similar to: Step error

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates <- as.Date(data.df[,1]) selection<-which(dates>=as.Date("1986-1-1") & dates<=as.Date("2007-12-31")) dep <- ts(data.df[selection,c("dep")]) indep.ret1

Hello R users, I am new to R and am wondering if anyone can help me out with the following issue: I wrote a function to build ts models using different inputs, but when R displays the call for a model, I cannot tell which variables it is using because it shows the arguments instead of the real variables passed to the function. (e.g Call: lm(formula = dyn(dep ~ lag(dep, -1) + indep)) --->

Hello everybody, I've been practicing with some data in R and SPSS and I noticed that there are some differences in ANOVA results. For example with : y<-c(1,2,34,2,3,45,2,1,67,3,2,67,2,2,98,4,4,23,1,1,23,2,3,45) and x<-rep(c(1,2,3),8) I get in R ( with summary(aov(y~x)) ) MSres=350.7 df=22 while in SPSS I get MSres=221.9 df=21 Can enyone explain me what is the problem or

Hello, Is there a better way to use paste such as: a = paste(colnames(list.indep)[1],colnames(list.indep)[2],colnames(list.indep)[3],colnames(list.indep)[4],colnames(list.indep)[5],sep="+") > a [1] "aa+dummy1+dummy2+bb+cc" I tried a = paste(colnames(list.indep)[1:5],sep="+") > a [1] "aa" "dummy1" "dummy2"

I am using R on a Windows XP professional platform. The following code is part of a bigger one CODE press=function(y,x){ library(qpcR) models.press=numeric(0) cat("\n") dep=y print(dep) indep=log(x) print(indep) yfit=dep-PRESS(lm(dep~indep))[[2]] cat("\n yfit\n") print(yfit) yfit.orig=yfit presid=y-yfit.orig press=sum(presid^2)

Hi useRs, I'm experiencing something quite weird with glm() and weights, and maybe someone can explain what I'm doing wrong. I have a dataset where each row represents a single case, and I run glm(...,family="binomial") and get my coefficients. However, some of my cases have the exact same values for predictor variables, so I should be able to aggregate up my data frame and

Hello R community, I have a question about the logistic regression function. Specifically, when the predictor variable has not just 0's and 1's, but also fractional values (between zero and one). I get a warning when I use the "glm(formula = ... , family = binomial(link = "logit"))" which says: "In eval(expr, envir, enclos) : non-integer #successes in a binomial

Hello R users, I have been trying to output all my results (text, plots, etc) into the same postscript file as one document, but have been unable to...Can anyone help me improve my code below so that I can accomplish this? Currently I have to output them separately then piece them back together into one document.. Thanks in Advance for any help! options (scipen=999, digits=7)

Does anyone know how can I calculate power in wilcoxon.test, kruskal. test and oneway.test in an F? I have already found power.t.test for calculating power in t.test but it seems that there is nothing for the others. Any answer could be useful. Thanks for your interest. --------------------------------- [[alternative HTML version deleted]]

In a function, say foo.glm for glm objects I want to use the name of the object as a label for some output, but *only* if a glm object was passed as an argument, not a call to glm() producing that object. How can I distinguish these two cases? For example, I can use the following to get the name of the argument: foo.glm <- function(object) { oname <- as.character(sys.call())[2]

Hello world, I'm trying to do an anova on data in data.set, dependent variable is a column named "dep.var", grouping variable is in a column called "indep.var", and is.factor(indep.var) is TRUE... why can't I just do aov(dep.var ~ indep.var, data = data.set)? What have I done to deserve this?! What gives? Am I missing something totlly obvious? R-base-1.0.0-1,

I'm trying to develop some graphic methods for glm objects, but they only apply for models where all predictors are discrete factors. How can I test for this in a function, given the glm model object? That is, I want something that will serve as an equivalent of is.discrete.glm() in the following context: myplot.glm <- function(model, ...) { if (!inherits(model,"glm"))

Default arguments are evaluated in the function frame, not in the calling environment (nor in the same place as explicit arguments). > Which to me reads that a with statement as above is equivalent to > > > attach(data) ; aov.SS1(y=Obs) ; detach(data) > > Or is that just wishful thinking?? The latter. On Thu, 7 Oct 2004, RenE J.V. Bertin wrote: > Hello, > >

Dear list, I have recently encountered an odd error when running glm(dep~indep, quasipoisson): while, with a subset of my data, I could get a perfectly reasonable model, once I include all of my data (17K+ observations, 29 variables), I get the following error: Error in if (any(y < 0)) stop("negative values not allowed for the quasiPoisson family") : missing value where

i have a data frame(dat) which has many variables.and i use the following script to get the crosstable. >danx2<-c("x1.1","x1.2","x1.3","x1.4","x1.5","x2","x4","x5","x6","x7","x8.1","x8.2","x8.3","x8.4","x11",

I wonder if anyone out there has written a routine to solve the simple linear calibration problem? - fit regression of y vs x - estimate the value x0 (with 95% CI) that gives y0 Thanks for any help. Bill -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

Dear All, I'me having (much) trouble understanding why it happened and answering a referee's comment to part of a submitted manuscript. I've tried to google for help but... I'm really confident that although this is a R-Help list someone can help me! I used R to do an ANCOVA w/ RNA/DNA as the dep var, sl as the indep var and gut (a factor w/ levels: prey and empty) as the

Hi, I use lme to fit models like R> res1 <- lme(y~A+B, data=mydata, random=~1|subject) R> res2 <- lme(y~B+A, data=mydata, random=~1|subject) (only difference between these two models are the sequence in which the indep variables are written in formula) where y is continuous and A, B, and subject are factors. To get ANOVA table I used R> anova(res1) R> anova(res2) and found

In http://tolstoy.newcastle.edu.au/R/e2/help/06/10/3064.html a method was given for converting a frequency table to an expanded data frame representing each observation as a set of factors. A slightly modified version was later included in the NCStats package, only on http://rforge.net/ (and it has too many dependencies to be useful). I've tried to make it more general, allowing an input

I have a matrix called mat and y is the column number of my response and x is a vector of the column numbers of my terms. The variable name of y can change, so I don't want to hardcode it. I can find out the name as follows: > names(mat)[y] [1] "er12.l" Then I can run the regression by hard coding the variable name as follows: > mod <-