similar to: help server slow

I recently got a new machine, running RedHat Linux 7.0 (kernel 2.2.16, XFree864.0.1), and tried to install R-1.1.1 on it from the rpm file on CRAN. RPM complained about missing dependencies: libreadline.so.3 libtcl8.0 libtk8.0 The RH7.0 installation includes later versions of these libraries. I went ahead and forced an installation, and of course R wouldn't work. By linking

Is there any R code available for solving ODE's? I am considering building an interface to ODEPACK, but don't want to duplicate any existing effort. -- Tom Wainwright NOAA/NMFS/NWFSC 2030 S Marine Science Dr Newport, OR 97365 USA thomas.wainwright at noaa.gov -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

If I wanted to fit a logit model and account for clustering of observations, I would do something like: library(Design) f <- lrm(Y1 ~ X1 + X2, x=TRUE, y=TRUE, data=d) g <- robcov(f, d$st.year) What would I do if I wanted to do the same thing with a probit model? ?robcov says the input model must come from the Design package, but the Design package appears not to do probit? Thanks very

Hi there, I have a set of data that looks like this: As1988<-c(1254.0, 22.0, 4.2, 1081.0, 35.0, 6.0, 1772.0, 192.0, 7.6) The mean of this (as calculated by R) is: 485.9778 The median of this (as calculated by R) is: 35 If I then make a beanplot(As1988), I find that the beanline (average) is now 77.68561 while the beanline (median) is 35.39739 (using the locator function to check the

Dear all, I am a R beginner, and I am looking for a way to do the same thing for all levels of a column in a table. Basically, I have a bunch of protein sequences composed of different amino acid residues, and each residue is represented by an uppercase letter. I want to calculate the ratio of different amino acid residues at each position of the proteins. Here is an example table: Proteins

Hello R users, For the last few days I am struggling with the following task: my data.frame: A1 A2 A3 B1 B2 B3 58.81 53.292 54.501 13.013 17.39 19.407 56.02 56.251 54.033 20.099 13.15 10.411 55.376 53.099 57.625 13.396 21.031 13.22 58.584 53.194 54.218 13.038 16.854 19.289 55.7 55.921 53.847 19.942 13.153 9.828 55.093 52.934

Hi, I want to make a plot similar to sm1 (attached). The code I tried is: dcols <- densCols(x,y) smoothScatter(x,y, col = dcols, pch=20,xlab="A",ylab="B") abline(h=0, col="red") But it turned out to be s1 (attached) with big dots. I was wondering if anything wrong with my code. Thanks,Zhengyu -------------- next part -------------- A non-text

Hi people, I'm having problems of connection with a Linksys SPA IP PHONE 942 when I use the WAN port, most of the time when I try to connect to the network or restart the IP Phone I can't get internet connectivity . I tried using both static IP and DHCP, but the problem is the same. Some have had similar problem with this brand of IP Phone? Thanks for the help and attention. Hugs

Hi, My Fedora 11 linux (version 2.6.30.9-96.fc11.x86_64 #1 SMP Wed Nov 4 00:02:04 EST 2009) recently updated me to R-2.10.0, and help.start() is no longer working. I see in the release notes that R now uses an internal web server to autogenerate help pages. Here's what I'm getting: IN R: > help.start() starting httpd help server ... done If

Hi all, I have two vectors (columns) called "efinal" and "efinal 2". I want to plot them on the same plot and "draw" a shaded area beween the two lines using function polygon I have tried all but I don ?t understand the polygon area, can you help me with examples? plot(efinal,type="l",ylim=range(min(efinal2),

My nightly logwatch report had a never before seen section last night, "barracuda spam firewall". I have not problem with the emails it noted as being rejected. But I've always thought of "barracuda" as a commercial product. I have neither configured nor enabled any barracuda software and "yum list '*barrac*'" comes up empty. What is this? Jon -- Jon

I am lousy at simple regex and I have not found a solution to a simple problem. I have a vector with some character values that I want to split. Sample data dd1 <- c( "XXY (mat harry)","XXY (jim bob)", "CAMP (joe blow)", "ALP (max jack)") Desired result dd2 <- data.frame( xx = c("XXY", "XXY", "CAMP",

Dear Johnny, Yes, I have tried much larger numbers than 60 and 3 for the above two parameters respectively. And I am sure it is using ?protocol 2? because it?s uncommented in sshd_config. Is there a way to catch what?s happing before quit? I couldn?t see anything except for the line ?write failed, broken pipe?. Thanks. Hua At 2015-10-08 19:37:24, "Johnny Hughes" <johnny at

hi i try to do a graph of a time series which shows in red the values > -0.05, in blue the values >0.05 and in white the values between -0.05 and 0.05 for un exemple :http://www.appinsys.com/globalwarming/enso.htm thanks !!!!! denisse -- View this message in context: http://r.789695.n4.nabble.com/how-to-do-a-graph-with-tree-different-colors-tp4635206.html Sent from the R help mailing

(First time post. Please be kind!) I've done some googling, read up on speex, found this mailing list, and have been following it for a few weeks. Now I really have to ask: Can it really be true that there are as yet no hardware phones which support the speex codec? Hard to believe (unless perhaps there's a good technology reason?) I'm not a developer - I'm a purchaser. I

Hi! Trying to make a forecast, and get the following error message: Error in NextMethod("[<-") : subscript out of bounds The script is as follows: > Forecast.A <- ts(matrix(NA, nrow=25, ncol = 1, + dimnames = list(c(), c("Outcome"))), + start = c(2006, 10), frequency = 12) > for (i in 1:25) { + j <- i + 321 + Data <- window(omxr, end = time(omxr)[j]) +

Hi, Kbytes RSS rNo 2689632 1450876 1433788 2689632 1450876 1433788 2689632 1450876 1433788 2689632 1450876 1433788 2689632 1450876 1433788 2689632 1450876 1433788 2689632 1450876 1433788 2689632 1460168 1443084 plot(data$rNo,data$RSS,pch=0,type="b",col=36,axes=FALSE, ylab="", xlab="",las=2,lwd=2.5) title(" PID -

Dear All, I have a data sets as shown below A (Patient ID ), B and C are the Concentration of drug in blood on day 1 and day 4, D is the difference in conc. To do this in R I have written a code as follows, identified the number of patients who have more concentration on day 4 . Here I want to identify specifically the patient ID (is he patient 1 or 2 or 5 and 7), whose concentration is more. How

I would like to plot two data sets (frequency (y-axis) of mean values for 0-1(x=axis)) on a single histogram for comparison. The hist() only allow the overlay of two histograms, and although barplot() allows beside=TRUE, it does not show frequency values (like hist) but rather all of the values. Is there any way that I can use the hist() to plot two data sets similar to the barplot(). Any help or

I am following a document teaching how to use regression and right at the onset I get an R error. I understand that variables "conc" and "signal" should have the same length but I am using the what R manual suggests to drop the error and it is not cooperating. What gives? The manual says that the default is na.ombit by the way, not that it's addition does anything How