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G'Day Tanya, Is it too late to bring in the following patches to fix some major brokenness in the AuroraUX tool chain for 2.6? http://llvm.org/viewvc/llvm-project/cfe/trunk/lib/Driver/Tools.cpp?r1=84468&r2=84469&view=diff&pathrev=84469 http://llvm.org/viewvc/llvm-project/cfe/trunk/lib/Driver/Tools.cpp?r1=84265&r2=84266&view=diff&pathrev=84266

Hi Tanya, > 1) Compile llvm from source and untar the llvm-test in the projects > directory (name it llvm-test or test-suite). Choose to use a > pre-compiled llvm-gcc or re-compile it yourself. I compiled llvm and llvm-gcc with separate objects directories. Platform is x86_64-linux-gnu. > 2) Run make check, report any failures (FAIL or unexpected pass). Note > that you need to

On Oct 20, 2009, at 6:02 AM, Duncan Sands wrote: > Hi Tanya, > >> 1) Compile llvm from source and untar the llvm-test in the projects >> directory (name it llvm-test or test-suite). Choose to use a pre- >> compiled llvm-gcc or re-compile it yourself. > > I compiled llvm and llvm-gcc with separate objects directories. > Platform is x86_64-linux-gnu. > Ok.

LLVMers, 2.6 pre-release2 is ready to be tested by the community. http://llvm.org/prereleases/2.6/ If you have time, I'd appreciate anyone who can help test the release. To test llvm-gcc: 1) Compile llvm from source and untar the llvm-test in the projects directory (name it llvm-test or test-suite). Choose to use a pre- compiled llvm-gcc or re-compile it yourself. 2) Run make check,

Hello, I have two data frames, X and Y, with two columns each and different numbers of rows. # creation of data frame X Loc1.alleles <- c(1,5,6,7,8) Loc1.Freq <- c(0.35, 0.15, 0.05, 0.10, 0.35) Loc1 <- cbind( Loc1.alleles,Loc1.Freq) X <- data.frame(Loc1) #creation of data frame Y Loc2.alleles <- c(1,4,6,8) Loc2.Freq <- c(0.35, 0.35,

I have a fairly large set of data with the following attributes: >str(raw.data) `data.frame': 1429 obs. of 16 variables: $ TStamp :`POSIXlt', format: chr "2001-11-25 02:00:00" "2001-11-25 01:55:00" "2001-11-25 01:50:00" "2001-11-25 01:45:00" ... $ iPDT.AHU14.14: num 0.0122 0.0125 0.0120 0.0120 0.0122 ... $ iPDT.AHU14.15: num 0.0121

Hello, I get > summary(E) level nodes ave_nodes time Min. : 1 Min. : 1.00 Min. : 10.71 Min. : 0.0000 1st Qu.: 237414 1st Qu.: 2.00 1st Qu.: 19.70 1st Qu.: 0.0100 Median : 749229 Median : 3.00 Median : 27.01 Median : 0.0100 Mean : 767902 Mean : 49.85 Mean : 98.89 Mean : 0.2296 3rd

Hello, My name is Craig and I need help plotting a "line" for a multiple linear regression in R. Here is my sample data (filename: convis.txt) Output of convis.txt is (vis and density being predictors of either avoidance or entrance): vis den avoid entrance 1 10 1 0.0000 0.0000 2 10 3 0.8750 0.0000 3 8 3 0.8180 0.0300 4 8 3 0.6670 0.0667 5 8 1

I just upgraded our optimizer to LLVM 3.0 from 2.8 and noticed that the scalar replacement of aggregates pass takes a lot longer for some code. Has there been a performance regression in this pass, or does it do more work? LLVM 3.0: Total Execution Time: 1.0600 seconds (1.0526 wall clock) ---User Time--- --System Time-- --User+System-- ---Wall Time--- --- Name --- 0.5100

Hi guys... I have problem with this excersise... Consider the pressure data frame again. (a) Plot pressure against temperature, and use the following command to pass a curve through these data: > curve((0.168 + 0.007*x)?(20/3), from=0, to=400, add=TRUE) (b) Now, apply the power transformation y3/20 to the pressure data values. Plot these transformed values against temperature. Is a linear

f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN", "119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos", "119OrgP", "119Phos", "119TKN", "119TOC", "148DAmm", "148DN", "148DNN", "148DO",

On Thu, 30 Jun 2005, Tanu Sharma wrote: > I am compiling SPEC 2000 benchmarks with llvm .Got stuck with > calculating "execution time" of all the .bc and native files. > > The log for nightly test itself gives execution times but I am passing > the bytecode files to my pass which gives another bytecode file.I have > to calculate execution time of such bytecode and

Dear R users, I have a response variable in a csv file called "y" and a matrix of predictor variables in a csv file called "mat". I have used the function "nls" I have specified the nonlinear relation between these variable.The code I have witten is called Rprog which begins with the phrase: L.minor.m1<-nls(Y~a ....etc.. The program when I execute the program, I

Hello , I am compiling SPEC 2000 benchmarks with llvm .Got stuck with calculating "execution time" of all the .bc and native files. The log for nightly test itself gives execution times but I am passing the bytecode files to my pass which gives another bytecode file.I have to calculate execution time of such bytecode and native files as well.If i simply do this: time lli

Hello All, Thanks for the reply.I can generate the reports by compiling Spec through llvm, but that couldn't resolve my problem. I m trying to determine execution time for the bytecode and native files , which are obtained as a result of running my pass over the original bytecode .I am running these experiments on spec benchmark. In SPEC we have command line tools such as runspec where

Dear R users, I am doing Cox regression using coxph(Survival) or cph(Design). I have time varying effects (diagnosed with schoenfeld residuals Chi2 test and graph) so I first want to split time into 2 separate intervals : t<6months and t>=6months, to estimate one hazard ratio (hr) for each interval. I am analysing Overall survival according to 3 prognostic factors (age,deep,ldh).

Dear R members, I am new in using frailty models in survival analyses and am getting some contrasting results when I compare the Wald and likelihood ratio tests provided by the r output. I am testing the survivorship of different sunflower interspecific crosses using cytoplasm (Cyt), Pollen and the interaction Cyt*Pollen as fixed effects, and sub-block as a random effect. I stratified

Hi, I am trying to use optim to solve a heavy calibration problem. I supply the parameters in vector form. But before entering my target The call is simply: optim(par = parameters, fn = SumLSQ, method = "Nelder-Mead") the function SumLSQ is simply: SumLSQ<-function(parameters, data = timeseries){ print("sumLSQ") nbseries =

Hi List, I have spent more than 30 minutes, but failed to read in this file using the read.table() function. I could not figure out how to fix the following error. Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 6 elements Any help would be be appreciated. Thanks, Pradip Muhuri ####### below is the reproducible example xd1 <-

This may not actually be an R/Splus problem, but it started off that way ..... ===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+=== Executive summary: ================== Simulations involving extreme value distributions seem to ``work'' when the underlying distribution is exponential(1) or exponential(2) == chi-squared_2, but NOT when the underlying distribution is