similar to: arima fails when called from command line

Awesome, thanks for the advice. Using the following now works... passdb { driver = static args = proxy=y password=doesnotmatter } Cheers. On Feb 15 2018, at 2:40 pm, Aki Tuomi <aki.tuomi at dovecot.fi> wrote: > > On 15 February 2018 at 20:22 Travis Dolan <travis.dolan at gmail.com> wrote: > > > Hello, > > I have Director setup to proxy

Hello, I would like to take the string "mylist = list(a = 5, b = 7)" and evaluate it as a list. I have attempted to use parse and several other functions with no success. Thanks for your time. -brian dolan ~~~ may all your sequences converge [[alternative HTML version deleted]]

Full_Name: Peter Dolan Version: 2.5.1 OS: Windows Submission from: (NULL) (128.193.227.43) gregexpr does not find all matching substrings if the substrings overlap: > gregexpr("abab","ababab") [[1]] [1] 1 attr(,"match.length") [1] 4 It does work correctly in Version 2.3.1 under linux.

> On 15 Feb 2018, at 22.16, Travis Dolan <travis.dolan at gmail.com> wrote: > > It would look as though the changes have now negatively affected a "normal" user from logging in. > > > telnet host 143 > > a login username password > > > a NO [AUTHENTICATIONFAILED] Authentication failed. > > > telnet host 143 > > 1 login

Hello, I have Director setup to proxy requests to backend servers. This works fine when using "standard" username/passwords. I am not try to enable the use of the Dovecot Master user through Director into the backend servers. a.) username is being sent as masteruser*username b.) request hits the proxy and authenticates, and then is passed to the backend servers and fails auth. - logs

I am puzzled by the warning message in the output below. It appears whether or not I fit the seasonal term (but the precise point of doing this was to fit what is effectively a second seasonal term). Is there some deep reason why AR parameters ("Warning message: some AR parameters were fixed: ...") should somehow intrude into the fitting of a model that has only MA terms? >

Hello, Does anybody know if the function arima with an external regressor (xreg) applies the auto correlation on the dependant variable or on the residuals. In comparison with SAS (proc autoreg), it seems that the auto correlation applies on the residuals but i'd like to have the confirmation. I want to estimate: Y[t] = a[1]*X[t] + a[2] + E[t] with E[t]=b[1]*E[t-1] Should I use : arima(Y,

Colleagues I am a new to R but already love it. I have the following problem: I fitted arima model to my time series like this (please ignore modeling parameters as they are not important now): x = scan("C:/data.txt") x = ts(x, start=1, frequency=1) x.fit<-arima(x, order = c(1,0,0), seasonal = list(order=c(0,0,1))) Now I want to use this model for forecasting and backtesting (!).

Hi users. I'm new user in R. I'm workiing with Time series and I would like to know how can I do to undo the command DIFF(X), for exemple: If I have the model: m=arima(X, order=c(0,1,1), seasonal=list(order=c(0,0,1))) (note that have d=1 one difference), to find, in the same scale, the original numbers (like one "unDiff"), after the forecast, I need to develop some function or in

I've touched the directory one level above the directory with the I\O issue as the one above that is the one showing as dirty. It hasn't healed. Should the self heal daemon automatically kick in here? Is there anything else I can do? Thanks David On Tue, 14 Feb 2023 at 07:03, Strahil Nikolov <hunter86_bg at yahoo.com> wrote: > You can always mount it locally on any of the

I guess you didn't receive my last e-mail. Use getfattr and identify if the gfid mismatch. If yes, move away the mismatched one. In order a dir to heal, you have to fix all files inside it before it can be healed. Best Regards, Strahil Nikolov ? ???????, 14 ???????? 2023 ?., 14:04:31 ?. ???????+2, David Dolan <daithidolan at gmail.com> ??????: I've touched the directory one

Dear colleagues, I am relatively new to R and time series and so I am experiencing difficulties in interpreting the output of "arima" in MA models (but not in AR models). I cannot make sense of the 1st innovations returned by "arima". In an AR(1) model I expect data[t]=phi1*data[t-1]+a[t] and in a MA(1) model data[t]=a[t]+theta1*a[t-1]. My interpretation from R-help is

Hello, I am currently running 2.2.31. I have a cluster of dovecot instances running in AWS. - Two instances running Director/Proxy - Two instances running Dovecot. -- Storage for the Dovecot instances is EFS. My setup works, requests for 143 hit my Director cluster and are proxied to an appropriate dovecot backend. My question is how often if at all the Director service looks up the addresses

Hi, I've been trying for the past few days to come up with a method for checking a FreeBSD system to see if it is vulnerable to an issue described by a FreeBSD security advisory in some automated way, similar to the way portaudit can use VuXML to check for vulnerabilities in ports. Right now, I'm a bit stuck--there seem to be fairly major issues with all the methods I've come up

Hi, I know how to use SPSS and JMP, and have quite a few JMP files I would like to use in R. I converted them to .xpt files, downloaded the 'foreign' library then tried this command: >read.xport("D:\\Databases\nameoffile.xpt") to which I get: >Error in lookup.xport(file) : unable to open file I have read FAQ lists and Google searched and cannot figure out what I'm

Hello, I have a question about function arima.sim I tried to somulate a AR(1) process, with no innovation, no error term. I used this code: library(forecast) e=rnorm(100,mean=0,sd=0) series=arima.sim(model=list(ar=0.75),n=100,innov=e)+20 Then I tried to applicate ti this series auto.arima function: mod1<-auto.arima(series,stepwise=FALSE,trace=TRUE,ic='aicc') The best model returned

Dear all, When I run an arima(1,1,1) on an I(1) variable, y, I get different estimates to when I first difference the variable myself, e.g y2<-diff(y), and then run arima(1,0,1) on y2. Shouldn't these two approaches give the same output? Any help will be much appreciated. george

Dear all, When I run an arima(1,1,1) on an I(1) variable, e.g. y, I get different estimates to when I first difference the variable myself, e.g y2<-diff(y), and then run arima(1,0,1) on y2. Shouldn't these two approaches give the same output? Any help will be much appreciated. george -- View this message in context:

I've lurked for a while, but I think this is one of my first "pleas" for help. I'm having issues where a parked call using the macro below is getting "stuck". Users park the call via a blfxfer key on an Aastra phone. If the call is a blind transfer, it tries to park the call. If it isn't a blind transfer, it tries to unpark the call. Only 2 extensions (2759 and

HI, there, When i use R, i tried to use function arima(), it complains: Error: couldn't find function "arima" But when I type "help.search("arima") ", I got arima() poped up.. arima(ts) ARIMA Modelling of Time Series arima.sim(ts) Simulate from an ARIMA Model arima0(ts) ARIMA Modelling of Time Series -- Preliminary