similar to: Running *slow*

R-help is all about solving R problems. So here ya go: http://www.portfolioprobe.com/2011/09/12/solve-your-r-problems/ -- Patrick Burns pburns at pburns.seanet.com twitter: @portfolioprobe http://www.portfolioprobe.com/blog http://www.burns-stat.com (home of 'Some hints for the R beginner' and 'The R Inferno')

I think the rule is that you can do anything as long as you don't complain. If you want to complain, you must follow the instructions. -- Jari Oksanen in Re: [Rd] Keeping up to date with R-devel -- Patrick Burns pburns at pburns.seanet.com twitter: @burnsstat @portfolioprobe http://www.portfolioprobe.com/blog http://www.burns-stat.com (home of: 'Impatient R' 'The R

I've just realized that it could be handy to have a 'decreasing' argument in 'is.unsorted'. And I'm cheekily hoping someone else will implement it. It is easy enough to work around (with 'rev'), but would be less hassle with an argument. The case I have in mind uses 'is.unsorted' in 'stopifnot'. Pat -- Patrick Burns pburns at pburns.seanet.com

Hello! I am optimizing my code in R and for this I need to know a bit more about the internals. It would help tremendously if someone could link me to a page with O()-complexities of all the operations. In this particular case, I need something like a linked list with O(1) insertLast/First ability. I can't preallocate a vector since I do not know the final size of the list ahead of time. The

Hi, I'm trying to set up R to run a simulation of two populations in which every 3.5 days, the initial value of one of the populations is reset to 1.5. I'm simulation an experiment we did in which we fed Daphnia populations twice a week with algae, so I want the initial value of the algal population to reset to 1.5 twice a week to simulate that feeding. I've use for loops and if/else

It seems obvious to me that the empty string "" is length 0. cheers Worik [[alternative HTML version deleted]]

I know that [] is used for indexing. I know that [[]] is used for reference to a property of a COM object. I cannot find any explanation of what [[1]] does or, more pertinently, where it should be used. Thank you. [[alternative HTML version deleted]]

Dear all, I would like to ask you if there is a way to define a global variable in R? I am having a bunch of function and I think the easier would be to have a global function defined at some point... Would that be possible? Regards Alex [[alternative HTML version deleted]]

Hi, Wondering if anyone could help me out with this error.Im trying to fill a matrix with random numbers taken from an exponential distribution using a loop: x.3<-matrix(rep(0,3000),nrow=1000,byrow=T)for(i in 1:1000){x[i,]<-rexp(3,rate=2/3)} I get the error message: Error in x[i, ] <- rexp(3, rate = 2/3) : incorrect number of subscripts on matrix Any ideas??? Appreciate any thoughts.

Hi everyone, I'm using the following code to go over every element of a data frame (row wise). The problem I am facing is that the outer 'x' variable is not incrementing itself, thus, only one row of values is obtained, and the program does not proceed to the next row. This is the code: while(x<=coln) { while(y<=rown) { n<-as.numeric(df[[y]][x]);

I noticed that the argument names after the dots argument are not partially matched. foo <- function(one, two, ...){ one + two } > foo(o=1, t=2) [1] 3 foo <- function(one, ..., two){ one + two } > foo(o=1, t=2) Fehler in one + two : 'two' fehlt Can someone explain me the reason for this behavior? THX Mark ???????????????????????????????????? Mark Heckmann Blog:

Dear friends, hope I could be able to explain my problem through following example. Please consider this: > set.seed(1) > input <- rnorm(10) > input [1] -0.6264538 0.1836433 -0.8356286 1.5952808 0.3295078 -0.8204684 0.4874291 0.7383247 0.5757814 -0.3053884 > tag <- vector(length=10) for(i in 1:10) # if there is any ****error**** in evaluating

Dear R forum helpers,I have following datatrans <- data.frame(currency_transacted = c("EURO", "USD", "USD", "GBP", "USD", "AUD"), position_amt = c(10000, 25000, 20000, 15000, 22000, 30000))date <- c("12/31/2010", "12/30/2010", "12/29/2010", "12/28/2010", "12/27/2010",

Hi, I have limited experience on R and recently started using REXcel. Although I have been able to run both simple functions (like mean etc) and some complex ones (like Principal Component analysis, PCA) using RExcel, I am facing some problems while running EGARCH model. For this I have downloaded the 'betategarch' package for R to run EGARCH with student t dist. Although the package has

Hello, I want to do fisher test for the rows in data file which has value less than 5 otherwise chi square test .The p values from both test should be stored in one resulted file. but there is some problem with bold if statement. I don't know how implement this line properly. x = cbind(obs1,obs2,exp1,exp2) a = matrix(c(0,0,0,0), ncol=2, byrow =TRUE) #matrix with initialized values

Hello- Sorry to ask a basic question, but I've spent many hours on this now and seem to be missing something. I have a loop that looks like this: mainmat=data.frame(matrix(data=0, ncol=92, nrow=length(predata$Words_MH))) for(i in 1:length(predata$Words_MH)){ for(j in 1:92){ mainmat[i,j]=ifelse(j %in% as.numeric(unlist(strsplit(predata$Words_MH[i], split=","))),

Dear list; How can I speed up the run of following code (illustrative) #======================================================================== con<-vector("numeric") for (i in 1:limit) { if(matched data for the ith item found) { if(i==1) {con<-RowOfMatchedData } else {con<-rbind(con,matchedData)} } }

Hello, I'm very interested in using financial time series data, but I'm a beginner of R programming. I'd like to fully understand internal logics on several time-series related packages such as quantmod, xts, zoo, chron, etc. So, I read some books, 'R Cookbook' and 'Art of R Programming' and another simple tutorials. But I still can't understand grammars of the

Hello everyone, Considering the following code sample : ---- indexes <- function(vec) { vec <- which(vec==TRUE) return(vec) } mat <- matrix(FALSE, nrow=10, ncol=10) mat[1,3] <- mat[3,1] <- TRUE ---- Issuing apply(mat, 1, indexes) returns a 10-cell list, as expected. Now if I do: ---- mat[1,3] <- mat[3,1] <- FALSE apply(mat, 1, indexes) ---- I would expect a

After a lot of processing I get a matrix into M. I expected each row and column to be a vector. But it is a list. R-Inferno says... "Arrays (including matrices) can be subscripted with a matrix of positive numbers. The subscripting matrix has as many columns as there are dimensions in the array—so two columns for a matrix. The result is a vector (not an array) containing the selected