similar to: Intervals in function cut

Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> writes: > I've rolled up R-1.0.0.tgz a short while ago. > I've build R-1.0.0 on my >uname -a Linux pchoel 2.2.14 #3 Mit Jan 5 08:57:39 MET 2000 i686 unknown box. Calling "make check" fails with ... make[4]: Entering directory `/opt/src/devel/R-1.0.0/tests/Examples' updating base-Ex.R ... make[5]: Entering

Hi! I had a database with some variables in sequence. Let me say: TX01, TX02, TX03 and TX04. But I need to run some regressions changing the variables... so: variable <- paste("TX0", 1:4, sep="") for(i in 1:4){ test[i] <- lm(variable[i] ~ INCOME, data=database) } But doesn't work... lm tries to find a variable inside database named variable[i] ... Suggestions?

I read a table as follows: > F1 <- read.table("Rtext3.txt") > F1 Price Floor Area Rooms Age Cent.heat a 52.00 111 830 5 6.2 no b 54.75 128 710 5 7.5 no c 57.50 101 1000 5 4.2 no d 57.50 131 690 6 8.8 no e 59.75 93 900 5 1.9 yes As it is seen, the rows have a name. However I don't know how to access a

Hi, I have a data set with 999 observations, for each of them I have data on four variables: site, colony, gender (quite a few NA values), and cohort. This is how the data set looks like: > str(dispersal) 'data.frame': 999 obs. of 4 variables: $ site : Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 2 2 ... $ gender: Factor w/ 2 levels "0","1":

I'm studying lists and I came to an example where > L $name [1] "Fred" $wife [1] "Mary" $no.children [1] 4 $child.ages [1] 4 7 9 then following the instructions to extend the list with a new component, I executed: > L[5] <-list(NewName="something") and the new list I got was: > L $name [1] "Fred" $wife [1] "Mary"

Hi, it looks like the IntervalPartition does not work as expected when constructed from another interval partition. Say, I have built an interval partition from function, and the first interval has two basic blocks. When I create second order partition and print all intervals, the second basic block of the function is not seen anywhere. Here's what's going on in IntervalIterator.h:

bhoel@server.python.net (Berthold Höllmann) writes: > Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> writes: > > > I've rolled up R-1.0.0.tgz a short while ago. > > > I've build R-1.0.0 on my > > >uname -a > Linux pchoel 2.2.14 #3 Mit Jan 5 08:57:39 MET 2000 i686 unknown > > box. Calling "make check" fails with .... > >

bhoel@server.python.net (Berthold Höllmann) writes: > Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> writes: > > > I've rolled up R-1.0.0.tgz a short while ago. > > > I've build R-1.0.0 on my > > >uname -a > Linux pchoel 2.2.14 #3 Mit Jan 5 08:57:39 MET 2000 i686 unknown > > box. Calling "make check" fails with .... > >

On Fri, 29 Apr 2005, Vladimir Prus wrote: > it looks like the IntervalPartition does not work as expected when constructed > from another interval partition. > > Say, I have built an interval partition from function, and the first interval > has two basic blocks. When I create second order partition and print all > intervals, the second basic block of the function is not seen

I thought I've understood the 'order' function, using simple examples like: order(c(5,4,-2)) [1] 3 2 1 However, I arrived to the following example: order(c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045)) [1] 8 9 10 7 11 6 5 4 3 2 1 and I was completely perplexed! Shouldn't the output vector be 11 10 9 8 7 6 4 1 2 3 5 ? Do I have a damaged

Hi, when I generated a survfit() object, I can get number of patients at risk at various time points by using summary(): fit<-survfit(Surv(time,status)~class,data=mtdata) summary(fit) class=1 time n.risk n.event survival std.err lower 95% CI upper 95% CI 9.9 78 1 0.987 0.0127 0.963 1 41.5 77 1 0.974 0.0179 0.940 1 54.0 76

I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply

Hello, not understanding the output of prcomp, I reduce the number of components and the output continues to show cumulative 100% of the variance explained, which can't be the case dropping from 8 components to 3. How do i get the output in terms of the cumulative % of the total variance, so when i go from total solution of 8 (8 variables in the data set), to a reduced number of

Hello. I know that it's usually possible to write cookies to a cookie file by removing the curl handle and doing a gc() call. I can do this with getURL(), but I just can't obtain the same results with postForm(). If I use: curlHandle <- getCurlHandle(cookiefile=FILE, cookiejar=FILE) and then do: getURL(http://example.com/script.cgi, curl=curlHandle) rm(curlHandle) gc() it's

Dear R-helpers, To get curves for a pseudo cohort other than the one centered at the mean of the covariates, I have been trying to use the newdata argument to survfit with no success. Here is my model statement, the newdata and the ensuing error. What am I doing wrong? > summary(fit) Call: coxph(formula = Surv(Start, Stop, Event, type = "counting") ~ Week + LagAOO + Prior.f +

In order to see how llvm-gcc-4.2 svn performs in code generation compared to the upcoming gcc 4.4, I ran the Polyhedron 2005 benchmarks on a MacPro with the -ffast-math -funroll-loops -msse3 -O3 optimization flags for both compilers. The results are summarized below. Ave Run (secs) Ave Run (secs) llvm-gcc-4.2/ Benchmark llvm-gcc-4.2 svn gcc trunk gcc trunk

I've rolled up R-1.0.0.tgz a short while ago. You can get it from ftp://cvs.r-project.org/pub/CRAN/src/base/R-1.0.0.tgz or http://cvs.r-project.org/pub/CRAN/src/base/R-1.0.0.tgz or wait for it to be mirrored at a CRAN site near you within a day or two. It should get to the CRAN master site within a few hours. There's also a version split in three for floppies if you prefer that.

I've rolled up R-1.0.0.tgz a short while ago. You can get it from ftp://cvs.r-project.org/pub/CRAN/src/base/R-1.0.0.tgz or http://cvs.r-project.org/pub/CRAN/src/base/R-1.0.0.tgz or wait for it to be mirrored at a CRAN site near you within a day or two. It should get to the CRAN master site within a few hours. There's also a version split in three for floppies if you prefer that.

Could you please help me on the following ANCOVA issue? This is a part of my dataset: sampling dist h 1 wi 200 0.8687212 2 wi 200 0.8812909 3 wi 200 0.8267464 4 wi 0 0.8554508 5 wi 0 0.9506721 6 wi 0 0.8112781 7 wi 400 0.8687212 8 wi 400 0.8414646 9 wi 400 0.7601675 10 wi 900 0.6577048 11 wi 900

I'm trying to figure out about factors, however the on-line documentation is rather sparse. I guess, factors are intended for grouping arrays members into categories, which R names "Levels". And so we have: * state <- c("tas", "sa", "qld", "nsw", "nsw", "nt", "wa", "wa",