similar to: scope of a function + lazy evaluation

Hi RUsers, Suppose I want to see the data on the website url <- "http://www.nseindia.com/content/indices/ind_histvalues.htm" for the index "S&P CNX NIFTY" for dates "FromDate"="01-11-2010","ToDate"="02-11-2010" then read the html table from the page using readHTMLtable() I am using this code webpage <-

Hi folks, I am confused whether the following is a bug or it is fine Here is the explanation a <- zoo(c(NA,1:9),1:10) Now If I do rollapply(a,FUN=mean,width=3,align="right") I get > rollapply(a,FUN=mean,width=3,align="right") 3 4 5 6 7 8 9 10 NA NA NA NA NA NA NA NA But I shouldn't be getting NA right ? i.e for index 10 I should get (1/3)*(9+8+7)

hello all I am using the R package Zelig for some tobit regression with robust standard errors. I have got R version 2.9.2 (2009-08-24) and Zelig Version: 3.4-5 when i do demo(robust) It ends like this way data(coalition) > # Fit the model with robust standard error > user.prompt() Press <return> to continue: > z.out3 <- zelig(Surv(duration, ciep12) ~ polar + numst2 +

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 been going back to old difficult R-list "evaluation" emails that I save in order to understand evaluation better and below still confuses me. Could someone explain why A) works and B) doesn't. A variant of below is in the Pat's Inferno book also but I'm still not clear on what is happening. Thanks. f <- function() { # FORCING i here doesn't help

* What were your biggest misconceptions or stumbling blocks to getting up and running with R? * What documents helped you the most in this initial phase? I especially want to hear from people who are lazy and impatient. Feel free to write to me off-list. Definitely write off-list if you are just confirming what has been said on-list. -- Patrick Burns pburns at pburns.seanet.com

"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

I'm interested in using a data frame as if it were a hash table. For instance if I had the following, > (d <- data.frame(key=seq(0.5, 3, 0.5), value=rnorm(6))) key value 1 0.5 -1.118665122 2 1.0 0.465122921 3 1.5 -0.529239211 4 2.0 -0.147324638 5 2.5 -1.531503795 6 3.0 -0.002720434 Then I'd like to be able to quickly retrieve the "value" of "key" 1.5

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

Is there a reason that 'which.min' and 'which.max' don't have an 'arr.ind' argument? The context in which I wanted that was a grid search optimization, which seems like it would be reasonably common to me. -- Patrick Burns pburns at pburns.seanet.com http://www.burns-stat.com (home of 'Some hints for the R beginner' and 'The R Inferno')

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, 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

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

Pat Burns makes a good point. -Peter -------- Original Message -------- Subject: Re: [R] Using seq_len() vs 1:n Date: Fri, 12 Feb 2010 09:01:20 +0000 From: Patrick Burns <pburns at pburns.seanet.com> To: Peter Ehlers <ehlers at ucalgary.ca> References: <4B746AEF.10900 at ucalgary.ca> If you want your code to be compatible with S+, then 'seq_len' isn't going to work.

Hello! I am trying to write a function with vector and data.frame parameters that uses the sum() function and values from the rows of the data.frame. I need to pass this function as a parameter to optim(). My starting point is: observs <- data.frame(y, x1, x2, x3) Fn <- function(par, observs) { sum( (y - (par[1] * (x1 + 1) * x2^(-par[2]) * x3^par[3])^2 ) }

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]);

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

Dear R experts: this is probably correct behavior, but I do want to point out that it is unexpected to someone not too well versed: > test=factor("A","B","C","A") > ifelse(test=="A", as.factor("A"), test) [1] 1 2 3 1 ok, my factor was just coerced into integers, even though I have a logical vector as my condition and factors as