similar to: Time data

Hey! I have some RHEL6 hypervisors and the VMs are in raw qemu image files in a local raid array linux raid + lvm + ext3. When a kernel update is installed a reboot is necessary, usually it has been more than 180 days since the last reboot and the file system is fsck'd and this takes 2-3 hours. I am curious to know if there is any documentation that addresses the pro's and con's of

Dear All, I would be grateful if you can help me. My problem is the following: I have a data set like: ID time X1 X2 1 1 x111 x211 1 2 x112 x212 2 1 x121 x221 2 2 x122 x222 2 3 x123 x223 where X1 and X2 are 2 covariates and "time" is the time of observation and ID indicates the

I want to use one scsi disk with two redondant servers, mounted r/w on server A and r/o on server B. Everything works, except that when I write data from server A, I can't read it on server B before un-mounting the disk ant re-mounting it... (with AND without soft-updates) Is there any way to mount a disk r/o keeping in sync with real disk data ? Thank you. -- Geoffroy DESVERNAY

Daer All, I have a factor variable, X with 5 levels. When I type tables(X) it gives me: table(X) 1 2 3 4 5 10 5 0 0 0 How to drop the levels with zeros such that when I will type: table(X) it will give me: table(X) 1 2 10 5 Thank a lot, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I have a vector x = (1,1,1,2,2,2,3,3,3,3) I am looking for a function to return a vector containing the distinct elements of x i,e y = (1,2,3) The following code gives the desired results: as.numeric(levels(as.factor(x))) Is there any other elegant way? Thanks, B --------------------------------- [[alternative HTML version deleted]]

Dear all, How can I plot the number of observations per box in a boxplot. Many thanks, Bernard --------------------------------- [[alternative HTML version deleted]]

Hello, I would like to be able to calculate the age of someone at a particular date. Both dates are date objects. Here is what I have come up with: floor(as.numeric(sampleInfo$Date.of.DIAGNOSIS-sampleInfo$Date.of.birth)/365.25) Is this the best approach? or is there an inbuilt function? I have looked at difftime but that does not seem to allow output in years. Many thanks Dan --

Dear All, I have written the following programs to find a non-singular (10*10) covariance matrix. Here is the program: nitems <- 10 x <- array(rnorm(5*nitems,3,3), c(5,nitems)) sigma <- t(x)%*%x inverse <- try(solve(sigma), TRUE) while(inherits(inverse, "try-error")) { x <- array(rnorm(5*nitems,3,3), c(5,nitems)) sigma <- t(x)%*%x inverse <-

Dear All, I wonder if there is any R function to generate a sequence of vectors from existing ones. For example: x 1<- c(1,2,3) x2 <- c(4,5) x3 <- c(6,7,8) The desired output is a list of all 3*2*3 = 18 possible combinations of elements of x1,x2 and x3. One element for example is (1,4,6). Many thanks in advance, Bernard

All, I would like to apply a set of rules to each row of the sample data set below. The rule sets are the guidelines for determining an individual's date for retirement eligibility. The rules are found in this document, http://www.opm.gov/feddata/RetirementPaperFinal_v4.pdf. I am only interested in the top two categories for retirement eligibility, the CSRS and FERS plans. The data set has

Dear R list: I modelled changes in a variable (mconc) over time (d) for individuals (replicate) given one of three treatments (treatment) using: mconc.lme <- lme(mconc~treatment*poly(d,2), random=~poly(d,2)|replicate, data=my.data) summary(mconc.lme) shows that the linear coefficient of one of the treatments is significantly different to zero, viz. Value Std.Error

Just surfing Net for 2-3 hr concerning "dracut" on Fedora 13. Several "warnings"  at the end of dmesg log. Seems like reaction on some system events scheduled to run on regular basis, but not user''s activity. Boris. --- On Tue, 8/17/10, Boris Derzhavets <bderzhavets@yahoo.com> wrote: From: Boris Derzhavets <bderzhavets@yahoo.com> Subject: Re: [Xen-devel]

Hi, I wrote a little helper that calulates someone''s age based on his/her date of birth. def age(dob) diff = Date.today - dob age = (diff / 365.25).floor age.to_s end It works fine, but it''s not completely accurate as it just takes the average days in a year. It should be able to calculate this more accurately, right? I can''t work it out

> Dear list, > > I have some data on a geneaology, here is a subset: > warmerge[1:11,c(1,6,25)] > Warrior SibID birth.year > 1100 3793 2013 1926 > 4 2013 2024 1934 > 1094 3769 2024 1918 > 632 2747 2037 1928 > 176 2083 2039 1944 > 187 2085 2039 1949 > 192 2086 2039 NA > 495

Dear All, I am wondering on how to use the abline.xyplot with xyplot such that I will have different vertical lines for each panel. More sepcifically, suppose that the xyplot generates 4 panels defined by the combination of two binary variables: X_1 and X_2. i.e. xyplot(Y ~ Z | X_1*X_2, data = df) I want something like: abline(v = 5) if X_1=0 and X_2 = 0 abline(v =

Dear All, Is there any R function to test if two data frames, say df1 and df2, having the same row names and the same column names are identiques? Thanks in advance, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I want to create one numeric variable from a factor one. here is a simple example: x1 x2 factor x11 x21 f1 x12 x22 f1 x13 x23 f2 x14 x24 f3 . . . . . . Suppose that the variable factor has 3 levels (f1,f2,f3). I want to add to this data frame a numeric variable x4 in (v1,v2,v3) such that x4=v1 if factor = f1, x4 = v2 if

Dear All, I wonder how to re-code a categorical variable without using the ifelse loops. For example: Let X be a categorical variable with 6 levels, levels(X) = c(1,2,3,4,5,6) How to create a new categorical variable Y with levels, 10,11,12, say , such that: Y = 10 if X =1 or X=2 Y = 11 if X =3 or 4 Y =11 if X = 5 or 6 many thanks, Bernard

Dear All, I am wondering if there is an R function to convert a covariance matrix to a correlation matrix. I have a covariance matrix sigma and I want to compute the corresponding correlation matrix R from sigma. Thank you very much, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I would be grateful if you can tell how can I suppress the legend (automatically created) when I plot a groupedData. suppose that I have data farme df_0 which contains the following columns: ID represents the clusters, t represents time of observation, Y: the variable of interest and X a binary covariate. Then: df_1 <- groupedData(Y ~ t | ID) plot(df_1,