similar to: Calculating AIC for the whole model in VAR

Hello! I installed rJava and am trying to load it. library(rJava) Error : .onLoad failed in loadNamespace() for 'rJava', details: call: fun(libname, pkgname) error: No CurrentVersion entry in Software/JavaSoft registry! Try re-installing Java and make sure R and Java have matching architectures. Error: package or namespace load failed for ‘rJava’ Any idea why? Background info:

Hi all, I am trying to write a new function based on an existing function in "stats". This function requires to call a c program "C_BinDist". The new function couldn't work without load this c script. How can I find this "C_BinDist" function and load it to make my new function work through? Thanks, Dan [[alternative HTML version deleted]]

I am running R 2.0.0, GOstats 1.1.1 and GO 1.7.0, and when I use the function GOHyperG, I have the following error: w1<-as.list(hgu95av2LOCUSID) w2<-unique(unlist(w1)) set.seed(123) myLL<-sample(w2,100) xx <- GOHyperG(myLL) Error in mget(x, env = GOTERM, ifnotfound = NA) : recursive default argument reference In fact first I tried this function with my locusId ' list (with

Hi So carrying on my use of analysis of variance to check for the effects of two factors. It's made simpler by the fact that both my factors have only two levels each, creating four unique groups. I have a highly significant interaction term. In the context of the experiment, this makes sense. I can visualise the data graphically, and sure enough I can see that both factors have different

I am writing a report using Sweave with specific R output incorporated into the text using the Sexpr{} command. I have run into two specific issues: 1) If the result inside the Sexpr{} command is an integer less than 10 I would like to print the ?word? for that number ? e.g., ?seven? for 7. 2) If a p-value is ?equal to zero? to some arbitrary number of decimal places -- e.g.,

Hi all, Initially I thought this was a bug in the built-in mocking framework(and it still may be), but I better hash it out on the mailing list before I file/reopen the ticket: http://rspec.lighthouseapp.com/projects/5645/tickets/478-mocks-on-constants#ticket-478-6 I thought my example illustrated my problem, but obviously I was passing the wrong arguments to the mock. I revised my example to

I think you should be more suspicious of yourself, Dimitri. A letter T variable can easily arise in the problem domain when you are not thinking of logical values at all, at which point your cavalier use of T as a synonym for TRUE can suddenly become a bug. -- Sent from my phone. Please excuse my brevity. On July 27, 2017 8:18:03 AM PDT, Dimitri Liakhovitski <dimitri.liakhovitski at

Dear R-list, maybe some of you could point me in the right direction: Are you aware of any FREE Fortran or Java libraries/actual pieces of code that are VERY efficient (time-wise) in running the regular linear least-squares multiple regression? More specifically, I have to run small regression models (between 1 and 15 predictors) on samples of up to N=700 but thousands and thousands of them. I

Dear all, I have been trying to fit my data (only right censored) with gumbel distribution using fitdistrplus. I am getting very high standard error. I have been wondering why. The followings are the outputs: fit1=fitdistcens(dr0, "gumbel", start=list(a=99, b=0.6), optim.method= "L-BFGS-B", lower = 0.0, upper = Inf) > summary(fit1) FITTING OF THE DISTRIBUTION ' gumbel

Hello! I have a column in my data frame that I have to split: I have to distill the numbers from the text. Below is my example and my solution. x<-data.frame(x=c("aaa1_bbb1_ccc3","aaa2_bbb3_ccc2","aaa3_bbb2_ccc1")) x library(stringr) out<-as.data.frame(str_split_fixed(x$x,"aaa",2)) out2<-as.data.frame(str_split_fixed(out$V2,"_bbb",2))

Hello! I am trying to understand how ggplot2's geom_bar treats NAs. The help file says: library(ggplot2) ?geom_bar na.rm: If FALSE, the default, missing values are removed with a warning. If TRUE, missing values are silently removed. I am trying it out: md <- data.frame(a = c(letters[1:5], letters[1:4], letters[1:3], rep(NA, 3))) str(md); levels(md$a) ggplot(data = md, mapping = aes(x =

Just a thought: Did you try na.rm = TRUE in case you have an object named "T" in scope? -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Thu, Jul 27, 2017 at 7:49 AM, Dimitri Liakhovitski <dimitri.liakhovitski at

Hello! s<-"start"; e<-"end" middle<-as.character(c(1,2,3)) I would like to get the following result: "start 123 end" or "start 1 2 3 end" or "start 1,2,3 end" How can I avoide this (undesired) result: paste(s,middle,e,sep=" ") Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

?hanks for the advice, Jeff. Will keep it in mind. But I am anal - I shy away from using letters and words that "look familiar" to me in R (such as mean, sd, T, etc.) But still, it's a good advice. On Thu, Jul 27, 2017 at 11:53 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > I think you should be more suspicious of yourself, Dimitri. A letter T > variable can

To clarify: my question is not about "who could I exclude NAs from being counted" - I know how to do that. My question is: Why na.rm = T is not working for geom_bar in this case? On Thu, Jul 27, 2017 at 8:24 AM, Dimitri Liakhovitski < dimitri.liakhovitski at gmail.com> wrote: > Hello! > > I am trying to understand how ggplot2's geom_bar treats NAs. > The help file

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

Hello everyone! I have a list X with 3 elements, each of which is a data frame, for example: a<-data.frame(a=1,b=2,c=3) b<-data.frame(a=c(4,7),b=c(5,8),c=c(6,9)) c<-data.frame(a=c(10,13,16),b=c(11,14,17),c=c(12,15,18)) X<-list() X[[1]]<-a X[[2]]<-b X[[3]]<-c (X) How can I most effectively transform X into a data frame with columns a, b, and c? I would love to find a generic

I have dose response data analyzed with the package 'drc'. 'summary(mymodel)' prints my kinetic parameters. I want that text in an ASCII text file. I want to get exactly what I would get if I copied and pasted from the terminal window. I've read the documentation on data export to text files here:

Hello! I am using function "curve" to create a line graph. I was wondering, if it's possible to "turn off" the default tick marks and introduce those tick marks in specific locations. For example, currently in my X axis tick marks are (automatically) at 10, 11, 12, 13 but I want them to be in 5 specific locations, like 9.89, 10.34, etc. Any hint would be greatly

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com