similar to: legend outside plotting area

Hi, I am trying to generate a figure of 9 plots that are contained in one device by using par(mfrow = c(3,3,)) I would like to have 1 common legend for all 9 plots somewhere outside of the plotting area (as opposed to one legend inside each of the 9 plots, which the function legend() seems to generate by default). Any hint how to do this? Best, Georg

something like "sprintf" in C? so I can do: print(sprintf("the correct result is %3.4f\n", myresult)); ------- Also, I am desperately looking for a "clear console screen" function in R... thanks a lot! [[alternative HTML version deleted]]

Is there a function in R for doing Greenhouse-Geiser correction in ANOVA models? Is it already available in the aov function? How do we use it? Best, Darren [[alternative HTML version deleted]]

Hello. Some months age an e-mail was posted in which a comparison between Glimmix and glm was discussed. I have not been able to find that e-mail on the R archive. Does anyone recall the date of the above e-mail? Thank you very much. ******************************************* Antonio Paredes USDA- Center for Veterinary Biologics Biometrics Unit 510 South 17th Street, Suite 104 Ames, IA 50010

When I use xyplot (from package lattice) to produce a multi-series graph: xyplot(linear+quadratic+sqrt~x, data=df, main="complexity of different functions", ylab="y", col=c("red", "black", "orange"), type="b", lty=c(1,2,3), pch=c(1,2,3), auto.key=TRUE) where I changed the default colour, line type and data point type so that each series

Hello. I am using R version 2.1.1 on Windows 2000. I am using a par(mfrow=c(2,2)) statement to produce 4 plots on one screen. I want a single horizontal legend to appear at the top of the four plots. My code is something like this: par(mfrow=c(2,2)) plot(x,y1) lines(x,y2) lines(x,y3) plot(x,z1) lines(x,z2) lines(x,z3) plot(x,t1) lines(x,t2) lines(x,t3) plot(x,w1) lines(x,w2) lines(x,w3)

Hi all I have a question regarding differences in the way gregpexr works in R 2.3.0 and R 2.4.0. In R 2.3.0, this is what happens: > gregexpr(" [a-z] [a-z] ", " a b c d e f ", perl=T) [[1]] [1] 1 3 5 7 9 attr(,"match.length") [1] 5 5 5 5 5 ... while in R 2.4.0, this is what happens: > gregexpr(" [a-z] [a-z] ", " a b c d e f ", perl=T)

Hello- It has been several years since anyone has asked, so i am asking again- has anyone created a routine to estimate confidence intervals for predictions from nls models (ala Bates and Watts 1988)? Thanks - Alice Shelly [[alternative HTML version deleted]]

Dear R-Experts, I was wondering how to fit a cumulative gaussian to a set of empirical data using R. On the R website as well as in the mail archives, I found a lot of help on how to fit a normal density function to empirical data, but unfortunately no advice on how to obtain reasonable estimates of m and sd for a gaussian ogive function. Specifically, I have data from a psychometric function

Dear R community, I am a beginner to R and have a question concerning mget, about I could not find anything in the various documentation. I have a column in a dataframe x for which I want to get values in y: mget(x[,1], env=y, ifnotfound=NA) I receive an error mesage: Error in mget(x[, 1], env = y, ifnotfound = NA) : attempt to use zero-length variable name this is probably due to

Hi, I have an R programming problem and I havent found anything in the documentation yet: I have a data matrix, in which two neighbouring columns represent replicates of the same experiment, e.g. something like this: A A B B C C row1 1 1 1 2 2 2 row2 1 1 1 1 1 2 I would like to test, if the values for the two replicates in a row are the same or if they differ and generate a new

Hello: Could anyone point me to a nice example where someone has created methods for "[" on a user defined Class? I looked at the package Matrix but that was a little daunting. I'm looking for someone a little more introductory. I've tried to search the help section and the web but its difficult since "[" isn't searchable. Thanks in advance! Greg [[alternative

Inspired by the functions that Barry Rawlingson and Dave Forrest posted for searching Rwiki and R-help archive, I've made up a function that does the search on Prof. Baron's site (Thanks to Prof. Baron's help on setting up the query string!): RSiteSearch <- function(string, restrict="Rhelp", format="long", sortby="score",

In http://finzi.psych.upenn.edu/R/Rhelp02a/archive/14637.html some rudimentary R functions were given for drawing proportional area venn diagrams with area of each intersection ~ the count in a 2 x 2 x 2 table. I'm interested in this, for another application: showing the correlations among Y, X1, X2 using area ~ r^2 of each pair (sometimes called a Ballantine diagram). Before I attempt

Hello useRs and helpRs, Some time ago, in a gallaxy far away (here is the thread : http://finzi.psych.upenn.edu/R/Rhelp02a/archive/46532.html ) we discussed about a graph gallery showing the power of R in that domain. I did some work around that, and there is a (pretty advanced) draft here : http://addictedtor.free.fr/graphiques/displayGallery.php For instance, there are some of my graphs,

This below is not solvable with uniroot to find "a": fn=function(a){ b=(0.7/a)-a (1/(a+b+1))-0.0025 } uniroot(fn,c(-500,500)) gives "Error in uniroot(fn, c(-500, 500)) : f() values at end points not of opposite sign" I read R-help posts and someone wrote a function: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/92407.html but it is not very precise. Is there any

Suppose I have the following data: y x id 44 0 104 48 58 104 48 55 204 47 105 204 41 275 206 18 67 209 ....... I fit the model >fit=lme(y~x+I(x^2),random=~1|id) Now I want to make a prediction plot: >time=seq(0,300,len=100) >plot(predict(fit,data.frame(x=time),level=0)) Very fine. It gives me the prediction curve based on the model. My further request is to make a confidence bands

Hi all, Is it possoble to include an R function in a Perl program? I would like to use the "limma" library for microarray analysis, and have no clue how this can be done. Thanks in Advance -Tarun

hie all i am trying to carry out a categorical data analysis but my problem is that when in i use the chi squared test some of my expected values are less than 5. is there a test that can handle this situation. the data is not a 2*2 table. its more from the social sciences where you have from strongly agree to strongly disagree. i know i can collapse vthe tables but there is a loss of

The CRAN host in Tampa, FL (cran.hostingzero.net) isn't responding, and hasn't responded in quite a while -- at least problems were reported more than a year ago (Oct 2007) http://finzi.psych.upenn.edu/R/Rhelp02a/archive/116706.html although someone apparently succeeded in July 2008 http://finzi.psych.upenn.edu/R/Rhelp02a/archive/137451.html Perhaps it's just flaky, and not