similar to: 'skewing' a normal random variable

Dear All: I would like to center the data in 'x' by 'group'. The following code scale the data and I have not been able to figure out how to change it so I get the centered data. x <- c(1, 2, 3, 4, 5, 6, 7, 8) group <- c(1,1,1,2,2,2,2,2) unsplit(lapply(split(x,group),scale),group) I would appreciate your help. Ashraf

Hi All: I would like to generate a binomial random variable that correlates with a normal random variables with a specified correlation. Off course, the correlation coefficient would not be same at each run because of randomness. I greatly appreciate your input. Ashraf

Dear All: lme(sbp ~ cengirth, data = family, random= ~ 1 | familyid) converges but lme(sbp ~ cengirth, data = family, random= ~ cengirth | familyid) does not. I get the following message: Error in lme.formula(sbp ~ cengirth, data = family, random = ~cengirth | : iteration limit reached without convergence (9) The data has 488 rows and 154 familyid levels. For

Hi All: I have to compute bootstrap confidence intervals, the statistic (incremental cost effectiveness ratio) is computed from two samples (intervention and control) of different sizes. All the bootstrap functions that I have seen use one dataset as argument. I may go ahead and get the desired number of bootstrap replications separately. I would appreciate if you could point me to a source of a

I am trying to find out if R can recognize specific criteria for removing rows (i.e. a prexisting function) I have a matrix myMatrix that is 12000 by 20 I would like to remove rows from myMatrix that have: -999 across all columns -999 across all columns but one -999 across all columns but two -999 across all columns but three -999 across all columns but four -999 across all columns but five

Hi all: I get the following error message that I am not able to resolve. Error in if (const(t, min(1e-08, mean(t)/1e+06))) { : missing value where TRUE/FALSE needed It appears right before the last data.frame statement. Below is the program that simulates data from one way random effects model and then computes normality and bootstrap confidence interval for

I am producing 2X2 correlation matrices by a class variable. I need to extract a vector of correlation coefficients only. I am doing that in a loop (see below) but I am sure there would be a simpler way. Please help! > by(d1[,c(2,3)],d1[,1],cor) d1[, 1]: 1 c e c 1.0000000 0.1972309 e 0.1972309 1.0000000

> -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch]On Behalf Of Caio Lucidius > Naberezny Azevedo > Sent: 01 September 2005 12:09 > To: Help mailing list - R > Subject: [R] Multivariate Skew Normal distribution > > > Hi all, > > Could anyone tell me if there is any package (or function)

Dear All: I have been struggling to run the conditional logistic regression on a dataset. I am using clogit function in survival package. Here is how the data is generated. I tried to shorted the data by using the second count variable but still the clogit did not run. Why I am not able to run clogit on my dataset. I appreciate any input. Regards, Ashraf age <- c(rep(0,8),rep(1,8)) scc <-

Fasiha Ashraf --- On Sat, 22/8/09, Fasiha Ashraf <feehapk@yahoo.co.in> wrote: From: Fasiha Ashraf <feehapk@yahoo.co.in> Subject: Re: [Xen-users] my bootlog To: "Boris Derzhavets" <bderzhavets@yahoo.com> Date: Saturday, 22 August, 2009, 11:12 AM Please check what wrong here grub.conf        title Fedora (2.6.30-rc6-tip)         root (hd0,6)         kernel

Hi Eric, Thanks for the help.But this will not solve the problem as it will generate a list and what I need is an object of class sp using SpatialLine function from sp package.So, I need to convert each matrix to coordinates and then to a line and then to a spatial line as figured in the code. My data structure is a list of 141 matrices.Each matrix represents coordinates of the river lines

Hi Ashraf, It is not obvious to me what your structures are but one problem in your function is the assignment tt1 <- SpatialLines(list(tt[[i]])). This will set tt1 to just have one item. Consider the following test.func <- function(x) { tt1 <- list() for ( i in ... ) { ... tt1[[i]] <- SpatialLines(tt[[i]]) } return(tt1) } HTH, Eric On Mon, Sep 25,

Hi, I need to create a plot (type = "h") and then overlay a skewed-normal curve on this distribution, but I'm not finding a procedure to accomplish this. I want to use the plot function here in order to control the bin distributions. I have explored the sn library and found the dsn function. dsn uses known location, scaling and shape parameters associated with a given input

Hello all: Is there a way to detect in R if a dataset is normally distributed or skewed without graphically seeing it? The reason I want to be able to do this is because I have developed and application with Visual Basic where Word,Access and Excel "talk" to each other and I want to integrate R to this application to estimate confidence intervals on fish sizes (mm). I basically want to

Hello everyone, Does anyone know what would cause the skewness() function (from e1071), as well as skew() from psych, to return a value of NaN? I have a vector of positively-skewed data (https://docs.google.com/file/d/0B6-m45Jvl3ZmYzlHRVRHRURzbVk/edit?usp=sharing) which these functions return a value for like normal: > skewness( data ) # returns 1.400405 but when I instead give those

Hi all,I'm trying to build a SpatialLines object from a list that contains 124 river segments. Each segment in the list contains the x,y coordinates. I'm using the following code to create the SpatialLines object, but it just retrieves one segment. Any suggestions? test.func = function(x){ ??? for (i in 1:length(x)) {??????? tt[[i]] <- x[i]; tt[[i]]? = Line(tt[[i]]); tt[[i]]? =

Does xen virtualize wireless networks, and if not, why not? What is so difficult about it? Regards! Fasiha Ashraf Unlimited freedom, unlimited storage. Get it now, on http://help.yahoo.com/l/in/yahoo/mail/yahoomail/tools/tools-08.html/ _______________________________________________ Xen-users mailing list Xen-users@lists.xensource.com http://lists.xensource.com/xen-users

I have set-up xen on my Intel quad core server. Now running different experiment to measure network throughput in virtualized environment. these are some of the results: Netperf-4.5 results for inter-domain communication. Sr.No.              Client               Server               Time(sec)        Throughput(Mbps)    1                   Guest-1              Dom0                   

We i have settup it like this it giveme caller id agent id and date-time on gsm file but i want them to be in folder on every day basis datewise. exten => _1NXXNXXXXXX,2,Set(CALLFILENAME=${ACCOUNTCODE}-${EXTEN}-${TIMESTAMP}) exten => _1NXXNXXXXXX,3,Monitor(gsm,/rec/asterisk/apr07/${CALLFILENAME},mb) Any Idea ? Faisal > ------------------------------ > > Message: 16 >

Hi, The link given in your http://syslinux.zytor.com/pxe.php#config page points to the http://www.argontechnology.ca/rbfg/index.shtml site. This site talks only of Win2k and .Net server. There is no mention of GNU/Linux. Is it possible to use rbfg.exe to create a pxelinux bootable stack on a floppy. If so how ? Rom-O-Matic has a pxe loadable ROM image. Is there a way to use this to