search for: ferraz

Hi, If I have a vector: junk <- c(2,0,0,3,0) and want to access, say, all the elements that are greater than zero. I just do: junk[which(junk>0)] Now, If I have a list: jlist <- list(NULL,c(1,0),NULL,c(1,2,3), NULL) and want to access all the elements that have length greater than zero, I know how to find the elements with: which(sapply(jlist,length)>0) But how do I get a

...the pt_BR team I wrote a script to summarize the status of the translations. It wasn't difficult to extend it to the other languages so I decided to set up a page with the summaries of the translation for all languages for which currently exist a translation. http://www.ime.usp.br/~feferraz/en/rtransstat.html If any of you find it useful I can keep it updated on a regular basis (daily or weekly). Thank you, -- Fernando Henrique Ferraz P. da Rosa http://www.feferraz.net

...a sound choice in adopting R. Thanks in advance for your attention, Fernando Rosa -- "Though this be randomness, yet there is structure in't." Rosa, F.H.F.P Instituto de Matemática e Estatística Universidade de São Paulo Fernando Henrique Ferraz P. da Rosa http://www.feferraz.net [[alternative HTML version deleted]]

...a sound choice in adopting R. Thanks in advance for your attention, Fernando Rosa -- "Though this be randomness, yet there is structure in't." Rosa, F.H.F.P Instituto de Matemática e Estatística Universidade de São Paulo Fernando Henrique Ferraz P. da Rosa http://www.feferraz.net [[alternative HTML version deleted]]

Hi, Imagine a vector x with elements (1,2,1,1,3,5,3,3,1) and a vector y with elements (2,3). I need to find out what elements of x match any of the elements of y. Is there a simple command that will return a vector with elements (F,T,F,F,T,F,T,T,F). Ideally, I would like a solution that works with dataframe colums as well. I have tried x==y and it doesn't work. x==any(y) doesn't work

...g pchisq). I''ve read a lot of documentation, also tried googling for ''goodness of fit R'' but it was helpless, most of it is only about ''regression analysis''. Does anyone know if there is a simpler way to do this? Thank you, <Fernando Henrique Ferraz Pereira da Rosa - mentus at gmx.de>

Has anyone got vim to have syntax highlighting with R function codes? I know there's something similar that works with emacs (ESS or something like that), but I was wondering if anyone knew an equivalent that worked with vim. Thank you, -- []'s mentus at gmx.de Bitte l?cheln! Fotogalerie online mit GMX ohne eigene Homepage!

Full_Name: Fernando Henrique Ferraz Pereira da Rosa Version: 1.8.0 OS: Linux 2.4.21 Submission from: (NULL) (200.206.211.169) After installing R 1.8.0, the R DBI interface stopped working. I tracked it down as a problem in the 'methods' package, that comes in the default installation. Somehow the function '.va...

Hi, I am writing a for loop that creates one object, say 'outn' on every round of the loop. I would like the name of each object to include the index of the loop as in, for example: out1, out2, out3, ... And I would like the naming of the object to take place automatically as the loop moves through? Similarly, I would like to be able to call different objects (in1, in2, in3,

Hi, I have a vector 'data' of 58 probability values (bounded between 0 and 1) and want to draw a probability density function of these values. For this, I used the commands: data <- runif(58) a <- density(data, from=0, to=1) plot(a, type="l",lwd=3) But then, when I try to approximate the area under the plotted curve with the command: area <- sum(a$y)*(a$x[1]-a$y[2])

...1 0 0 0 12 1 -1 -1 0 0 1 0 0 0 13 1 -1 -1 0 0 -1 0 0 0 14 1 -1 -1 0 0 -1 0 0 0 (notice that in the second matrix all collumns sum to 0, in the first they don't). Thank you, -- Fernando Henrique Ferraz P. da Rosa http://www.ime.usp.br/~feferraz

I noticed some very weird behaviour of the function: options(digits=n), where n is the number of digits you would expect to get in R calculations. Let's take a example: > options(digits=4) > getdata(caso.pool.k3.r3.e2) [1] 6.053 2.641 -3.639 14.259 6.082 Which works fine... now, trying again, with different data: > options(digits=4) > getdata(controle.pool.k3.r3.e2)

...("dbConnect"), "DBIConnection", "dbConnect") <environment: 0x91201a8> Methods may be defined for arguments: drv Anyone has a clue on what might be going on? Is it DBI working for any of you using R 1.8.0? Thank you, -- []'s Fernando Henrique Ferraz P. da Rosa

Hi, Anyone knows what would be a short way of extracting a month from a date in numeric or integer format? months("1979-12-20") returns "December" in character format. How could I get 12 in numeric or integer format? Thanks! G. [[alternative HTML version deleted]]

Hi Gon?alo Ferraz <gferraz29 at gmail.com> napsal dne 11.09.2007 16:35:57: > Thanks! What is the easiest way of declaring an empty data frame. Is there > anything similar to: > matrix(0,x,y) > In the help files, I am only finding longer commands that ask me to specify > each individual colu...

Hi, I'd like to get away from SAS, but I don't really know R well enough at this point to know if it would be good for this project. I tried to describe the essence of the project below without getting bogged down in details. It starts when I receive a data flat file. There's lots of columns, but the relevant ones are: custid (customer ID number) saledt (date of sale)

Hi, I have a matrix BEE and want to find the row and column numbers of the minimum value in that matrix. The command which(BEE==min(BEE)) returns only one value which, I take, is the position of the minimum in a vector with as many elements as the matrix. Is there a quick and simple way of getting row and column numbers? Thanks, Gonçalo [[alternative HTML version deleted]]

Hi, my doubt is very simple. I'm sure I've seen someone using something like this before, but unfortunatelly my searches in the archives were useless. Well, I have some objects called after a name that has a number attached to it, varying. Let's say I have: > ls poly1 poly2 poly3 poly4 poly5 poly6 ... I would like to access these objects using a for(), in which I could do

Hi, I have a vector of strings (class character) with 6 elements (length 6). I call it 'names'. "Graham Chapman" "John Cleese" "Terry Gilliam" "Eric Idle" "Terry Jones" "Michael Palin" And I want to turn it into another vector of strings called 'shortnames' with the same length. The new vector should look like:

Is it possible in R to subset a dataframe by more than one factor, all at once? For instance, I have the dataframe: >data p1 p2 p3 p4 p5 p6 p7 p8 p9 p10 pred 1 0 1 0 0 0 0 0 0 0 0 0.5862069 4 0 0 0 0 0 0 0 0 0 1 0.5862069 5 0 0 0 0 0 0 1 0 0 0 0.5862069 6 0 0 0 0 0 0 0 1 0 0 0.5862069 7 0 0 1 0 0 0 0 0 0