similar to: "Global" Variable in R

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

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]]

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.

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

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 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 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

"The R Inferno" is now on the Burns Statistics website at http://www.burns-stat.com/pages/Tutor/R_inferno.pdf Abstract: If you are using R and you think you're in hell, this is a map for you. Also, I've expanded the outline concerning R on the Burns Statistics 'Links' page. Suggestions (off-list) for additional items are encouraged. Patrick Burns patrick at

Hi there, I hope to modify values in a vector or matrix in the following code: for (i in 1:9) { assign(paste("a.", i, sep = ""), 1:i) get(paste("a.", i, sep = ""))[i] <- i+50 } I get the following error message: Error in get(paste("a.", i, sep = ""))[i] <- i + 50 : target of assignment expands to non-language object

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

---------- Forwarded message ---------- From: Emiliano Zapata <ezapataika@gmail.com> Date: Sun, May 20, 2012 at 12:09 PM Subject: To: R-help@r-project.org Hi, I have a 64 bits machine (Windows) with a total of 192GB of physical memory (RAM), and total of 8 CPU. I wanted to ask how can I make R make use of all the memory. I recently ran a script requiring approximately 92 GB of memory to

Hi Everyone, Can anyone help why the errors are coming and rectify it? invalid.ids <- c(1,3,5) if (length(invalid.ids)==0) { cat("No Errors found") } else {

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

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')

Hello, I am trying to understand why the FUNCTION used in several codes, won't create the object after it finishes running the code. For instance, look at the following: Number<- function(x) {MyNumberIs<-x} When I run Number(5) Everything goes well, except that if I try to call the object MyNumberIs, I won't find it. I understand that this function can assume many parameters,

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

Hi, What is quickest way to learn R? I am unnecessarily having fear of learning R. rgds Parag Kulkarni Haridwar,India -- View this message in context: http://r.789695.n4.nabble.com/R-learning-tp4631814.html Sent from the R help mailing list archive at Nabble.com.

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

Anyone got any hints on how to make this code more efficient? An early version (which to be fair did more than this one is) ran for 330 hours and produced no output. I have a two column table, Dat, with 12,000,000 rows and I want to produce a lookup table, ltable, in a 1 dimensional matrix with one copy of each of the values in Dat: for (i in 1:nrow(Dat)) { for (j in 1:2) { #If next

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]);