similar to: legend color problems

Hello R-folks, I don't get the color of the legend in a lattice-plot right. I select a palette from RColorBrewer and use (with a col = mypalette argument) it in the barchart plot. The resulting graph shows the new palette in the graph, but uses the standard palette in the legend rectangles. Adding a col argument into auto.key uses the new palette with the legend text, but not in the

The following code demonstrates an annoyance with plot.lm(): library(DAAGxtras) x11(width=3.75, height=4) nihills.lm <- lm(log(time) ~ log(dist) + log(climb), data = nihills) plot(nihills.lm, which=5) OR try the following xy <- data.frame(x=c(3,1:5), y=c(-2, 1:5)) plot(lm(y ~ x, data=xy), which=5) The "Cook's distance" text overplots the label for the point with the

I'm using ssplot for drawing a map of Austria and colour the nine provinces regarding their share of employment. Now I wanted to add the figures in each province and failed miserably. Using the locator() and text() function caused the error message "invalid graphics state". I try to show you what I have done below, maybe you can find a general fault in my codes. I know that it's

I'm new to the R language and still struggling with compactness of R. I haven't got the right compass into the Documentation, too. So, please apologize the possibly stupidiy of my question. I have the following problem: I have two data sets combined in a data.frame x y 1.3 2.2 2.5 3.4 3.1 3.7 8.2 9.5 7.5 8.3 For the analyses of the data I want to group one column (like the classes in a

Der R-Gurus, first apologies if this is a FAQ, but I due to lack of R-knowledge and terminology I wasn't able to find it. I have the following problem in aggregating results of a model calculation: The results are yearly values of several parameters with several hierarchical spatial factors taken from a database as a data frame with the following structure value | year | spatial1 | spatial2

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I am working on a script that takes numeric performance indicators and runs them against a series of regressors (dummy regressors, yes\no stuff via 0 and 1, e.g. Was is Christmas this week 0=no, 1=yes). The script is as follows (Written as a function): -- Begin Script -- doEnv <- function(HOUR,ENVNAME,REPORTNAME) { library(RODBC) library(forecast) library("geneplotter")

Hallo all I realise this might be a "too much to ask" question, but can I improve the Lattice plot produced by the following code? The type of figure I would like to produce in each segment of the plot appears below the code. library(reserving) # http://toolkit.pbwiki.com/RToolkit <http://toolkit.pbwiki.com/RToolkit> data(GenIns) GenInsInc <- incrementalTriangle(GenIns)

Full_Name: John Maindonald Version: R-2.8.1 OS: MacOS X 10.5.6 Submission from: (NULL) (203.173.3.75) The following code demonstrates an annoyance with plot.lm(): library(DAAGxtras) x11(width=3.75, height=4) nihills.lm <- lm(log(time) ~ log(dist) + log(climb), data = nihills) plot(nihills.lm, which=5) OR try the following xy <- data.frame(x=c(3,1:5), y=c(-2, 1:5)) plot(lm(y ~ x,

Dear list-members, I am trying to use R to conduct a meta-analysis, i.e. I'd like to use a multi-level model to integrate the findings of a number of primary research studies. I set up a simple two level-model (only summary statistics are provided by each study) as follows: sapp.lme <- lme(D ~ 1, data = sapp.frame, random = ~ 1 | STUDYNR, weights=varFixed(~-1+STDERR_D),na.action =

Dear list users, I modeled the probability of occurrence of one species: "Cyperus dilatatus". I modeled the species using three different approaches: c("random","target","index") What I want to achieve is to make a plot of all prediction maps in a row with to conditional variables, that is, with the species and the approach I prepared a data.frame to try

Hello again, everytime I think I got something to work, the next issue comes up... I have the following data.frame, I want to visualize: > data_rb tld spam1 spam2 share 1 ca 826436 73452 0.0889 2 org 470550 25740 0.0547 3 de 156042 15531 0.0995 4 com 140753 7527 0.0535 5 edu 34845 2507 0.0719 6 net 12781 382 0.0299 7 ru 7648 18 0.0024

Hii, Can anybody help me, I don't know how to print the "median". Below is my code snipplet... x <-read.table(file="D:/Uni/Diplom/Diplom/Grafiken/R/BATMAN/Kabel/Batman1hop/Standardabweichung__output_30_1_Kabel(30m)_b.txt") png(filename = "D:/Grafiken/R/Standardabweichung/Kopie.png", width = 640, height = 480,pointsize = 12, bg = "white", res =

Dear all, Myself Vijaykumar Muley working as senior research fellow. By training I am a computational biologist with not a strong knowledge of statistics. I have done some analysis which is explained as follows, I have 10340 (X) profiles of binary vectors with same length(N=845), I will call then "gene profiles" for example... v1 v2 v3 v4.....vN a 1 0 1 0 1 b 0

Hi, I'd like to run regsubsets for model selection by exhaustive search. I have a list with 20 potential explanatory variables, which represent the real and the imaginary parts of 10 "kinds" of complex numbers: x <- list(r1=r1, r2=r2, r3=r3, ..., r10=r10, i1=i1, i2=i2, i3=i3, ..., i10=i10) Is there an easy way to constrain the model search so that "r"s and

Hi there, How can I superscript the "2" of "r2 =..." in the legend below? legend(210, 110, paste("r2 = ", format(summary(regression)$adj.r.squared,digits=3), sep="")) I usually use "expression(paste(...", but it won't work this time because "format(summary(..." needs to be evaluated. Thanks in advance! -- View this message in