similar to: Data-mining using R

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)

I have a dataset which I want to model using a Poisson distribution, with a given parameter. I would like to know what is the proper way to do a ''goodness of fit'' test using R. I know the steps I''d take if I were to do it ''manually'': grouping the numbers into classes, calculating the expected frequencies using ''ppois'', then

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!

Dear R list, I have been trying to do a linear model, extracting the effect of a covariate.... and the results do not match, when I do it with other programs (e.g. minitab).... so it is obvious that I was doing something wrong. Whan I do it with minitab, I have this results: (sector is a factor and depth is the covariate): Source DF Seq SS Adj SS Adj MS F P

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

Hmm. Chuck: I don't see how this example represents incomplete/incommensurate recycling. Doesn't TRUE replicate from length-1 to length-3 in this case (mat[c(TRUE,FALSE),2] would be an example of incomplete recycling)? William: clever, but maybe too clever unless you really need the speed? (The clever way is 8 times faster in the following case ...) x <- rep(1,1e6)

First, thanks to all who helped me with my question about rescaling axes on the fly. Using unlist() and range() to set the axis ranges in advance worked well. I've since plotted about 300 datasets with relative ease. Now I'm trying to fit a lossy oscillator resonance to (the square root of) a lorentzian (testframe$y is oscillator amplitude, testframe$x is drive frequency): lorentz

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

Hi, 2 questions: Question 1: example of what I currently do: for(i in 1:6){sink("temp.txt",append=TRUE) dput(i+0) sink()} x=scan(file="temp.txt") print(prod(x)) file.remove("C:/R-2.5.0/temp.txt") But how to convert the output of the loop to a vector that I can manipulate (by prod or sum etc), without having to write and append to a file? Question 2: >

Hi, I'm fitting a standard nonlinear model to the luminances measured from the red, green and blue guns of a TV display, using nls. The call is: dd.nls <- nls(Lum ~ Blev + beta[Gun] * GL^gamm, data = dd, start = st) where st was initally estimated using optim() st $Blev [1] -0.06551802 $beta [1] 1.509686e-05 4.555250e-05 7.322720e-06 $gamm [1] 2.511870 This works fine but I

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1"

Hello! I work with: R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) On Windows XP Professional (Version 2002) SP2 I think there is a bug in the conditional execution if (expr1) {expr2} else {expr3} If I try: "if (expr1) expr2 else expr3" it works well but when I put the expression expr2 and expr3 between {} I receive an error message

This may have been asked before, but is there an elegant way to check whether an variable/argument passed to a function is a "parse tree" for an (unevaluated) expression or not, *without* evaluating it if not? Currently, I do various rather ad hoc eval()+substitute() tricks for this that most likely only work under certain circumstances. Ideally, I'm looking for a isParseTree()

Hi How do I install "gregmisc" packages? I did- % sudo R > install.packages("gregmisc") . . > barplot2() but, Error: couldn't find function "barplot2" -- atuya Mac OSX 10.2.6 R 1.7.1

Hi, Is there any equivalent for ifelse (except if (cond) expr1 else expr2) which takes an atomic element as argument but returns vector since ifelse returns an object of the same length as its argument? x = c(1,2,3) y = c(4,5,6,7) z = 3 ifelse(z <= 3,x,y) would return x and not 1 thanks

Hello, I have a function for reading a data-frame from a file, which contains E = read.table(file = filename, header = T, colClasses = c(rep("integer",6),"numeric","integer",rep("numeric",8)), ...) Now a small variation arose, where colClasses =

Dear list # I have a function ff <- function(a,b=2,c=4){a+b+c} # which I programmatically want to modify to a more specialized function in which a is replaced by 1 ff1 <- function(b=2,c=4){1+b+c} # I do as follows: vals <- list(a=1) (expr1 <- as.expression(body(ff))) expression({ a + b + c }) (expr2 <- do.call("substitute", list(expr1[[1]], vals))) { 1 +

I am more than a little puzzled by your question. In the construct {expr1; expr2; expr3} all of the expressions expr1, expr2, and expr3 are evaluated, in that order. That's what curly braces are FOR. When you want some expressions evaluated in a specific order, that's why and when you use curly braces. If that's not what you want, don't use them. Complaining about it is like

Apologies for cross posting.... --------------------------------------------------------------------- CART Data Mining'04: First International CART(R) Conferences Focusing on the Data Mining technology of Leo Breiman, Jerome Friedman, Richard Olshen, Charles Stone (CART, MARS(R), TreeNet(tm), PRIM(tm)...) First Call For submissions

hi , this can be done easily if (cond) expr ex:  > for (i in 1: 4)+ {+ if(i==2) print("a")+ if(i==2) print("b")+ } output : [1] "a"[1] "b" but i want this  if (cond) expr1 expr 2 i tried this :  > for (i in 1: 4)+ {+ if(i==2) (print("b") && print("a"))+ } output : [1] "b"Error in print("b") &&