similar to: moving correlation coef ?

Here are another three problems with logarithmic scales: 1) segments() does not work with logarithmic scales. I suggest to change lines 962-973 in "plot.c": for (i = 0; i < n; i++) { if (FINITE(xt(x0[i%nx0])) && FINITE(yt(y0[i%ny0])) && FINITE(xt(x1[i%nx1])) && FINITE(yt(y1[i%ny1]))) { GP->col = INTEGER(col)[i % ncol];

Dear R users, I still feel new to R so please apologize if I am doing something stupid here. My use of the polygon() function produces a result that I cannot comprehend: In a plot, I would like to shade the area below a normal distribution. However, I do not want the entire area to be shaded, but just the area on the right side of a vertical line that I draw through the distribution (in

I'm having trouble creating a looping variable and i can't see wher ethe problem arises from any hep gratfully appreciated First create a table x<-table(SURVEY$n_0,exposed) > x exposed False True Under 16 24 1 16-19 68 9 20-24 190 37 25-34 555 204 35-44 330 87 45-54 198 65 55-64 67 35 65+

Hi all, I want to estimate parameters from a VARMA(p,q)-Modell. The equations of the model or the model structures is given by: Xt=beta1+beta2*Xt-1+beta3*Yt-1+epsilon1 Yt=beta4+beta5*Yt-1+espilon2 epsilon1 and espilon2 are white noise. Xt is given by a vector of n elements e.g. (2, 4, 7, 9, …,n)’ and Yt is given by a vector of n elements e.g. (4,9,12,17,…,n)’. The lineVar from

Hi all, Can I run a cointegrating regression, for example delta Xt=a1(Yt-1-cXt-1)+E1t and delta Yt=-b1(Yt-1-cXt-1)+E2t with R were Xt and Yt are non stationary time series at t a,b,c are parameters and E1t and E2t are error terms at t. Yt-Xt is stationary Any suggestions are welcome. Best regards, /fb -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing

Hi, are there any plans to add a colorbar legend to image()? Or such a possibility already implemented which I just haven't discovered yet. Anyway, I will be willing to spent some time on the implementation if there isn't anyone working on that already. Thanks Thomas -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read

Dear R users, i am using locfit package developed by loader in R software, my problem is that as i am doing independont forecast using locfit object , i am able to do independont forecast for more than one years simultaniously. But when i am doing one year forecast(single) this code is giving following error... "Warning message: 'newdata' had 1 rows but variable(s) found have 24

Dear WizaRds, dear Thomas, First of all, I want to tell you how grateful I am for all your support. I wish I will be able to help others along one day the same way you do. Thank you so much. I am struggling with a multistage sampling design: library(survey) multi3 <- data.frame(cluster=c(1,1,1,1 ,2,2,2, 3,3), id=c(1,2,3,4, 1,2,3, 1,2), nl=c(4,4,4,4, 3,3,3, 2,2), Nl=c(100,100,100,100,

Hello all, * * I'm having some difficulty, and I think the problem is with how I'm using append() nested inside a for loop. The data are: y,x 237537.61,873 5007.148438,227 17705.77306,400 12396.64369,427 228703.4021,1173 350181.9752,1538 59967.79376,630 140322.7774,710 42650.07251,630 5382.858702,264 34405.82429,637 92261.34614,980 144927.1713,1094 362998.7355,1420 203313.6442,1070

Is it possible to fit a lagged regression, "y[t]=b0+b1*x[t-1]+e", using the function "lag"? If so, how? If not, of what use is the function "lag"? I get the same answer from y~x as y~lag(x), whether using lm or arima. I found it using y~c(NA, x[-length(x)])). Consider the following: > set.seed(1) > x <- rep(c(rep(0, 4), 9), len=9) > y <-

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible Url : https://stat.ethz.ch/pipermail/r-help/attachments/20061107/01cea378/attachment.pl

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 would like to regress y (dependent variable) on x (independent variable) and y(-1). I have create the y(-1) variable in this way: ly<-lag(y, -1) Now if I do the following regression lm (y ~ x + ly) the results I obtain are not correct. Can someone tell me the code to use in R in order to perform a regression using as explanatory variable a lagged dependent variable? My best regards,

Seriously, if a method gives a wrong result, it's wrong. line() does NOT implement the algorithm of Tukey, even not after the patch. We're not discussing Excel here, are we? The method of Tukey is rather clear, and it is NOT using the default quantile definition from the quantile function. Actually, it doesn't even use quantiles to define the groups. It just says that the groups

I could not find a way to specify the border color of the boxes drawn in a legend, so that the legend can match exactly the colors of the actual plot (e.g. in the case of two superimposed histograms which have different shading and different borders). Indeed, the legend function seems to hard code the color "black" for the borders in this call: rect2(left = xt, top = yt +

On Sun, 27 Jul 2008, S Ellison wrote: > Looking at the legend() source the filled box line colour is hardcoded : > if (mfill) { > if (plot) { > fill <- rep(fill, length.out = n.leg) > rect2(left = xt, top = yt + ybox/2, dx = xbox, dy = ybox, > col = fill, density = density, angle = angle, > border =

Full_Name: Jerome Asselin Version: 1.6.2 OS: RedHat Linux 7.2 Submission from: (NULL) (142.103.173.179) Should be an easy fix... Consider the examble below: plot(0,0) legend(0,0,c("Hello!","Hi!"),pch=1:2,lty=1:2,trace=T) It gives the following trace: > plot(0,0) > legend(0,0,c("Hello!","Hi!"),pch=1:2,lty=1:2,trace=T) xchar= 0.05178 ;

Hi, Drawing a legend I would like to be able to specify the color of boxes which are drawn if fill or density is specified. eg. legend(0,40000,c("raw","LR/PR-TPS"),fill=c(1,2),col=c(1,2),density=c(20,20),angle=c(-20,45),bty="n") Currently the color of the boxes -- border is always black and can *not* be changed. To get this option only a *minimal* change is

Dear R-Users My problem is quite simple: I need to use a fitted model to predict the next point (that is, just one single point in a curve). The data was divided in two parts: identification (x and y - class matrix) and validation (xt and yt - class matrix). I don't use all values in x and y but only the 10 nearest points (x[b,] and y[b,]) for each regression (b is a vector with the

Hi guys, I am afraid I am stuck with an estimation problem. I have two variables, X and Y. Y is explained by the weighted sum of n lagged values of X. My aim is to estimate the two parameters c(alpha0,alpha1) in: Yt = Sum from j=1 to n of ( ( alpha0 + alpha1 * j ) * Xt-j ) Where Xt-j denotes the jth lag of X. I came up with this approach because I thought it would be a good idea to estimate