similar to: Searching for option explanations

Is there a way to append to the outfile when using R CMD BATCH? My code, right now, is: R CMD BATCH --slave --vanilla '--args place .2 -.1 .9 .6' StratificationSimulation example.output Everything else is working the way I'd like it. The first few lines of code of my script file are: options(echo=FALSE) cmd_args = commandArgs() print (cmd_args) #d <-

Hi all, I'll preface this with saying I've gone through the archives, and am still in need of some help. I've been using this likelihood model with mean = 0 and s.d. = sqrt( (c + ( 1 / N1 ) + ( 1 / N2 ) ) * x * ( 1 - x )), where c is a genetic drift parameter (usually very small, like between .005 - .001), N1 and N2 are my population sizes (~200), and x is a value between 0 and 1.

Hi, I am having difficulty entering a 'programmable' argument into the multinom function from the nnet package. Interactively, I can get the function to work fine by calling it this way: z1=multinom(formula = class.ind(grp[-outgroup])~ (PC1 + PC2 + PC3), data=data.frame(scores)) However I need to be able to change the number of variables I am looking for in 'scores' and so am

> On Jun 6, 2019, at 2:04 PM, Richard O'Keefe <raoknz at gmail.com> wrote: > > How can expanding tildes anywhere but the beginning of a file name NOT be > considered a bug? > > I think that that IS what libreadline is doing if one allows a whitespace separated list of file names. As reported in R-help, https://www.mail-archive.com/r-help at

Hi all, I was wondering whether it is possible to use the lapply() function to alter the value of the input, something in the spirit of : a1<-runif(100) a2<-function(i){ a1[i]<-a1[i-1]*a1[i];a1[i] } a3<-lapply(2:100,a2) Something akin to a for() loop, but using the lapply() infrastructure. I haven't been able to get rapply() to do this. The reason is that the "real"

Hi there, I have a binary matrix (dim 100x100) filled with values 0 and 1. I need select a record "n" positions of that matrix when values are 1. How can I do that? Thanks for all, Miltinho Brazil --------------------------------- [[alternative HTML version deleted]]

Hi there, I run two regressions: y = a1 + b1 * x + e1 y = a2 + b2 * z + e2 I want to test against the null hypothesis: b1 = b2. How do I design the test? I think I can add two equations together and divide both sides by 2: y = 0.5*(a1+a2) + 0.5*b1 * x + 0.5*b2 * z + e3, where e3 = 0.5*(e1 + e2). or just y = a3 + 0.5*b1 * x + 0.5*b2 * z + e3 If I run this new regression, I can test against

Hi, When qqnorm on a vector of length 10M+ I get a huge pdf file which cannot be loaded by acroread or evince. Any suggestions? (apart from sampling the data). Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://mideasttruth.com http://honestreporting.com http://camera.org http://openvotingconsortium.org http://pmw.org.il

I have a problem with expressions. I am trying to create a title where the parameter of interest is displayed as a Greek character. Which parameter is being considered is stored in a character variable. As an example, if I have param <- "alpha" and then do plot(0, 0, main = bquote(Parameter==.(param))) then in the title I get "Parameter = alpha", whereas I want the

I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should

Hi, How can I check if a certain ... argument has been passed on to my user-defined function or not? foo <- function(data, ...) { ### here I want to check whether xlab was passed with the ... arguments ### or if the ... arguments did not contain an xlab argument } I tried missing(xlab) , exists(xlab) and several other things but did not find a solution. TIA, Mark

I was playing around a bit to see how I could find the two points in a set of points (or ordered pairs) furthest from each other. Here's what I did: 1) created a Nrow by 2col matrix, so each row contains an x,y coordinate pair. 2) fed the matrix to a nested mapply (cv is my matrix): mapply(function(k,l) mapply(function(x,y,a,b) +

Using R version 2.4.1 (2006-12-18) on Windows, I have a dataset which resembles this: id att1 att2 att3 1 1 1 0 2 1 0 0 3 0 1 1 4 1 1 1 ratings <- data.frame(id = c(1,2,3,4), att1 = c(1,1,0,1), att2 = c(1,0,0,1), att3 = c(0,1,1,1)) I would like to get a cross tab of counts of co-ocurrence, which might resemble this:

Hi, I have a question regarding the image function. Is there a way to specify a separate color set for each column in the matrix? TIA, ajar. [[alternative HTML version deleted]]

I am doing cluster analysis [hclust(Dist, method="average")] on data that potentially contains redundant objects. As expected, the inclusion of redundant objects affects the clustering result, i.e., the data a1, = a2, = a3, b, c, d, e1, = e2 is likely to cluster differently from the same data without the redundancy, i.e., a1, b, c, d, e1. This is apparent when the outcome is visualized

I have hundreds of humans who have undergone SNP genotyping at hundreds of loci. Some have even undergone the procedure twice or thrice (kind of an internal control). So obviously I need to find those replications, and confirm that the results are the same. If there is discordance then I need to address it. I tried to use the aggregate function nr.attempts

Hello R users, I have gene expression data of two groups of genes (large and small). Gene expression intensities of those genes are classified into 1 to 10 levels. What I want is to make a random set of genes that have the same levels as the small group from large group using sample(). I used smallvec to hold the number of genes in each levels (1 to 10) for small group, largevec for large group.

Hi all, So far I only know one way to get the confidence limit for the Poisson distribution is to use the look-up table given by the 2 parameter (the number of observation x and the confidence level, e.g. 95%) and the table is limit by the maximum number of observations (x <= 50). I know the formula to compute the CI, however, mathematically it is not easy to do it. So, anyone know an R

Hello all again, I want to do bitwise addition in R. I am trying to generate a matrix 0000 0001 0010 .... .... 1111 I know the other ways of generating this matrix but I need to look at bitwise addition. Any suggestions??? thanks a lot Nameeta ------------------------------------------------- This email is intended only for the use of the individual or...{{dropped}}

?"=", ?"==", ?"!=", ?">=", and ?"<=" sends me to the documentation for ?help on Windows, while returning the correct documentation on Linux. Robert > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor