similar to: Suitable test for ordinal variable vs continuous variable trend

Hi R-users,   I wrote a code to evaluate double sum as follows:   ff2 <- function(bb,eta,z,k) { r <- length(z) for (i in 1:r) { sm1 <- sum((z[i]*bb/2)*(psigamma((0:k)+eta+1,deriv=0)/(factorial(0:k)*gamma((0:k)+eta+1))))  sm2 <- sum((besselI(z[i]*bb,eta)*log(z[i]*bb/2) - sm1)/besselI(z[i]*bb,eta))  sm2 } ff2(bb,eta,z,10)     but it gave me the following message:   >

Hello, R community, I have been using the lmer and mcmcsamp functions in R with some difficulty. I do not believe this is my code or data, however, because my attempts to use the sample code and 'sleepstudy' data provided with the lme4 packaged (and used on several R-Wiki pages) do not return the same results as those indicated in the help pages. For instance: > sessionInfo() R

Sure there is, Install debian, follow my howto and you will have success. Just, your using an .local domain, and thats a reserved name for apples mDNS (zeroconf) And should not be used. ( same for .lan ) https://wiki.samba.org/index.php/FAQ#Can_I_Use_the_.local_Top-level_Domain_for_My_AD_DNS_Zone.3F So the info is good, thats not the problem, finding it, is. Can you post your /etc/hosts

Hello to everyone I''ve downloaded the version sm2 for smoothing methods and when I try the simple code y <- rnorm(50) sm.density(y, model = "Normal") I get the error message Error in if (any(omit)) { : missing value where logical needed I''m running R 1.1.1. on windows 98 Anybody can help? -------------- next part -------------- An HTML attachment was

Hi, I'm attempting to add a "Julian Day" column to a data frame. Here is my code and the resulting data frame: vic.data <- read.table("C:/VIC/data/vic.data.csv", header=F) names(vic.data) <- c("year", "month", "day", "precip", "evap", "runoff", "baseflow", "Tsup",

Dear list users How is it possible to visualise both a linear trend line and a quadratic trend line on a plot of two variables? Here my almost working exsample. data(Duncan) attach(Duncan) plot(prestige ~ income) abline(lm(prestige ~ income), col=2, lwd=2) Now I would like to add yet another trend line, but this time a quadratic one. So I have two trend lines. One linear trend line

I've been trying to get some linear classifiers (LiblineaR, kernlab, e1071) to work with a sparse matrix of feature data. In the case of LiblineaR and kernlab, it seems I have to coerce my data into a dense matrix in order to train a model. I've done a number of searches, read through the manuals and vignettes, but I can't seem to see how to use either of these packages with sparse

Hi Louis, Thanks for information, find it sometimes is a real challenge. Would you please share your how to link? I wish to read it. For the .local domain I suppose I have nothing to do. This is a running windows Active Directory and it is not possible to change domain suffix. Here is my /etc/hosts 127.0.0.1 localhost.localdomain localhost 10.254.104.8 wdc04.aa.local wdc04 10.254.105.208

I have a 3d scatter plot and I need to add a trend line. I can add a regression plane, but I need a line, not a plane. Thanks. -- View this message in context: http://n4.nabble.com/3d-trend-line-tp1053686p1053686.html Sent from the R help mailing list archive at Nabble.com.

Hi everyone, I'm trying to determine the significance of a trendline. From my internet search months ago, I came across the following post. I modified tim and dat for simiplicity. tim <- 1:10 dat <- c(0.17, 1.09 ,0.11, 0.82, 0.23, 0.38 ,2.47 ,0.41 ,0.75, 1.44) fstat <- summary(lm(dat~tim))$fstatistic p.val <-

Dear All, My question is: how can I estimate VECM system with "unrestricted trend" (aka "case 5") option as a deterministic term? As far as I know, ca.jo in urca package allows for "restricted trend" only [vecm <- ca.jo(data, type = "trace"/"eigen", ecdet = "trend", K = n, spec = "transitory"/"longrun")].

hej! can anyone help me with a hint how to find a function to fit a linear trend line into a scatter plot. as well as testing it for significance. thanx/thisse

Hi! I am searching for the Cochran-Armitage-trend-test. Is it included in an R-package? Thank you! --

Hi R Users, I wish to estimate if there is a trend in my relative risk obtained by coxph. The best i've found is a method from Greenland and Longnecker ( Am J Epidemiology 1992 vol 135 (11) - p 1301-9 ). But all I have is my relative risk and their std error. Is there a command in R that can do this ? If there is someone with a better method ? Thanking you in advance, I look forward your

To the R-help list, In the documentation of the prop.trend.test function in the stats package, no bibliography has been provided which would allow one to find out the theoretical basis of that function and/or details of its implementation. May I suggest that some bibliography be included, as it generally happens with other statistical functions. I currently use R 2.8.0 running on Windows XP.

Hello, I used glsm.mcmc and likfit.glsm to create model. Now I want to predict at different locations, but I can't get glsm.krige to work. I keep getting the error that trend.d and trend.l must have similar specifications. I have tried numerous ways to include the covariates in the glsm.krige model, and I keep getting the same error message. The bolded part is the part that doesn't work.

Hi R-help, I'd like to extract the seasonal component of a short timeseries, and was hoping to use stats::decompose. I don't want to decompose the 'trend' component so I thought I should call decompose(x,filter=0). I think I've either misunderstood the filter argument or come upon a bug/feature in decompose. # EXAMPLE

I partly answered my question since independence_test() function in coin package apparently do Cochran-Armitage trend test just like Eric Lecoutre's function tabletrend() - slightly modified here: > independence_test(pheno ~ geno, data = dat2, teststat = "quad", scores = list(geno = c(0, 1, 2))) Asymptotic General Independence Test data: pheno by groups 1 < 2

Hi, I have a time series (say "x") of 13 years showing an evident increase. I want to exclude two observations (the fourth and 10th), so I do: > trend.test(x[-c(4,10)]) where: > x[-c(4,10)] [1] 7 37 79 72 197 385 636 705 700 1500 1900 and I get: Spearman's rank correlation rho data: x[-c(4, 10)] and time(x[-c(4, 10)]) S = 4, p-value < 2.2e-16

Hello! I am applying the ADF.test function from package uroot to a time series of data. When I apply the full test, incorporating drift and trend terms, the regressor estimate of the drift term is not significantly different from zero. So I apply the test to a model without drift term, with deterministic trend only. But then I always get the following error: