similar to: nth root

Hi all, I need some guidance to get my problem fixed. I believe there is an issue with the ''nth'' patch from the patch-o-matic, which is labeled as status ''works''. I have tunnels back and forth across the internet, using ''nth'' to balance packets between different public networks (over the tunnels). I need to access some networks over two

Curious to know if anyone's created something similar to the following, if so and you'd care to share an AGI or dialplan, much appreciated. I will be eventually write a script to answer the nTH call. (if I can't find it (why reinvent wheels). Looking to do some testing sending anywhere between 50-200 calls to a machine. I'd like a Snom/Polycom/whatever to pick up after the nTh

First of all, I''d like to thank Andy Furniss for his direction and for helping me get a working example up and running. For the following set up to work, you will need a linux computer at the ISP (server), a linux computer at the client location (client), and some a public range of IP''s you plan to send down to your client. (as this configuration involves patching the linux

Hi, Here i have an matrix like this, ABC PQR XYZ MNO ------ ------- ------ -------- 3 6 7 15 2 12 24 15 20 5 1 2 25 50 15 35 i need to get the "MODE" - for each column-wise "VARIANCE" - for

Hi, I would like to produce a plot with a symbol on every nth point in a time series data, like the one in the following: http://www.phon.ucl.ac.uk/home/yi/ProsodyPro/EnglishFocus.png x <- seq(-100,1000,25) plot(x,type="l") Could someone help me out with the above example? Thanks.... [[alternative HTML version deleted]]

Dear R users, how do I e.g. square each second element of a vector with an even number of elements? Or more generally to apply a function to every 'nth' element of a vector. I looked into the apply functions, but found no hint. For example: v <- c(1, 2, 3, 4) mysquare <- function (x) { return (x*x) } w <- applyfun(v, mysquare, 2) then w should be c(1, 4, 3, 16) Thanks for

Hello, I have a question in regards to markov chains and transition probabilities. I am trying to figure out a way to calculate the "kth-step transition matrix" of a given matrix. Say for example I have a single step 2x2 matrix: 1 2 P= 1 .95 .05 2 .01. 99 If I were to convert this matrix to a 2-step transition probability matrix I would get:

R-helpers: Is there a simple way to extract every nth element from a very long vector? [The vector I want to sample from is an object returned by lowess(). It is so long (11,628 pairs of elements) that it is causing non-R-related memory problems elsewhere. I don't see a better way other than sampling the output returned from lowess.] Thanks in advance. Barry Cooke

Hello! I am working with signals and a plot of several signals on the same axes can get quite messy. With lines that are very fractured, distinction by only the linestyle is not very clear. If I add symbols to the plot however, there are so many symbols, that they overplot and the whole plot is unreadable once again. I am looking for advice on how to make a plot with continuous lines and symbols

Hi, Sorry if this is an obvious one, but is there a simple way to modify the lm function to test whether a slope coefficient is significantly different from 1 instead of different from 0? Thanks, Martin Martin Biuw SEA MAMMAL RESEARCH UNIT Gatty Marine Laboratory School of Environmental and Evolutionary Biology University of St Andrews Fife, Scotland KY16 8LB UK phone +44 (0)1334 462630 fax

Hello, Once again, I posted a message without a subject line. Sorry.... here is the question again. Is there a simple way to modify the circ.mean function in the CircStats package to include a vector of weights to obtain a weighted average angle? Thanks! Martin -- Martin Biuw Sea Mammal Research Unit Gatty Marine Laboratory, University of St Andrews St Andrews, Fife KY16 8PA Scotland Ph:

Hello, I''m trying to fit a linear mixed effects model of the form: lme(y ~ x * Sex * Year, random=x|subject) where Sex and Year are factors with two and three levels respectively. I want to compare the fixed effects for each level to the overall mean, but the default in R is to compare to the first level. This can be changed by adding the term -1 to the righthand side of the model

I've written the function below to simulate the mean 1st through nth nearest neighbor distances for a random spatial pattern using the functions nndist() and runifpoint() from spatsat. It works, but runs relatively slowly - would appreciate suggestions on how to speed up this function. Thanks. --Dale library(spatstat) sim.nth.mdist <- function(nth,nsim) { D <- matrix(ncol=nth,

Dear R-tisans, I am trying to calculate the 12th root of a transition (square) matrix, but can't seem to obtain an accurate result. I realize that this post is laced with intimations of quantitative finance, but the question is both R-related and broadly mathematical. That said, I'm happy to post this to R-SIG-Finance if I've erred in posting this to the general list. I've

Hi I have a small problem, I'm new in using R, so I hope you can help me... I'm running a logistic regression model and want to do a nice plot. The plot I have made is done a plot with the command jitter: plot(jitter(overto$age[overto$sex=="F"]),jitter(overto$neg.pos[overto$sex=="F"]),xlab="age",ylab="neg and pos") my responsvariable is positive

I want to make a glm and then use predict. I have a fairly small sample (4000 cases) and I want to train on 90% and test on 10% but I want to do it in slices so I test on every 10th case and train on the others. Is there some simple way to get these elements? Stephen -- 21/10/2005 [[alternative HTML version deleted]]

Hello everyone, I need some help with Dovecot+LDAP config and local delivery. We're using Dovecot 2.0.9 on Centos 6.4. Before, we used Qmail with the same LDAP. Our LDAP accounts have the attribute deliveryMode which we used for mailing lists (OK, they are in fact mailing groups rather than lists). When the attribute is set to deliveryMode = nolocal Qmail did not deliver mail to the

Hello, I'm using the nls function and would like to increase the number of iterations. According to the documentation as well as other postings on R-help, I've tried to do this using the "control" argument: nls(y ~ SSfpl(x, A, B, xmid, scal), data=my.data, control=nls.control(maxiter=200)) but no matter how much I increase "maxiter", I get the following error

Dear All, I have to get the value of say 90th percentile of precipitation time series.. The series is of daily precipitation value of 96 years, I have to to get 90the percentile value of daily precipitation each year. If you know the R code or command for this please let me know. I would appreciate your early response. Thanking you, Sincerely, Ajay.

Hello, I have 1-minute time series stock data and I'd like to calculate mean of every n-th candle data of m-days. result = c(mean of 1th data, mean of 2nd data, ...) mean of 1th data = (1th data of 2012-1-1 + 1th data of 2012-1-2 + 1th data of 2012-1-3) / 3 mean of 2nd data = (2nd data of 2012-1-1 + 2nd data of 2012-1-2 + 2nd data of 2012-1-3) / 3 ... Could you let me know the fastest