similar to: aggregate.formula

Hi all, I I need to manipulate the x axis in a stripchart. I will use one of the data sets included in R to explain what I need to do. attach(ToothGrowth) stripchart(len[supp=='VC']~dose[supp=='VC'], vertical=TRUE, group.names=c('A','A','A')) stripchart(len[supp=='OJ']~dose[supp=='OJ'], add=TRUE, vertical=TRUE, at=c(1:3)+.1,

Thank you for your guys reply for my previous question. But I got one more question about the boxplot. With the code in the R-help: boxplot(len ~ dose, data = ToothGrowth, boxwex = 0.25, at = 1:3 - 0.2, subset = supp == "VC", col = "yellow", main = "Guinea Pigs' Tooth Growth", xlab = "Vitamin C dose mg", ylab =

Kaiya Liu wrote: > > Hi, > > I encountered a problem when I did interaction plots using the > interaction.plot() function. The labels for the trace factor appeared on > the plot when I did the plots in separate windows, but disappeared/cut off > when multiple plots were placed in one window. Can anybody help me? >From the example of ?interaction.plot: data(ToothGrowth)

Dear List, ? I am creating a boxplot with two subsets, very similar to the example by Roger Bivand at ?boxplot (reproduced below).? I am trying to change the labels on the x-axis to have one number to cover both subsets.? I can do this in other plots by using axis=FALSE followed by a separate axis() command.? I have also tried variations in the names= argument but can't get it to work.?

termplot() does not carry subsetting over to carriers that are in the environment but not in the data frame. This generates a "subscript out of bounds" error. > data(ToothGrowth) > logdose <- log(ToothGrowth$dose) > tooth.lm <- lm(len ~ logdose, data=ToothGrowth) > termplot(tooth.lm) ## Works fine > toothVC2.lm <- lm(len ~ poly(dose,2),

I'm trying to plot a scatterplot of two variables using pch to plot different characters based on a third factor. Here is my example > data("ToothGrowth") > attach(ToothGrowth) > levels(supp) [1] "OJ" "VC" > plot(len ~ dose,pch=as.numeric(supp)) > legend(locator(1),pch=as.numeric(supp),legend=levels(supp)) The command as.numeric(supp) returns 2 2

Hi All, As a follow-on from the mail below, it is possible to have boxplots based on means and variance for more than one input, i.e., bxp(input1,input2,....) Thanks in Advance, Dermot ******** Previous Mail on this subject *********** >I have a couple more questions about boxplots. In the books I've read on >statistics (I'm not a statistics expert but just a poor engineer

Hi to all, Does anybody knows why this is giving an error? data(ToothGrowth) # Two-way design with options bargraph.CI(dose, len, group = supp, data = ToothGrowth, xlab = "Dose", ylab = "Growth", cex.lab = 1.5, x.leg = 1, col = "black", angle = 45, cex.names = 1.25, density = c(0,20), legend = TRUE) Error in dn.call[[1]] :

Dear list, How can I add the mean and standard deviation to each of the boxplots using the example provided in the boxplot function? boxplot(len ~ dose, data = ToothGrowth, boxwex = 0.25, at = 1:3 - 0.2, subset = supp == "VC", col = "yellow", main = "Guinea Pigs' Tooth Growth", xlab = "Vitamin C dose mg",

Hi, sorry for bothering your guys. I will trying to make some nice graph using boxplot. when I check the help file of boxplot, there is a sample code as: boxplot(len ~ dose, data = ToothGrowth, add = TRUE, boxwex = 0.25, at = 1:3 + 0.2, subset = supp == "OJ", col = "orange") legend(2, 9, c("Ascorbic acid", "Orange juice"),

hi List and Manuel, I have encounter the following problem with the function "lineplot.CI".? I'm running R 2.10.1, sciplot 1.0-7 on Win XP.? It seems like it's a scoping issue, but I couldn't figure it out. Thanks! ...Tao > lineplot.CI(x.factor = dose, response = len, data = ToothGrowth)??? ## fine > lineplot.CI(x.factor = dose, response = len, data = ToothGrowth,

From:r-devel-admin@stat.math.ethz.ch on 02/22/2003 12:00 PM CET Sent by: r-devel-admin@stat.math.ethz.ch Please respond to r-devel@stat.math.ethz.ch To: r-devel@stat.math.ethz.ch cc: (bcc: Asheka Rahman/arahma1/LSU) Subject: R-devel digest, Vol 1 #101 - 10 msgs Send R-devel mailing list submissions to r-devel@stat.math.ethz.ch To subscribe or unsubscribe via the World Wide Web,

I present you with a function that solves a problem that has bugged me for many years. I think the problem may be general enough to at least consider adding this function, or a revamped version of it, to the 'stats' package, with the other multinomial functions reside. I'm using R to export data to text files, which are input data for an external model written in C++. Parts of the

Under help(matrix) it is written: 'is.matrix' tests if its argument is a (strict) matrix. It is generic: you can write methods to handle specific classes of objects, see InternalMethods. Further down, under "Details", the meaning of "strict" is explained more explicitly: 'is.matrix' returns 'TRUE' if 'x' is a matrix

Full_Name: Arni Magnusson Version: 1.6.2 OS: Windows XP Submission from: (NULL) (210.48.49.68) Hi there. I'm experiencing unexpected behaviour from as.POSIXct: > as.POSIXct("1969-12-24") [1] "1969-12-23 23:00:00 New Zealand Standard Time" > as.POSIXlt("1969-12-24") [1] "1969-12-24" > as.POSIXlt("1969-12-24")+1 [1] "1969-12-23

Dear r-devel, Among the R functions I have written and later shared with colleagues, there are five that I hope will become a part of the R base package. The tasks are neither specific nor marginal, so rather than creating one more 'misc' package, I would be happy if the R Development Core Team would adopt these functions and hammer them into shape. The functions are available from

Hello, When using bargraph.CI in package sciplot can the bars for each group be different colors? How do I select the color for each group? When I use this instead of the default (SD vs SE): bargraph.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun= function(x) c(mean(x)-sd(x), mean(x) + sd(x)) ) Am I getting 95% CI bars? Thank you kindly, Michael Just

Hi folks Run; > ToothGrowth > attach(ToothGrowth) > toothgrowth=lm(len~dose) adding abline: > abline(toothgrowth) I got it done adding single abline. How to add more ablines on the same diagram? I found following thread, applying "mapply" command; Plotting multiple ablines http://www.mail-archive.com/r-help@r-project.org/msg51543.html mapply(abline, (converge$kY +

Dear Martin, Thank you for the well-reasoned response. I realized I was rather late to make this suggestion for 4.0.0, changing a somewhat low-level function that can indeed affect packages. I was just reviewing some R user scripts that were using type.convert(), mainly on data frames. In all cases, people were passing as.is=TRUE, so I was reminded that I would not be the only user who would

Dear all How does one resolve the variance between the actual availability of data sets in the default implementation with those mentioned in the documentation? I am unable to attach some of the datasets, even though help() is available for the same datasets. For example, "ToothGrowth" is a dataset that is widely mentioned in the documentation as being part of the base install.