points and the predictions from the regression. Sigma <- matrix(c(1,.6,1,.6), 2) mu <- c(0,0) dat <- mvrnorm(5000, mu, Sigma) x <- dat[,1] * 50 + 200 y <- dat[,2] * 50 + 200 fm <- lm(y ~ x) ### This gives the regression line, but not the data xyplot(y ~ x, type = c('g', 'p'),

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

I am writing a program where non-technical R users will read in a config file and the config file will then parse the arguments found within the config and pass them to respective functions. I'm having trouble (efficiently) writing a piece of code to retain quotation marks around the argument which requires it as input, as found in the example function below, myFuncton1. Below is a minimal,

First off, Marc Schwartz posted this link earlier today, read it. http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-are-p_002dvalues-not-di splayed-when-using-lmer_0028_0029_003f Second, your email is not really descriptive enough. I have no idea what OR is, so I have no reaction. Third, you're comparing estimates from different methods of estimation. lmer will give standard errors that