similar to: Sensibility analysis

Hi all, I am running Samba 3.0.20B on an AIX server with SECURITY=USER using an LDAP backend. I am looking for a way to capture the actual username(%USER_NAME%) that the client user is logged onto his individual workstation with and compare it with the user(%u) they are connecting with the share as. If they are different, I want to reject the user's connection. After doing some

Dear R users, I'm using optim to optimize a pretty complicated function. This function takes the parameter vector "theta" and within its body I use instructions like sigma<-theta[a:b]; computations with sigma... out<-c() for (i in 1:d){ a<-theta[(3*d+i):c] out[i]<-evaluation of an expression involving 'a' (I use symbolic differentiation) } Unfortunately

The PDF file on the web, which is an appendix on nonlinear regression associated with the CAR book, is very nice. When I ran through the code presented there, I found something odd. The code does a certain model in 3 ways: Vanilla NLS (using numerical differentation), Analytical derivatives (where the user supplies the derivatives) and analytical derivatives (using automatic differentiation). The

Dear list, I am currently working with a rather large data set on body temperature regulation in wintering birds. My original model contains quite a few dependent variables, but I do not (of course) wish to keep them all in my final model. I've fitted the following model to the data: >

Hello, maybe that my Question is a "beginner"-Question, but up to now, my research didn't bring any useful result. I'm trying to fit a distribution (e.g. lognormal) to a given set of data (ML-Estimation). I KNOW about my data that there is a truncation for all data below a well known threshold. Is there an R-solution for an ML-estimation for this kind of data-problem? As

Dear R-users, I would like to calculate elasticities and sensitivities of each parameters involved in the following transition matrix: A <- matrix(c( sigma*s0*f1, sigma*s0*f2, s, v ), nrow=2, byrow=TRUE,dimnames=list(stage,stage)) The command "eigen.analysis" avaliable in package "popbio" provides sensibility matrix and elasticity matrix

Greetings, It is possible to manually adjust the sensibility of the VAD module in the Opus encoder? Thanks, ML www.eid.pt EID, S.A. Capital Social 1.100.000 N de matricula unico 501 400 699 DISCLAIMER art_001_01d0fc50.d7ccaf08" -------------- next part -------------- An HTML attachment was scrubbed... URL:

Hi, is there a possibility to tell zaptel or Asterisk to modify the DTMF sensibility? The problem what i have is that the Asterisk don't get all Numbers which the analog-FAX dial, let say the FAX dial 123456789 the Asterisk get to number 24679. I think that can be to DTMF Tone duration or the Frequenzy. so you got yna idea what it could be? Thx for helping me. Bye Timm

Hello, I had noticed strange crackling sound on my phone calls going through my zaptel device (TDM400P), so i decided to check on possible timer issue, and found lots of issues on forums concerning the sensibility of zaptel with IRQs, and tried about everything: moving PCI slots, noapic and acpi=off boot options, play with different kernel options: iosched/preemption/timer/..., play with

Hello all, I'm very new to R and am having some trouble with the results of the interp function. I'm trying to produce a chart roughly akin to a weather map with natural looking filled contours over a large region of the south pacific. I've got a list of points and values to be mapped to those points and I use the interp function as follows: tab<-read.table("data.txt")

Thanks for the reply. We are currently using AMBE (4Kbps) for our Traffic/Weather Channels. If you have ever had a chance to hear the service, you will know that AMBE does not do us well. I understand that 2Kbps is low quality, but any poorer than AMBE? If can get a decent quality for other low bandwidth talk channels, such as about 10-16Kbps and have it sound rather clean, then I would be

> My understanding is that this represents bivariate normal > approximation of the data which uses the kernel density function to > test for inclusion within a level set. (please correct me) You can fit a bivariate normal distribution by computing five parameters. Two means, two standard deviations (or two variances) and one correlation (or covariance) coefficient. The bivariate normal

We have two time series: the first is a series of weekly counts of isolates of RSV (respiratory syncytial virus) by pathology laboratories, and the second is a series of weekly counts of cases of bronchiolitis in young children presenting to hospital emergency departments. Bronchiolitis in young children is usually caused by RSV infection, and simple visual inspection reveals a very close

Hello, I have a data frame like this: > head(SNP) mean var sd FQC.10090295 0.0327 0.002678 0.0517 FQC.10119363 0.0220 0.000978 0.0313 FQC.10132112 0.0275 0.002088 0.0457 FQC.10201128 0.0169 0.000289 0.0170 FQC.10208432 0.0443 0.004081 0.0639 FQC.10218466 0.0116 0.000131 0.0115 ... and I am creating plot like this: s <- ggplot(SNP, mapping = aes(x = mean, y = var))

I recommend that you consult with a local statistical expert. Much of what you say (outliers?!?) seems to make little sense, and your statistical knowledge seems minimal. Perhaps more to the point, none of your questions can be properly answered without subject matter context, which this list is not designed to provide. That's why I believe you need local expertise. Bert Gunter "The

My understanding is that this represents bivariate normal approximation of the data which uses the kernel density function to test for inclusion within a level set. (please correct me) In order to exclude the outlier to these ellipses/contours is it advisable to do something like this: SNP$density <- get_density(SNP$mean, SNP$var) > summary(SNP$density) Min. 1st Qu. Median Mean 3rd

Hi, many people are having problems using SFTP with ChrootDirectory when the jail directory (or the path above) is not owned by root. The question is if chroot'ing to usual home directories can be allowed, even though they are owned by regular users. I know that this topic has been discussed on the list several times now, so I searched the list archives for posts that invalidate the

Hi, I am trying a two way anova with unequal sample sizes but results are not as expected: I take the example from Applied Linear Statistical Models (Neter et al. pp889-897, 1996) growth rate gender bone development 1.4 1 1 2.4 1 1 2.2 1 1 2.4 1 2 2.1 2 1 1.7 2 1 2.5 2 2 1.8 2 2 2 2 2 0.7 3 1 1.1 3 1 0.5 3 2 0.9 3 2 1.3 3 2 expected results are

Hi Abby, thank you for getting back to me and for this useful information. I'm trying to detect the outliers in my distribution based of mean and variance. Can I see that from the plot I provided? Would outliers be outside of ellipses? If so how do I extract those from my data frame, based on which parameter? So I am trying to connect outliers based on what the plot is showing: s <-

I have a column in a data frame that has a class of "Date" and a mode of "numeric". When I: max(df$Date) My output stays in Date format, i.e. "2006-01-03". However, when I run the following statment: tapply(df$Date, df$SomeFactor, max) my output looks like this: 9129 9493 9861 10226 10591 10956 11320 11687 12052 12417 The returned object is of