similar to: lm coefficients output confusing

Hello, I'm somewhat new to R. I've searched the archive for the last year and tried to consult the manual pages for the following problem, but did not find an answer. I want to sort an array by the index values. Here is the array >acc.gp.bl.wtmn Gp17 Gp4 1 0.5703125 0.6406250 10 0.7812500 0.7109375 11 0.8046875 0.7343750 12 0.8359375 0.7890625 13 0.8515625 0.7109375

Hello all, I want to define a vector like w[k+1]=w[k]*a/(b+k) for k=1,...,N-1 without use loop. Is it posible to do in R? Regards khazaei

Everyone, I'm working with scatter plots with different colored symbols (via lattice). I'm currently using these colors for points and lines: col1 <- c(rgb(1, 0, 0), rgb(0, 0, 1), rgb(0, 1, 0), rgb(0.55482458, 0.40350876, 0.04166666), rgb(0, 0, 0)) plot(seq(along = col1), pch = 16, col = col1, cex = 1.5) I'm also using these with transparency (alpha

Hi, Could you tell me, if it's possible to create ifelse and put function inside, for example: code{ ifelse ((is.na(vek)), call_fun_1(arguments), call_fun_2(arguments)) call_fun_1 <- function(arguments) { sth... } } -- View this message in context: http://www.nabble.com/function-inside-ifelse-tp24035739p24035739.html Sent from the R help mailing list archive at Nabble.com.

Hi, I sometimes play around with extreme parameters for distributions and found that qbeta is not always monotone as the following example shows. I don't know whether this is serious enough to submit a bug report (as this example is near to the limitations of floating point arithmetic). Josef > x <- qbeta((0:100)/100,0.01,5) > x [1] 0.000000e+00 1.253990e-201 1.589622e-171

I'm sory for my weak english. I need to analyze this subject : x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 y 0 0 1 0 0 1 0 0 1 0 czarne 1 1 0 0 0 0 1 0 0 0 rude 0 0 1 0 0 1 1 0 0 0 braz 0 0 1 0 1 0 1 0 0 0 blond 1 0 0 0 0 1 0 0 0 1 rude 1 1 0 0 0 0 0 0 0 1 blond 0 0 1 1 0 0 0 0 1 0 czarne 1 0 0 1 0 0 1 0 0 0 blond 0 0 1 0 0 1 1 0 0 0 blond 1 0 0 0 0 1 1 0 0 0 czarne 0 0 1 0 0 1 0 0 0 1 czarne 1 0 1 0 0 0

Hi all, I have 2 data frames the first contains a list with repeats of words and an associated response time (RT) measure for each word. The second is a tabulation of each unique word and other information such as the amount and of responses for each word. I need to determine the mean RT for each word and add that as a column in the second data frame. Any help would be appreciated Cheers

The "genetics" package for handling single-locus genetic data is now available on CRAN in both source and Windows binary formats. The purpose of this package is to make it easy to create and manipulate genetic information, and to facility use of this information in statistical models. The library includes classes and methods for creating, representing, and manipulating genotypes

The "genetics" package for handling single-locus genetic data is now available on CRAN in both source and Windows binary formats. The purpose of this package is to make it easy to create and manipulate genetic information, and to facility use of this information in statistical models. The library includes classes and methods for creating, representing, and manipulating genotypes

Hi all, I have two sets: dig<-c("DAVID ADAMS","PIERS AKERMAN","SHERYLE BAGWELL","JULIAN BAJKOWSKI","CANDIDA BAKER") import<-c("by DAVID ADAMS","piersAKERMAN","SHERYLE BagWEL","JULIAN BAJKOWSKI with ","Cand BAKER","smith green")     I want to get the following result from

Hi, #working directory data1 #changed name data to data1.? Added some files in each of sub directories a1, a2, etc. ?indx1<- indx[indx!=""] lapply(indx1,function(x) list.files(x)) #[[1]] #[1] "a1.txt"??????? "mmmmm11kk.txt" #[[2]] #[1] "a2.txt"??????? "mmmmm11kk.txt" #[[3]] #[1] "a3.txt"??????? "mmmmm11kk.txt" #[[4]] #[1]

Dear all R-users, I wanted to fit a Garch(1,1) model to a dataset by: >garch1 = garch(na.omit(dat)) But I got a warning message while executing, which is: >Warning message: >NaNs produced in: sqrt(pred$e) The garch parameters that I got are: > garch1 Call: garch(x = na.omit(dat)) Coefficient(s): a0 a1 b1 1.212e-04 1.001e+00 1.111e-14 Can any one

See ?rpart.control. I get: > golf.rp = rpart(Outlook ~ ., golf, control=rpart.control(minsplit=1)) > golf.rp n= 14 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 14 9 rain (0.2857143 0.3571429 0.3571429) 2) Temperature< 71.5 6 2 rain (0.1666667 0.6666667 0.1666667) 4) Temperature< 64.5 1 0 overcast (1.0000000 0.0000000 0.0000000) * 5)

Hi, I am using package deSolve to run some ordinary differential equations (ODE) as part of a mathematical modeling project. I have solved for the following equilibrium states: Seq1<-a*(1-Neq1)/(f*Veq1+m+d) Ceq1<-(f*Seq1*Veq1+g*Ieq1+r*(1-Neq1)-b1*Veq1*Ieq1)/(b2+m+d+g) Ieq1<-(-b2*Ceq1)-r*(1-Neq1)/(b1*Veq1-g-u) Veq1<-o*(Ceq1+Ieq1)/e I want to take the differential of both sides of

I've searched the Web with Google and do not find what might cause this particular error from an invocation of cenboxplot: cenboxplot(cu.t$quant, cu.t$ceneq1, cu.t$era, range=1.5, main='Total Recoverable Copper', ylab='Concentration (mg/L)', xlab='Time Period') Error in if (n > 0) (1L:n - a)/(n + 1 - 2 * a) else numeric() : argument is of length zero I do

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >

Greetings, I got the below dmesg and the first bad commit is commit cf39c8e5352b4fb9efedfe7e9acb566a85ed847c Merge: 3398d25 23b7eaf Author: Linus Torvalds <torvalds@linux-foundation.org> Date: Wed Sep 4 17:45:39 2013 -0700 Merge tag ''stable/for-linus-3.12-rc0-tag'' of git://git.kernel.org/pub/scm/linux/kernel/git/xen/tip Pull Xen updates from Konrad