similar to: Extract subsets of different and unknown lengths from huge dataset

Dear useRs, Thanks for any advices # I do not know where are the examples how to mark groups # based on signal occurence in the additional variable: cf. variable c2, # How to calculate different calculations for groups defined by (split by occurence of c2 characteristic data) #First example of simple data #mexample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Hello, I really wonder how to distinguish between secs and mins in the example below. In other terms, how can I access the time unit in variable d ? start1 <- Sys.time(); stop1 <- Sys.time(); d <- stop1-start1; print(d); v<-unlist(strsplit(as.character(d), split=" ")); print(v) Time difference of 3.024054 secs [1] "3.02405381202698" stop1 <- Sys.time(); d

Hi, I was trying to get a graph in lattice with the following data frame (7 rows, 5 cols): chr start1 end1 meth positive 1 1 10 20 1.5 y 2 2 12 18 -0.7 n 3 3 22 34 2.0 y 4 1 35 70 3.0 y 5 1 120 140 -1.3 n 6 1 180 190 0.2 y 7 2 220 300 0.4 y I wanted the panels to be organized by 'chr' -

> Suppose you have four lifetime operations on the same address in memory, > with loads and stores all around them: > > start1--end1 .. start2--end2 > > If you remove start1 then you have a bare pointer, the memory came from > somewhere and you lose the optimization that loads before start1 become > undef, but you don't miscompile. This is assuming no looping after

On 12/27/2012 12:35 PM, Rafael Espíndola wrote: >>> Oh, I was reading "precedes/following" as having static (dominance) >>> meaning. That is, in the above example you could not delete the store >>> since it is not true that >>> llvm.lifetime.end dominates it. >>> >>> Nick, is this what you had in mind? If not, then we must delete a

Hello, I have a specific problem, I have a large dataframe, and after clustering I want to select certain colums, the elements of a subcluster. My dataframe looks like this : > colnames(data) [1] "101KF4319097339" "102KF4319101170" "103KF4319047549" "104KF4319046389" [5] "105KF4319013260" "106KF4319025582"

Can someone explain why I am getting the following error: in the r code below? Error in solve.default(diag(2) + ((1/currvar) * (XX1 %*% t(XX1)))) : system is computationally singular: reciprocal condition number = 0 In addition: There were 50 or more warnings (use warnings() to see the first 50) The R code is part of a bigger program. ##sample from full conditional

[Please keep r-help in the cc: list] I don't quite know how to interpret the difference between specifying effort as an offset vs. as weights; I would have to spend more time thinking about it/working through it than I have available at the moment. I don't know that specifying effort as weights is *wrong*, but I don't know that it's right or what it is doing: if I were

Hi, all I can not figure this out, please have a look and help me out. thank you! Note: this is in SPLUS, not R. I have following code *********************************** modfit<-function(yir,yew, ft) { n<-length(yew) yew<-yew[1:(n-1)] yy<-yir-ft xx<-yew-ft n<-length(xx) xx0<-xx[2:n] yy0 <-yy [2:n] xx1<-xx[1:(n-1)] fit <- garch(yy0~xx0 + xx1+var.in.mean,

Dear all, I am a biologist. I have two sets of distance P(start1, end1) and Q(start2, end2). The distance will be like this. P ------------------------ Q ---------------------------------------- I want to know whether P falls closely to the right end or left end of Q. P and Q are of different lengths for each data point. There are more than 10000 pairs of P and Q. Is there any test or

My data consists of a numeric (yy) variable and a categorical (xx) variable, as shown below. xx = c(rep("C", 5), rep("D",5)) yy = rnorm(10, 0, 4) xx1 = as.integer(as.factor(xx))   plot(xx1, yy, ylim = c(-13.5, 4), col="blue")   I wish to generate a scatter plot of the data such that instead of 1 it prints C, and instead of 2, it prints D on the x- axis.

I've been using the following code to plot using Sweave in version 1.9.1 library(RODBC) library(lattice) channel <-odbcConnectExcel("h:/water.xls") data <- sqlQuery(channel,"select * from `Sheet1$` where Test = 'TOC' and (Valve='5010-05' or Valve='8030-V26' or Valve='1180-08' or Valve='5040-08')") odbcClose(channel) srt <-

Hi All, I am trying to convert the following piece of matlab code to R: XX1 = sum(w(:,ones(1,N1)).*X1.*X1,1); #square the elements of X1, weight it and repeat this vector N1 times XX2 = sum(w(:,ones(1,N2)).*X2.*X2,1); #square the elements of X2, weigh and repeat this vector N2 times X1X2 = (w(:,ones(1,N1)).*X1)'*X2; #get the weighted 'covariance'

fb_do_apertures_overlap is returning wrong value when one aperture is completely whithin the other. Add generic ranges_overlap macro (probably kernel.h candidate) and use it here. Signed-off-by: Marcin Slusarz <marcin.slusarz at gmail.com> Cc: Dave Airlie <airlied at redhat.com> Cc: Peter Jones <pjones at redhat.com> Cc: Andrew Morton <akpm at linux-foundation.org> ---

Hello, I have a nasty loop that I have to do 11877 times. The only thing that slows it down really is this merge: xx1 = merge(dt,ua_rd,by.x=1,by.y= 'rt_date',all.x=T) Any ideas on how to speed it up? The output can't change materially (it works), but I'd like it to go faster. I'm looking at getting around the loop (not shown), but I'm trying to speed up the merge first.

On some systems (alpha), tick marks don't appear on plots. The easiest way to see the problem is something like: > plot(0:1,axes=FALSE) > axis(1,1:2) The problem is in X11_Line(...) from .../src/unix/devX11.c, which is so short I've included the whole function below: static void X11_Line(double x1, double y1, double x2, double y2, int coords, DevDesc *dd) {

Hi there: I have this configuration: |-----[Server 2] | [Internet]--------[Router]----------[Switch]------------ [Server 1] | |-----[PC1] | |-----[PC2] | |-----[PC3] Server 1 has IP 216.251.XXX.XX1 Server 2 has IP 216.251.XXX.XX2 PC1 has IP 216.251.XXX.XX3 PC2 has IP 192.168.XXX.1 PC3 has IP 192.168.XXX.2 How do I configure shorewall in SERVER 2 to block to/from the Internet Port 22

Dear all, I am stumped at what should be a painfully easy task: predicting from an lm object. A toy example would be this: XX <- matrix(runif(8),ncol=2) yy <- runif(4) model <- lm(yy~XX) XX.pred <- data.frame(matrix(runif(6),ncol=2)) colnames(XX.pred) <- c("XX1","XX2") predict(model,newdata=XX.pred) I would have expected the last line to give me the

I'm having problems with the iPhone client in the 1.1.1 version of the iPhone software. Things with other clients work fine, and the iPhone worked fine before it was upgraded to 1.1.1. Other phones on 1.1.1 failed. I'm still working on testing against the client with the iPhone 1.0.2 software on it. The problem is that the client simply doesn't connect to dovecot. I can see the

Dear R-users, Attach with is my data..what i want to do is finding a suitable distribution for my data..I want to run a few test like the poisson and the exponential distribution. Please help me on how to find the p-value for poisson as well as the exponential distribution without knowing the parameter. Is it possible?? Thanks in advance. love, Anisah -------------- next