similar to: legend - bug with argument angle (PR#1404)

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 =

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

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

In legend() is it possibly to mix lines and boxes? I would like to do something like legend(locator(1), legend = c("al", "ben", "chris", "dave"), lty = c(1, 2, 3, NA), fill = c(NA, NA, NA, "gray")) to indicate that "al", "ben" and "chris" have lines, and "dave" has a gray shading. Many thanks,

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];

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 R-help I am trying to mask the ocean from an image plot I have made. Here is some example code: library(mapdata) image(x=110:155, y =-40:-10, z = outer(1:45, 1:30, "+"), xlab = "lon", ylab = "lat") outline <- map("worldHires", plot=FALSE) # returns a list of x/y coords xrange <- range(outline$x, na.rm=TRUE) # get bounding box yrange

"contour" can't return a list of contours (it would be nice) "arrows" has different options from S-PLUS (no "open" or "rel" options, no "lwd" parameter or graphics parameter pass-through; "size" in S corresponds to "length" in R). [I know we needn't follow S-PLUS syntax slavishly, but it would be nice at least to

I have made a minor hack to "legend" (in R 1.0.0, but I didn't notice any changes to legend in the 1.0.1 NEWS) to allow the legend to be formatted in multiple columns, or horizontally (number of columns <- number of legend items). (I find this helpful when I have lots of legend items and not a lot of vertical space to squeeze the legend into.) (Another hack I've considered

Dear all, maybe I have misunderstood something but to me it seems like a minor error in the help for ?legend for the argument 'bg'. There it says: bg: the background color for the legend box. (Note that this is only used if 'bty = "n"'.) I think, however, that it should be changed to: bg: the background color for the legend box. (Note that this

Hi, I have done the interpolation for my data and I was able to create the contours in multipanel with the help of Pascal. Now, I want to clip the contour with the shapefile. I want only the portion of contour to be displayed which falls inside the boundary of the shapefile. The data mydata.csv can be found on https://www.dropbox.com/s/khi7nv0160hi68p/mydata.csv The data for shapefile can be

Problem: neither fg or bg nor color work properly in dotchart. version: R-1.7.1 for windows code which shows the errors: x <- matrix(rnorm(16),ncol=2,dimnames=list(paste("a",1:8,sep=""),c("before","afte r"))) dotchart(x,fg="blue",bg="lightgrey") dotchart(x,color=c("red","blue")) Dr Ian J Wilson Lecturer in

Full_Name: Mai Zhou Version: 1.9.1 OS: Win XP Professional Submission from: (NULL) (12.222.227.93) > power.t.test(n=25, delta=0.1, sig.level=1.1, strict=TRUE, type="one.sample") One-sample t test power calculation n = 25 delta = 0.1 sd = 1 sig.level = 1.1 power = 1.088311 alternative = two.sided ### power can never be

The package ipred is uploaded to CRAN. The main focus of the package is the calculation of improved predictors in classification tasks. Misclassification error can be improved by bootstrap aggregated classification trees and/or the framework of indirect classification. Furthermore, a unified interface for the estimation of misclassification error completes the features of ipred. We try to make

The package ipred is uploaded to CRAN. The main focus of the package is the calculation of improved predictors in classification tasks. Misclassification error can be improved by bootstrap aggregated classification trees and/or the framework of indirect classification. Furthermore, a unified interface for the estimation of misclassification error completes the features of ipred. We try to make

Code like df <- data.frame(x=1:10) y <- 20:29 eval(quote(x+y), env=df) does what you might expect: it looks for x and y in the data.frame, and when it doesn't find y there, it looks in the parent environment. However, sometimes I'd like to construct a single environment out of df, so that I can pass it to nested functions and get the same behaviour. Right now, I get the wrong

Code like df <- data.frame(x=1:10) y <- 20:29 eval(quote(x+y), env=df) does what you might expect: it looks for x and y in the data.frame, and when it doesn't find y there, it looks in the parent environment. However, sometimes I'd like to construct a single environment out of df, so that I can pass it to nested functions and get the same behaviour. Right now, I get the wrong

When finally got some time to do some coding, I started and stopped right after. The stationary test is a good starting point because it demonstrates how we should be able to move the very basic R matrices. I have a real- world small N data set with rows: id(n=1)---t1---variable1 ... id=(N=20)---T=21---variable1 Thus, a good test case. For first id I was considering something like this: lag

Full_Name: Tomas Larsson Version: 2.6.0 OS: Windows XP Submission from: (NULL) (198.208.251.24) This is not a bug, it is a performance issue but I think it should have an easy fix. I have a large matrix (about 2,000,000 by 20), when I type colnames(x) it takes a long time to get the result. However, if I select just the first couple of rows of the matrix I don't have to wait for the