similar to: Unusual slowing of R matrix multiplication version 2.12.1 (2010-10-15) vs 2.12.0

Hi, Would you help me solve the following question? Thanks. Question: If I try to set the probability=0.05 and find the approximate x. (The answer should be somewhere between 2.1782 and 2.1783.) I write about this R program as follows but I don¡¦t know how to get the value of x which is between 2.1782 and 2.1783. library(mvtnorm) value<-array(1000) a<-array(1000)

Hello everyone, I have a dataset with 5 colums (4 colums with thresholds of weather stations and one with month - data of 5 years). Now I would like to calculate the average for each month. I tried this unsuccessfully: lf.med <- sapply(LF[,1:4],mean,LF[,5]) Error in mean.default(X[[1L]], ...) : 'trim' must be numeric and have length 1 With lf.med <- by(LF[,1:4],LF[,5],mean)

Hi all, I'm trying to generate a Weibull distribution including four covariates in the model. Here is the code I used: T = rweibull(200, shape=1.3, scale=0.004*exp(-(-2.5*b1+2.5*b2+0.9*x1-1.3*x2)/1.3)) C = rweibull(n, shape=1.5, scale=0.008) #censoring time time = pmin(T,C) #observed time is min of censored and true event = time==T # set to 1 if event is observed

Hello friends. I'm wondering what happened to package uroot. I worked quite well with older versions of R, however with 2.12.1 version it simply doesn't work. The worst thing is that I look for it in the contributed packages and it simply doesn't appear. I want to develop ADF test with seasonal (centered) dummies, is there any other possible package that contains this test? Thanks

I've compiled R-2.12.1 on OpenBSD and everything works great from the console. However, if I load R with the --gui=Tk option, R opens in a window and will accept input, but will crash (90% of the time) after trying to execute the first command. For example, if I type: > mynumbers <- c(1,2,3,4) Then, when I press enter, I receive: *** caught segfault *** address 0x8ab88000, cause

This is to announce that we plan to release R version 2.12.1 on Thursday, December 16, 2010. Those directly involved should review the generic schedule at http://developer.r-project.org/release-checklist.html The source tarballs will be made available daily (barring build troubles) via http://cran.r-project.org/src/base-prerelease/ For the R Core Team Peter Dalgaard -- Peter Dalgaard

This is to announce that we plan to release R version 2.12.1 on Thursday, December 16, 2010. Those directly involved should review the generic schedule at http://developer.r-project.org/release-checklist.html The source tarballs will be made available daily (barring build troubles) via http://cran.r-project.org/src/base-prerelease/ For the R Core Team Peter Dalgaard -- Peter Dalgaard

R 2.12.1 is built now for Fedora and EPEL. It is in "updates-testing" (or it will be within the next 24 hours). In accordance with the new policies on Fedora Updates, these new packages will not be pushed as official updates until they either receive positive testing from users, or sit in updates-testing for two weeks. You can help us test these packages, and move them forward.

Hello, and thanks for your time reading this. I'm trying to test interactions of my dataset, in which the all of the factors are within the same column. Type Vol 1 CMass -4.598 2 BBack -4.605 3 BMass -4.602 4 CMass -4.601 5 CBack -4.605 6 CMass -4.604 7 CMass -4.602 8 CMass -4.604 9 CBack -4.605 10 BBack -4.503 11 CMass -4.605 Im attempting to determine the interaction effects of B or C

I have a factor vector of subject races (Asian, Black, Hispanic, White; n=30) that I want to plot with a Cleveland dotplot or dotchart. I tried the following in R2.12.1 : > dotchart(table(school$Race)) Error in plot.xy(xy.coords(x, y), type = type, ...) : invalid plot type Using the same data set in R2.11.1 the operation succeeded (I tried several variations to be sure): >

Hi, When using lapply (or sapply) to loop over a list, can I somehow access the index of the list from inside the function? A trivial example: df1 <- split( x=rnorm(n=100, sd=seq(from=1, to=10, each=10)), f=letters[seq(from=1, to=10, each=10)] ) str(df1) #List of 10 # $ a: num [1:10] -0.801 0.418 1.451 -0.554 -0.578 ... # $ b: num [1:10] -2.464 0.279 4.099 -2.483 1.921 ... # $ c: num

Dear R developers, I just compiled the latest version of R (2.12.1) and noticed that 'libRblas.so' is missing in the '/x86_64/src/extra/blas' subdirectory of my R-installation. Did I miss ongoing discussions on the Mailinglist about this or might it be a local problem? Thanks for this brilliant software. Best Christian -- Christian Kohler Institute of Functional Genomics ~

Hi, the following Code demonstrates an possibly Error in R (or you can explain me, why this happens, thanks in advance) Code: ##################################### testClass <- function( stackData= c()) { list( write= function( ...) { sChain= "" for( s in c( stackData, ...)) { sChain= paste( sChain, '"', sub( '"',

Dear R-HELP, We are compiling R-2.12.1 for 64 bits onto a Sun Sparc machine below: SunOS 5.10 Generic_142900-13 sun4u sparc SUNW,Netra-T12 The source was compiled successfully with "gcc version 3.4.6" by using the default configuration and has produced R-2.12.1 in 32 bits. Compiling R-2.12.1 for the 64 bits by configuring the config.site file as follow: CC="gcc -m64"

Dear List, With the R 2.12.0 addition of table methods for points(), dotchart() struggles with tables. I found several possible solutions, but it is beyond my skill to decide what is "best". Here is a small example: ############################################# x <- table(infert$education) y <- 1:3L dotchart(x) # error about incorrect plot type ## moving closer to the cause,

I've written a very simple bit of code to plot a trajectory using rgl: x <- (c(0,-5.947,-11.496,-16.665,-21.474,-25.947,-30.116,-34.017,-37.684,-41.148,-44.435,-47.568,-50.567,-53.448,-56.223,-58.906)) y <- (c(0,33.729,65.198,94.514,121.784,147.151,170.797,192.920,213.717,233.360,252.003,269.774,286.781,303.117,318.858,334.071)) z <-

Dear forum members, How can I force a negative slope in a linear regression even though the slope might be positive? I will need it for the purpose of determining the trend due reasons other than biological because the biological (genetic) trend is not positive for these data. Thanks. Julia Example of the data: [1] 1.254 1.235 1.261 0.952 1.202 1.152 0.801 0.424 0.330 0.251 0.229

Dear list, Please consider the following call of sort > sort(c("a","f")) [1] "a" "f" > sort(c("f","a")) [1] "a" "f" > > sort(c("aa","ff")) [1] "ff" "aa" > sort(c("ff","aa")) [1] "ff" "aa" The last two results look strange

Hi list. I was saving my modified pairs graphic using a custom panel from the R Graphics site, and I got an interesting difference in the final image when I save it as eps or png. This custom panel make possible to show at the left side of the pairs plot the p-value as symbol and the correlation r value with its cex proportional to the r value itself. Well, when saving as a png file

Hi everyone. I have have a dataset with daily measurement from January 1st of 1966 up to December 31th of 2011. Here's the first part of the data: Date SLEV 1/1/1966 1.086 1/2/1966 1.079 1/3/1966 1.133 1/4/1966 1.261 1/5/1966 1.391 1/6/1966 1.571 1/7/1966 1.728 1/8/1966 1.823 1/9/1966 1.97 1/10/1966 1.804 1/11/1966 2.02 1/12/1966 2.017 1/13/1966 1.86 1/14/1966 1.96 1/15/1966 1.813 1/16/1966