search for: 0.124

It's checked in, it seems to work, it didn't bloat the system significantly, but it makes life a lot easier... I just released klibc-0.124 with the 64-bit off_t and statfs64 changes. The API shouldn't have changed, but the ABI is now 64 bits in (hopefully) all the right places. It's probably buggy... beat me up or (better) send a patch... -hpa

I'm having trouble running MAME on Wine 9.60 it's to slow. :P

Dear list, I have a data frame where one of the columns are p values with scientific notation mixed with regular numbers with decimals. >a=data frame >a P OR N 0.50 0.7500 237 0.047 1.1030 237 0.124 0.7742 237 0.124 0.7742 237 0.0080 1.1590 237 0.50 0.7500 237 4.5e-07 1.2 237 5.6e-04 0.9 237 when I try to do >pval=a$P/2 R gives me an error saying "In Ops.factor(pval, 2)

dear R wizards: here is the strange question for the day. It seems to me that nrow() is very slow. Let me explain what I mean: ds= data.frame( NA, x=rnorm(10000) ) ## a sample data set > system.time( { for (i in 1:10000) NA } ) ## doing nothing takes virtually no time user system elapsed 0.000 0.000 0.001 ## this is something that should take time; we need to add 10,000

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

The confidence interval calculation in prop.test appears to be incorrect when alternative="greater". The upper limit is always set to 1.000. Am I missing something? > total=c(250,250) > success=c(55,31) > prop.test(success,total,alternative="greater",correct=TRUE) 2-sample test for equality of proportions with continuity correction data: success out of

Dear subscribers, I am looking for a function which would allow me to model the dependent variable as the number of successes in a series of Bernoulli trials. My data looks like this ID TRIALS SUCCESSESS INDEP1 INDEP2 INDEP3 1 4444 0 0.273 0.055 0.156 2 98170 74 0.123 0.456 0.789 3 145486 30 0.124

Full_Name: Richard Johnston Version: 2.4.0 OS: OS X Submission from: (NULL) (69.169.0.241) The confidence interval calculation for prop.test appears incorrect when alternative="greater" . The upper limit is always set to 1.0000. The lower limit appears to be correct. > total=c(250,250) > success=c(55,31) >

Hi, Couldn't is.unsorted() bail out immediately here (after comparing the first 2 elements): > x <- 20000000:1 > system.time(is.unsorted(x), gcFirst=TRUE) user system elapsed 0.084 0.040 0.124 > x <- 200000000:1 > system.time(is.unsorted(x), gcFirst=TRUE) user system elapsed 0.772 0.440 1.214 Thanks! H.

Hello, I hesitate to call this a bug, because I could have forgotten something important, but the MarkowitzPortfolio example in fPortfolio does not work for me. Here's my code: > library(fPortfolio) > >xmpPortfolio("\nStart: Load monthly data set of returns > ") > data(berndtInvest) > # Exclude Date, Market and Interest Rate columns from data

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90

I am doing some coxPH model fitting and would like to have some idea about how good the fits are. Someone suggested to use Frank Harrell's C-index measure. As I understand it, a C-index > 0.5 indicates a useful model. I am probably making an error here because I am getting values less than 0.5 on real datasets. Can someone tell me where I am going wrong please ? Here is an example using

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

I want to generate multiple multivariate normal samples with different mean vectors and common covariance matrix. I can do this with a loop, but can't quite figure out how to do it with apply and friends. In the example below, I want values to have 3 columns: group, x, y # number of groups, and group means x <- jitter(seq(2,10,by=2)) y <- x + rnorm(length(x), 0, .5) means <-

... and the mailing list is picky about attachments... whatever you attached did not conform to the stringent requirements mentioned in the Posting Guide. Pasting the code right into the email is usually safest, though you DO have to post using plain text (as the Posting Guide indicates) or your code may get mangled by the automatic html format removal. On May 15, 2018 7:04:31 AM PDT, Bert Gunter

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

Dear R community members, I think I am asking a very simple question, but I really looked up in the faqs and manuals and found nothing helpful. I am trying to plot a data frame with the following structure (this is just a small extract): glo conc odor line series X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 1 0 AIR LN1 UP -0.488

Unless there is good reason not to, always cc the list -- there are lots of smarter folks than I on it who can help. I may or may not have time to look at this. Hopefully someone else will. -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip

Dear R users I have a big matrix like 6021 1188 790 290 1174 1015 1990 6613 6288 100714 6021 1 0.658 0.688 0.474 0.262 0.163 0.137 0.32 0.252 0.206 1188 0.658 1 0.917 0.245 0.331 0.122 0.148 0.194 0.168 0.171 790 0.688 0.917 1 0.243 0.31 0.122 0.15 0.19 0.171 0.174 290 0.474

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need