similar to: Interpretation of 'swtich'

I noticed that when I fit a linear model using 'lm' there is an attribute called "factors" that is added to the "term". It doesn't seem to appear for 'model.matrix', just 'lm'. I have been unable to find where it gets constructed or what it means? It looks like a two dimensional array that I may be able to use so I would just like to get some

Can someone enlighten me on why the following doesn't work? setwd('C:/Temp/R') d <- rep(1:53,2) (s <- ts(d, frequency=53, start=c(2000,10))) n <- length(s) k <- n%/%3 for(i in (n-k):n) { st <- c(start(s)[1] + (start(s)[2] + i)%/%frequency(s), (start(s)[2] + i) %% frequency(s)) ed <- c(start(s)[1] +

I am trying to upgrade to R 2.14 from R 2.13.1 I have compied all the libraries from the 'library' directory in my existing installation (2.13.1) to the installed R 2.14. Now I want to uninstall the old installation (R 2.13.1) and I get the error: Internal Error: Cannot find utCompiledCode record for this version of the uninstaller. Any ideas? Kevin [[alternative HTML

I noticed that the text() command adds text to a plot. Is there a way to either make the plot blank or add text to a "blank sheet". I would like to "plot" a page that contains just text, no plot lines, labels, etc. Suggestions? Kevin [[alternative HTML version deleted]]

I can get multiple plots on a page like: op <- par(mfcol = c(3, 1)) What I was wondering is if there is a way to have a title for the whole page? I can specify the title for each individual plot like: plot(xxx, main=".") But I would like a 'title' for the group of plots. Is this possible? Thank you. Kevin [[alternative HTML version deleted]]

I have two time series a <- ts(1:10, start=c(1,6), end=c(2,5), frequency=10) b <- ts(1:5, start=c(2,1), end=c(2,5), frequency=10) Obviously 'b' is a subset of 'a'. I want a single index value indicating where that start of 'b' lines up with the start of 'a'. So in this simple example I would expect an index of 5. I was playing with 'merge'.

Hello, I asked a question about what the most likely process to follow if after a time-series fit is performed the residuals are found to be non-normal. One peron responded and offered to help if I supplied a sample data set. Unfortunately now that I have a sample I have lost the emai addressl. If you are that person or have some ideas please email me back at rkevinburton at charter.net. Thank

What is wrong with the following? x <- 1:2 if(x[1] > 0) { if(x[2] > 0) { print("1 & 2 > 0") } else { print("1 > 0") } } else { if(x[2] > 0) {

Thanks to Marc Schultz I found the documentation on the "factors" attribute under ?term.object. It stats: factors: A matrix of variables by terms showing which variables appear in which terms. The entries are 0 if the variable does not occur in the term, 1 if it does occur and should be coded by contrasts, and 2 if it occurs and should be coded via dummy

I have some data where the frequency is heavily weighted on the lower end. So I have lots of low values with very few higher values. I would like to find breakpoints that cover the data with as much detail as possible. I find that if I use hist() to automatically find the breaks for me it finds breaks that are too coarse for the low values. I have tried the other algorithms (like 'Scott'

I can put a legend on a plot with something like: legend('bottom', leg.txt, horiz = TRUE, fill = colors) But what if the arrays leg.txt and colors are too big? I would still like to provide a legend but to save space I would like to just show small boxes with the color filled in so it will still fit on the plot. If I could adjust the size of the boxes (at least in one dimension)

Hi there - I am using R1.5.1 on WinNT and the latest MASS (Venables and Ripley) library. Running the following code: >minimod<-glm(miniSF~gtbt*f.batch+log(mxjd),data=gtbt,family="binomial") >summary(minimod,cor=F) Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 0.91561 0.32655 2.804 0.005049 ** gtbtgt 0.47171

I started to check what I thought I knew with autocovariance and it doesn’t jive with the the calculations given by ‘R’. I was wondering if there is some scaling or something that I am not aware of. Take the example Ø d <- 1:10 Ø (a <- acf(d, type="covariance", demean=FALSE, plot=FALSE)) Autocovariances of series ‘d’, by lag 0 1 2 3 4 5 6

I can get an array of strings for the data that I want using 'paste()' as follows: paste('ma', 1:am$arma[2], '=', coef(am)[1:am$arma[2] + am$arma[1]], sep='') This results in a vector of strings like: [1] "ma1=1.17760133668255" "ma2=0.649795570407939" "ma3=0.329456750858276" What I would like is fixed.pars <-

I was wondering what the best approach is for missing data in a time series. I give an example using xts but I would like to know what seems to be the "best" method. Say I have library(xts) xts.ts <- xts(1:4,as.Date(c("1970-01-01", "1970-1-3", "1980-10-10", "2007-8-19")), frequency=52) I would like to turn this into a time series (still

Given a Date object or simply a year is there an R function to tell me if the it is a leap year or not? I was hoping for something like 'is.leapyear'. I probably can build my own function (year divisible by 4 etc.) but I would rather use an existing function if it is available. Thank you. Kevin

Has anyone come across the right combinations to print a limited number of digits? My trial and error approach is taking too much time. Here is what I have tried: > op <- options() > a <- c(1e-10,1,2,3,.5,.25) > names(a) <- c("A", "B", "C", "D", "E", "F") > # default > a A B C D

Why does: (shape/scale) * (1:365/scale)^(shape - 1) return a vector of numbers but calling a function hasard(1:365,shape,scale) defined like: hazard <- function(x,shape,scale) { return (shape/scale) * (x/scale)^(shape - 1) } Only return a single value? It is like x becomes a single value passed as an argument. Thank you. Kevin

I am running R 2.9.2 and creating a PDF that I am trying to open with Adobe Reader 9.2 but when I try to open it the reader responds with "There was an error opening this document. The file is damaged and cannot be repaired.: I am using the R command(s): pdf(file="cat.pdf", title="Historical Sales By Category") for(j in 1:length(master)) { d <-

I would like the text plotted with 'mtext' to be alighned like it is for printing on the console. Here is what I have: > print(emt) ME RMSE MAE MPE MAPE MASE original -1.034568e+07 1.097695e+08 2.433160e+07 -31.30554 37.47713 1.5100050 xreg 1.561235e+01 2.008599e+03 9.089473e+02 267.05490 280.66734