search for: 0.125

I have a file: x y z 0.025 0.025 1.65775 0.025 0.050 1.62602 0.025 0.075 1.63683 0.025 0.100 1.91847 0.025 0.125 2.00913 0.025 0.150 1.82222 0.025 0.175 1.70901 0.025 0.200 1.39759 0.025 0.225 1.39089 0.025 0.250 1.04762 If I read the file like this: data<-read.table("file.dat") How do I access the vectors x,y,z that are inside the dataframe data? I studied Venables and

Sorry if this is the wrong list, but doing a code checkout via Mercurial produces the following: #hg clone http://xenbits.xensource.com/xen-3.2-testing.hg destination directory: xen-3.2-testing.hg requesting all changes adding changesets abort: could not import module ELFCLASS32! transaction abort! rollback completed Is there a workaround for this right now? -- Joe Auty NetMusician: web

Is there a way to have printing data.frames with POSIXct to display milliseconds if digits.secs is set as a default? You can use the digits argument in print, such as print(df, digits = 3) to get the intended output, but I assumed it was done with the option digits.secs set. Tibbles by default do this printing, which is shown below, but I was unsure if digits.secs should affect printing

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hey guys, One of our developers DESTROYED both yum and rpm on one of the development machines with an rm command. And guess who's lucky enough to get to fix this mess? *raises hand* So I hope y'all don't mind me turning to you once again for some advice. You've been wonderful in the past. Pavlov's dogs as they say. At any rate, I've managed to fix rpm (yay) by copying

Hi, In the exactRankTest package, I've become aware that you can get anomalous p-values (i.e., above 1) from the wilcox.exact method, as in: > wilcox.exact(c(-0.6,0.8,-0.5)) Exact Wilcoxon signed rank test data: c(-0.6, 0.8, -0.5) V = 3, p-value = 1.25 alternative hypothesis: true mu is not equal to 0 This is disturbing. Has anyone encountered this before, and if so

Right now, all profile information is funneled through two analysis passes prior to any part of the optimizer using it. First, we have BranchProbabilityInfo, which provides a simple interface to the simplest form of profile information: local and relative branch probabilities. These merely express the likelihood of taking one of a mutually exclusive set of exit paths from a basic block. They are

I have a data.frame with three columns + the row number Sample MidPoint Count [1,] 1 0.025 1 [2,] 1 0.075 3 [3,] 1 0.125 15 [4,] 1 0.175 12 [5,] 1 0.225 5 [6,] 1 0.275 8 When I export the data to a csv the column names are shifted because the row numbers are being exported as well. Sample MidPoint

First I would really like to thank the mailing list for help I got in the past, as a new to R I am really needing some support on hoe to code the following problem. I am trying to sort some data I have in a big file. The file has 4 columns and 19000 rows. An example of it looks like this:- G 0.892 A 0.108 G 0.883 T 0.117 T 0.5 C

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Full_Name: Amos Waterland Version: 2.8.1 OS: Ubuntu Linux Submission from: (NULL) (68.175.8.163) I calculated the covariance for a small data set as follows: X <- c(1,2,3,4) Y <- c(3,3,4,3) cov(X,Y) [1] 0.1666667 But when doing the computation with pencil and paper I get: ((-1.5)*(-0.25) + (-0.5)*(-0.25) + (0.5)*(0.75) + (1.5)*(-0.25))/4 [1] 0.125 Microsoft Excel 2003 covar() also

I am tangled with a syntax question. I want to calculate basic statistics for a large dataset provided in weights and values and I can't figure out an elegant way to expand the data. For example here are the counts: > counts n4 n3 n2 n1 p0 p1 p2 p3 p4 1 0 0 0 1 1 3 16 55 24 2 0 0 0 0 2 8 28 47 15 3 1 17 17 13 4 5 12 24 8 ... and the values: > values

On Sun, Feb 2, 2014 at 6:18 PM, Andrew Trick <atrick at apple.com> wrote: > > On Feb 2, 2014, at 2:13 AM, Chandler Carruth <chandlerc at gmail.com> wrote: > > > Right now, all profile information is funneled through two analysis > passes prior to any part of the optimizer using it. > > > > First, we have BranchProbabilityInfo, which provides a simple

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,] 

Dear R helpers, I have been using the loadings function from the multiv library and I get the typical output (see below). When I try to export these results to a file using a write.table() I get the following error message "Error in as.data.frame.default(x[[i]], optional = TRUE) : can't coerce loadings into a data.frame" Any idea why write.table is doing that and any

On Feb 3, 2014, at 12:13 AM, Andrew Trick <atrick at apple.com> wrote: > On Feb 2, 2014, at 6:55 PM, Chandler Carruth <chandlerc at gmail.com> wrote: > >> On Sun, Feb 2, 2014 at 6:18 PM, Andrew Trick <atrick at apple.com> wrote: >> >> On Feb 2, 2014, at 2:13 AM, Chandler Carruth <chandlerc at gmail.com> wrote: >> >> > Right now,

Hi All I am getting the above mentioned error when I run the code below. I don't know why because I have implemented the function and I'm calling the function with a parameter. I'm obviously missing the plot ... Can someone perhaps point out the error of my ways? Error: > out<-ode(y=init, times=times, func=G1999, parms=parms, method="lsoda") Error in m_Na(v) : attempt

Hi I have the feeling, that the argument stringsAsFactors has no impact in the function expand.grid: a <- c("PR", "NC", "A2", "BS") b <- c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125) class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]]) [1] "factor" class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]]) [1] "factor" Also, when

Dear All, I?m using R package ?msm? to fit a multi state model to infection history data (counts of infections per month upto diagnosis of a particular disease (sink state is state 11). The observed transitions are as follows: to from 1 2 3 4 5 6 7 8 10 11 1 35192 3806 899 233 46 11 3 0 1 534 2 3801 790 249 69 15

Hi All, is there a function in R that allows me to work with fractions without transforming them to floats (or whatever) in between? Something that would calculate something like: (1/2 + 1/8) * 1/2 = 5/16 without ever transforming to 0.5 and 0.125? Best, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St Mary's Campus Norfolk Place,