similar to: Why is there no c.factor?

Hi, Given factors x and y, c(x,y) does not seem to return a useful result : > x [1] a b c d e Levels: a b c d e > y [1] d e f g h Levels: d e f g h > c(x,y) [1] 1 2 3 4 5 1 2 3 4 5 > Is there a case for a new method c.factor as follows? Does something similar exist already? Is there a better way to write the function? > c.factor = function(x,y) { newlevels =

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

Hi, my data looks this: id forma program kod obor rocnik 1 10001 kombinovan? Matematika M1101 matematika 1 2 10002 prezen?n? Informatika N1801 teoretick? informatika 1 3 10002 prezen?n? Informatika B1801 obecn? informatika 3 4 10003 prezen?n? Informatika M1801 softwarov?

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=","))),

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

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.

"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

* 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

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

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

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

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]]