similar to: difference between ns and bs in predict.glm

All, I am trying to include a day of week variable (1-7) in in a regression model. I would like to have the day of week treated as a categorical variable rather than a number the code looks like lm( dep ~ WKDY) I know this is a basic question, but help would be appreciated thanks spencer [[alternative HTML version deleted]]

hello, I am using arima to evaluate a time series regression model. I am using categorical variables such as day of week(Su-Sa), month(Jan-Dec), and holiday status (1/0) as my independent variables. There is evidence of multiplicative interaction between holiday status and weekday and I would like to add an interaction term to my arima model, but I am not sure about best way to go about doing it.

I have been trying to find a command similiar to bs or ns to add a periodic/or cyclical splie term to a regression model in glm.nb I have been using ns to fit a spline to day of the year (1:365): glm.nb(CNT ~ WKDY + ns(DY,df=6) + HOLIDAY_FLAG + Trend) but I think a periodic spline might be more appropriate. Any suggestions would be appreciated. best, Spencer Jones, M.Stat. NLM Fellow Dept.

Hello: I have a question about spline: ns (or bs) in R (library: splines). What's the exact analytic form of the B splines generated by ns (or bs)? I need to calculate the penalty (square of second order derivative) of the basis functions. Is there any shortcut or I need to program myself? thanks, Shuangge Ma ********************* Department of Statistics Phone: 608-263-4782(O) University

I have a function that fits polynomial models for the orders in n: lmn <- function(d,n){ models=list() for(i in n){ models[[i]]=lm(y~poly(x,i),data=d) } return(models) } My data is: > d=data.frame(x=1:10,y=runif(10)) So first just do it for a cubic: > mmn = lmn(d,3) > predict(mmn[[3]]) 1 2 3 4 5 6 7 8

Moi! A student here has been getting a bit irritated with some side effects of scale() (OS is Windows XP, the behaviour occurs in R 2.0.0, but not 1.7.1). The problem is that she scales a variable in a data frame, then does a regression, and tries to get some predictions for some new data. However, at this point she gets an error (see the example below). This seems to be because the

Hallo, I'll need a matrix with n rows of the an identical vector. > h [1] 3 3 3 3 2 2 2 The nmatrix should look like this: > x<-rbind(h,h,h) > x [,1] [,2] [,3] [,4] [,5] [,6] [,7] h 3 3 3 3 2 2 2 h 3 3 3 3 2 2 2 h 3 3 3 3 2 2 2 but I need n rows which must be variable. Can anyone help me? thanks Andreas

Hi all, I recently wrote some code to store a number of polynomial models in a list and return this list. The model is returned fine, but when I make a subsequent call to predict(), I have an error. The code for polyModel selection is listed at the end of the email. Here is an example of the error: > poly.fit <- polyModelSelection(x,y,10,F) > for (d in 1:4) { + lm.model <-

Dear All, I have a question regarding predict.rpart. I use rpart to build classification and regression trees and I deal with data with relatively large number of input variables (predictors). For example, I build an rpart model like this rpartModel <- rpart(Y ~ X, method="class", minsplit =1, minbucket=nMinBucket,cp=nCp); and get predictors used in building the model like

Hi, I work with hyperspectral remote sensing data and I try to built a pls model with this data. I already built the model but if I try to calculate the RMSEP and R2 with a test data set I get the following error message: Error: variable 'subX' was fitted with type "nmatrix.501" but type "nmatrix.73" was supplied The problem is that I don't get the message for

Dear R users, I'm working with 2 data sets which look like (for example) dx and dy in the next code: # Seed set.seed(4) # First data frame dx=matrix(rnorm(6*5),ncol=6) colnames(dx)=LETTERS[1:6] # Second data frame dy=matrix(rnorm(3*5),ncol=3) colnames(dy)=c('A','C','E') As you will notice, some columns in both data sets have the same names. At the end, what I need

I am trying to add 2 stdev error bars to lattice type plots: panel.ebar<-function(x,y,dy=NULL,...) { panel.xyplot(x,y,...) panel.segments(x,y-dy,x,y+dy,...) } Then: xyplot(y~x|fc,data=dat,dy=dat$dy,panel=panel.ebar) This adds error bars but they are not conditioned on the factor fc. xyplot(y+I(y-dy)+I(y+dy)~x|fc,data=dat) This produces 3 series of points in different colors, conditioned

Dear all, I have two coordinates vectors, say X and Y of length n. I want to generate for each couple of coordinates X1,Y1 X2,Y2 X3,Y3....Xn,Yn a random coordinate which is located in a square define as X +/- dx and Y +/- dy. I saw the runif function which can generate for just one value at a time what I want : runif(1, X - dx, X + dx) for X and runif(1, Y - dy, Y + dy) for Y. I would like

Hi! Could you tell me how I can draw a graph with error bars? Sorry, I don't use R that often and I couldn't find it easily in the documentation. TIA -- myriam -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",

Hello, Is this a bug or a feature? I am using R 2.7.1 on Apple OS X. > y <- matrix(1:3,nrow=3) # y is a single-column matrix > df <-data.frame(x=1:3,y=y) > sapply(df,data.class) x y "numeric" "numeric" > df$yy <- y > sapply(df,data.class) x y yy "numeric" "numeric" "matrix"

I want to implement the following algorithm in R: I want to split my data, use a t test to compare both means of the groups to see if they significantly differ from each other. If this is a yes (p < alpha) I want to split again (into 4 groups) and do the same procedure twice, and stop otherwise (here the problem arises). As a final result I would have different groups of data. I made some

Below is a SIMEX object that was generated with the "simex" function from the "simex" package applied to a logistic regression fit. From this mountain of information I would like to extract all of the values summarized in this line: .. ..$ variance.jackknife: num [1:5, 1:4] 1.684 1.144 0.85 0.624 0.519 ... Can someone suggest how to go about doing this? I can extract the

When legend() is used with the angle argument as follows, not only the boxes beside the legend text, but also the whole legend box is filled with shading lines. I think this is not intended: plot(1:10) legend(8, 4, c("A", "B"), angle=c(10, 80), fill=NULL, density=20) I'd suggest as a fix (legend.R of R-1.5.0): 25c25 < rect2 <- function(left, top, dx, dy,

Hi! I have some questions about R function. I try to write a function for multi-returns. The function code is as attachment. dgp.par<-function(ai, bi, t, n) { t0<-t+20 y0<-matrix(0, nr=t0, nc=n) y0[1,]<-ai/(1-bi) for(tt in 2:t0) { y0[tt,]<-ai+bi*y0[tt-1,]+rnorm(n, 0, 1) } y<-y0[21:t0,] x<-y0[20:t0-1,] z<-y0[19:t0-2,] z<-z[2:t,] dy<-y[2:t,]-y[1:t-1,]

I am hoping someone will be kind enough to have a look at the following piece of code and tell me if there is a way to run lme() so it is a lot faster. The inner loop, j in 1:15000, takes about 2 hrs on my 2.8GHz dual Xeon 4GB RAM machine. The timings I have done show the dominant execution time is in lme. options(contrasts=c("contr.sum", "contr.sum"))