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The Blue Book allows the 'type' argument to be used in matpoints and matlines. matlines(x, y, type="l", lty=1:5, pch=, col=1:4) R-1.0.0 does not. Thus, type="h", "b", must be invoked thru matplot( x, y, type = "h", add=TRUE) For the sake of consistency with S, it would be nice to have matlines defined as: "matlines" &...

Hi, I am trying to specify four ticks (at 0,1,2,3 for the y axis) in a persp plot but the defaults overrule my specification and I obtain seven of them. Is it possible to gain full control over them in such a plot? Here is my code: matlines=matrix(c(1:7,3:9,3:9,2:8),nrow=7,ncol=4) persp(x=seq(1,7,1),y=seq(0,3,1),z=matlines,theta=-30,phi=25,zlim=c(0,10),d=5,ticktype="detailed",xlab="x", ylab="y",zlab="z") [[alternative HTML version deleted]]

I'm using (Linux version) xfig() within a function and then simple matplot() and matline() plots. Although I do not define any bg or fg default color, sometimes all lines in the final fig file are green. The same code works as (I) expected if I use x11() or pdf(). This is what I'm doing: I open 2 devs: xfig() dev.set(2) layout(mat1) xfig() dev.set(3)...

...1, col2rgb) dimnames(colnames.rgb)[[2]] <- colnames.col baseline <- 1:32 offset2 <- 2*baseline offset3 <- 3*baseline offset4 <- 4*baseline offsets <- cbind(offset2, offset3, offset4) # Produces expected result X11() matplot(baseline, col = colnames.col[1], type = "l") matlines(offsets, col = colnames.col[-1]) # Displays a ??yellow?? line X11() matplot(baseline, col = as.matrix(colnames.rgb[,1]), type = "l") matlines(offsets, col = colnames.rgb[,-1]) ---------------------------------------------------------- SIGSIG -- signature too long (core dumped)

...11","12","62","9212","33","19","12529252","7111") however R won't let me use "10" (i.e. solid line) because it contains a zero. I could possibly add the total line in later (using something similar to lines or matlines) and then define the lty as lty=1, but then I have the problem of trying to get the new line into the legend. Does anyone have any ideas about how to combine custom and preset line types? Regards, James http://www.nabble.com/file/p21593760/CRA.gif -- View this message in context: htt...

...31 2679,140625 > plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab="CHAO", ylab="Depth", pch=15, las=2, main="Sep12-RNA", cex.main=1) > lmR <- lm(cd$Depth~cd$CHAOsep12RNA) > abline(lmR) pconfR <- predict(lmR,interval="confidence") matlines(cd$CHAOsep12RNA,pconfR[,c("lwr","upr")], col=1, lty=2) I also tried > newx <- seq(min(cd$CHAOsep12RNA), max(cd$CHAOsep12RNA), length.out=11) > a <- predict(lmR, newdata=data.frame(CHAO=newx), interval=c("confidence")) > plot(cd$CHAOsep12RNA,cd$Depth,...

...n issue with WMF's. I am working on WindowsXP and outputting WMF format images. I then take the WMF format images and insert them in PowerPoint. I take the PowerPoint and convert it to PDF. The WMFs are nothing special. Just the typical x-y plot with a gray dotted grid added and a few matlines. The WMFs, when produced look great! They show all the necessary detail. Unfortunately, after placed in the PowerPoint and the PowerPoint is converted to PDF via MS Office's built in conversion utility, the resulting image have diagonal streaks across them from the y-axis down to the...

I have a sample of lung capacities from a population measured against height. I need to know the 95% CI of the lung capacity of a person of average height. I have fitted a regression line. How do I get a minimum and maximum values of the 95% CI? My thinking was that this has something to do with covariance, but how? My other thinking was that I could derive the 0.975 (sqrt 0.95) CI for the

...I added the transformed variables to the dataframe pp <- predict(wt.mod, int = "p", newdata=pred.frame) # runs ok? pp <- predict(lm(wt.data$logS~wt.data$logA), int = "p", newdata=pred.frame) # lines are jagged?? # I am not sure how to get the lines to draw correctly here matlines(pred.frame$a,pp, lty=c(1,2,2),col="black") > sessionInfo() R version 2.8.1 (2008-12-22) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attac...

..."matplot" help page with examples > that worked when I tested them recently. Indeed. That's how I understood you well. matplot() has been one of my favorite plotting functions in S and Splus in the 1990s ... and that's why the R source code of matplot(), matpoints(), and matlines() still has a comment mentioning that I wrote (the first version of) it for R on June 27, 1997. By design (from S), matplot() {etc} has always been thought as a convenience wrapper to calling plot() and lines(), lines(), lines() or plot() and points(), points(), points() {plot() : for...

Is there a way to create Bollinger Bands without having to loop on the observations of a time serie? Any help appreciated Thanks

...ad to use a glm instead of my basic lm on some data due to unconstant variance. now, when I plot the model over the data, how can I easily get the 95% confidence interval that sormally coming from: > yv <- predict(modelVar,list(aveLength=xv),int="c") > matlines(xv,yv,lty=c(1,2,2)) There is no "interval" argument to pass to the predict function when using a glm, so I was wondering if I had to use an other function thanks -- View this message in context: http://r.789695.n4.nabble.com/95-confidence-intercal-with-glm-tp2716906p2716906.html Sen...

...t should instead be labeled the same as graphics::plot(x, y[,1], ...) in such cases.? To fix this problem, I made fda::matplot generic; graphics::matplot is not generic.? And a coded methods for x of class numeric, matrix, Date and POSIXct plus a default.? Each calls either graphics::matplot or matlines as appropriate after first setting up the horizontal axis properly if x is of class Date or POSIXct. ????? For specific examples, consider the following taken from fda::matplot.Rd: invasion1 <- as.Date('1775-09-04') invasion2 <- as.Date('1812-07-12') earlyUS.Canada &...

...11","12","62","9212","33","19","12529252","7111") however R won't let me use "10" (i.e. solid line) because it contains a zero. I could possibly add the total line in later (using something similar to lines or matlines) and then define the lty as lty=1, but then I have the problem of trying to get the new line into the legend. Does anyone have any ideas of how to combine custom linetypes with preset types? http://www.nabble.com/file/p21593757/CRA.gif -- View this message in context: http://www.nabble...

Dear All, I created a script to calculate averages - two groups: "parcel" and "date" - and, based on these averages, make a graph. The problem is that 'R' does not recognize the first column even if I try to insert one. A brief example Raw data: Data <- sample(1:100, 30, replace = FALSE, prob = NULL) Date <-

Dear R-Users I have 2 questions: Firstly, If I create a matplot and legend for multiple vectors and then tag another vector on using matlines (e.g. a 'total' of all vectors), is there anyway to add the new line to the legend without recreating it? I have created the plot this way because I would like to define the lty and lwd for the 'total' vector so that it can be differentiated from the rest. Secondly, in many of...

...lot x with the mean of each column of y as: plot (x,mean(y)). But, to do what I want, I had to write this pair of for loops: for (i in 1:dim(y)[1]) { for (j in 1:length(x)) { points(x[j],y[i,j]) } } to add the points. I think there must be a better way using some variant of apply or matlines. Even after reading the docs, I'm still scratching my head a bit... Or maybe I copy x out so that it is of size m by n?!? Any thoughts on what an R user would do here? Thanks! Ray

Hi R People: If I have a fitted values from a model, how do I plot the (1-alpha)100% confidence intervals along with the fitted values, please? Also, if the intervals are "shaded" gray, that would be nice too, please? I check confint, but that doesn't seem to do what I want. Thanks in advance, Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and

...orig Tue Jan 13 18:17:54 1998 --- matplot.new2 Tue Jan 13 18:16:49 1998 *************** *** 1,12 **** ###---- As S (just 'better' ...) ! matpoints <- function(x, y, lty=1:5, pch=NULL, col=1:6, ...) matplot(x=x, y=y, type = 'p', lty=lty, pch=pch, col=col, add=TRUE, ...) ! matlines <- function(x, y, lty=1:5, pch=NULL, col=1:6, ...) matplot(x=x, y=y, type = 'l', lty=lty, pch=pch, col=col, add=TRUE, ...) matplot <- function(x, y, type="p", ! lty=1:5, pch=NULL, col=1:6, xlab=NULL, ylab=NULL, xlim=NULL, ylim=NULL, ..., add=...

...ale = 1, type = ifelse(type == : calling predict.lm(<fake-lm-object>) ..." This isn't a geepack problem because I get the same error message using the gee package as well. The above simplified code is how I usually create and plot 95% CI bands for a linear model, followed by: matlines(x,y) I have read through the package PDFs and searched the web and archives of various listservs without success. Any help would be most appreciative. Seems like there should be a simple work-around. Thank you, Jason [[alternative HTML version deleted]]