similar to: SuSE 10 / No access to domU network services

Hi, I have installed debian 3.1 on an Intel Pentium D processor 920 ( including vanderpool.) The installation worked fine and i''m able to start guest-domains. When I try to start the 5.th Guest-Domain, I Get the following error: Error: Device 769 (vbd) could not be connected. Backend device not found. this problem exists still if i have shutdown or destroyed all guests, after

Hi, I'm banging my head on compiling bristuff modules for Suse 10.0 with kernel : Linux laps1 2.6.13-15.7-smp #1 SMP Tue Nov 29 14:32:29 UTC 2005 x86_64 x86_64 x86_64 GNU/Linux and Asterisk 1.2.1-BRIstuffed-0.3.0-PRE-1f . I get this : laps1:~/Voipy/1.2.1/bristuff-0.3.0-PRE-1f/zaphfc # make load make -C /usr/src/linux-2.6 SUBDIRS=/root/Voipy/1.2.1/bristuff-0.3.0-PRE-1f/zaphfc

Hi there, I'm using RH7.2 (with the 2.4.9-30 kernal and it's required components) as a base for a server system running kernel 2.4.18. I've gone to this version to get around non-performing aic7xxx drivers in the stock 7.2 kernels, and updated gigabit ethernet drivers. I have a raid unit (Medea) attached to an Adaptec 3916, coming up as sdb. It has 2kb blocks, but the fault

- UPS MGE Premium Ellipse Premium 1200 - SuSE Linux 10.0 OSS, kernel 2.6.13 smp - gcc version 4.0.2 20050901 (prerelease), glibc-2.3.5-40 - mgeups-psp-3.0.0-3, nut-2.0.2-4 - Dell Dimension 8400 with dual core Intel Xeon. 3.0GHz. lsusb shows: Bus 002 Device 003: ID 0463:ffff MGE UPS Systems UPS less /proc/bus/usb/devices shows for bus 2 device 3: T: Bus=02 Lev=01 Prnt=01 Port=00 Cnt=01 Dev#=

Hi, My name is Sebastian von Conrad and I''m not a recruiter; I''m a software developer with e-channel search, an industry leading search engine marketing agency with offices in Adelaide and Sydney. We''ve been in business for over 10 years and was recently named as one of Deloitte Technology Fast 50 Australia for 2009, which means that we''re one of the 50 fastest

Dear R-users, I wondered if anyone has extended GLMMGibbs to include correlated random effects, and if so, whether they would be willing to let me use their code? Jonathan Myles has no plans to extend glmm in this manner within the foreseeable future. With thanks, Patty -- -------------------------------------------------------------------------------- Assoc Prof Patty Solomon

Hello Everyone, I have three variables: Waist circumference (WC), serum triglyceride (TG) level and gender. Waist circumference and serum triglyceride is numeric and gender (male and female) is categorical. From these three variables, I want to calculate the "Lipid Accumulation Product (LAP) Index". The equation to calculate LAP is different for male and females. I am giving both

Dear all, Could you please help me how to get the output as I described in the following example? x<-c(543, 543, 543, 543, 551 , 551 ,1128 ,1197, 1197) diff<-x-lag(x) diff [1] NA 0 0 0 8 0 577 69 0 How to index the occasions in x repeatedly if the diff<15? if diff>=15, it will give a new index. I want the output be like y. y<-c(1,1,1,1,1,1,2,3,3) Thank you so

Dear all, Could you please help me how to get the output as I described in the following example? x<-c(543, 543, 543, 543, 551 , 551 ,1128 ,1197, 1197) diff<-x-lag(x) diff [1] NA 0 0 0 8 0 577 69 0 How to index the occasions in x repeatedly if the diff<15? if diff>=15, it will give a new index. I want the output be like y. y<-c(1,1,1,1,1,1,2,3,3) Thank you so

I thnk what you want is ?outer. e.g.: outer(Data -min(Data),value,FUN = "+") Whether this works for your real task, however, may depend on details and complexities that you have omitted. Cheers, 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

Well, something like: LAP <- ifelse(gender =='male', (WC-65)*TG, (WC-58)*TG) The exact code depends on whether your variables are in a data frame or list or whatever, which you failed to specify. If so, ?with may be useful. Cheers, Bert On Fri, Nov 3, 2023 at 3:43?AM Md. Kamruzzaman <mkzaman.m at gmail.com> wrote: > Hello Everyone, > I have three variables: Waist

Hello Everyone, I was analysing big survey data using survey packages on RStudio. Survey package allows survey data analysis with the design effect.The survey package included functions for all other statistical analysis except two-proportion z tests. I was trying to calculate the difference in prevalence of Diabetes and Prediabetes between the year 2011 and 2017 (with 95%CI). I was able to

df$LAP <- with(df, ifelse(G=='male', (WC-65)*TG, (WC-58)*TG)) That will do both calculations and merge the two vectors appropriately. It will use extra memory, but it should be much faster than a 'for' loop. Regards, Jorgen Harmse. ------------------------------ Message: 8 Date: Fri, 3 Nov 2023 11:10:49 +1030 From: "Md. Kamruzzaman" <mkzaman.m at gmail.com>

I'm trying to install * 1.2.5 for the first time on SUSE 10, 64bit (I've previously used Fedora or Centos). After I installed the kernel sources and followed some instructions on Novell's web site to copy the Module.symvers.gz file across from /boot and do a make modules_prepare, I was successfully able to complete 'make linux26' in my /usr/src/zaptel folder. I'm still

Dear Md Kamruzzaman, To answer your second question first, you could just use the svychisq() function. The difference-of-proportion test is equivalent to a chisquare test for the 2-by-2 table. You don't say how you computed the confidence intervals for the two separate proportions, but if you have their standard errors (and if not, you should be able to infer them from the confidence

Hi All, I have a vector of data on which I am operating. The equation with which I want to operate on the vector has a value k. I want to run the equation and output a new vector, each time replacing k with each value from the sequence I defined. I have thought about using for loops and such, but this seems like overkill. I am wondering if there is a simple solution that would allow me to

On 18.08.23 07:39, Darren Tucker wrote: > On Fri, 18 Aug 2023 at 15:25, Stuart Longland VK4MSL <me at vk4msl.com> wrote: > [...] >> The crux of this is that we cannot assume the local IPv4 address is >> unique, since it's not (and in many cases, not even static). > > If the IP address is not significant, you can tell ssh to not record > them ("CheckHostIP

# Here is the solution: require(fBasics) # Be sure that R is running in time zone GMT. # Set your Windows environment variable to "GMT" # Your PC Windows clock can still run in any other time zone! # My clock is now running in Zurich in Europe. Date = c("2003-10-09", "2003-10-10", "2003-10-13", "2003-10-14") Open = c(1.27, 1.25, 1.27,

Dear R users, Using the library(mgcv) and running R under MacOSX, I have fitted a generalised additive model with binomial errors in order to check the linearity of two continuous variables ap2mm and diffdaysm in a glm: > mymodel.gam <- gam(diedhos~ s(ap2mm) + Dweekm + s(diffdaysm) + Dweekm:diffdaysm + ap2mm:Dweekm, binomial) I would like postscript gam plots for the two smoothed

Just a minor point in the suggested solution: df$LAP <- with(df, ifelse(G=='male', (WC-65)*TG, (WC-58)*TG)) since WC and TG are not conditional, would this be a slight improvement? df$LAP <- with(df, TG*(WC - ifelse(G=='male', 65, 58))) -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Jorgen Harmse via R-help Sent: Friday,