similar to: boxplot grouped by two variables.

> On Tue, 21 Nov 2000, Vadik Kutsyy wrote: > > > Is there a quick way to make boxplots groups by two > variables? By that > > I mean, that if x axes have values ("A","B","C"), than at > each value > > there would be a few boxplots each for a value of second > variable (say > > ("1","2","3")). >

Dear Ashim, Try spreadLevelPlot(breaks ~ interaction(tension, wool), data=warpbreaks) . I hope this helps, John ----------------------------- John Fox, Professor Emeritus McMaster University Hamilton, Ontario, Canada Web: socialsciences.mcmaster.ca/jfox/ > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ashim > Kapoor > Sent:

Dear Sir, Many thanks for your reply. I have a query. I have a whole set of distributions which should be made normal / homoscedastic. Take for instance the warpbreaks data set. We have the following boxplots for the warpbreaks dataset: a. boxplot(breaks ~ wool) b. boxplot(breaks ~ tension) c. boxplot(breaks ~ interaction(wool,tension)) d. boxplot(breaks ~ wool @ each level of tension) e.

Dear Ashim, I?ll address your questions briefly but they?re really not appropriate for this list, which is for questions about using R, not general statistical questions. (1) The relevant distribution is within cells of the wool x tension cross-classification because it?s the deviations from the cell means that are supposed to be normally distributed with equal variance. In the warpbreaks data

Dear All, I want a transformation which will make the spread of the response at all combinations of 2 factors the same. See for example : boxplot(breaks ~ tension * wool, warpbreaks) The closest I can do is : spreadLevelPlot(breaks ~tension , warpbreaks) spreadLevelPlot(breaks ~ wool , warpbreaks) I want to do : spreadLevelPlot(breaks ~tension * wool, warpbreaks) But I get : >

Dear All, we need to do : library(car) for the spreadLevelPlot function I forgot to say that. Apologies, Ashim On Sun, Jan 7, 2018 at 10:37 AM, Ashim Kapoor <ashimkapoor at gmail.com> wrote: > Dear All, > > I want a transformation which will make the spread of the response at all > combinations > of 2 factors the same. > > See for example : > >

Hello to all users and wizards. I am regulary using 'boxplot' function or its analogue - 'bwplot' from the 'lattice' library. But they are, as far as I understand, totally flawed in functionality: they miss ability to select what they would draw 'in the middle' - median, mean. What the box means - standard error, 90% or something else. What the whiskers mean -

Hi, I am trying to put some graphics I have generated from R on a webpage using dev2bitmap to create a bitmap, .BMP, file. When I look at my notes from 2 or 3 months ago I was successfully able to put the ACTIVE device plot result into a bitmap file using a command such as: dev2bitmap("InsectSpray.BMP") Job done! Now when I try the same command, I get the following: >

I am attempting to construct a conditioning plot with a pair of boxplots within each panel. The resulting plot has panels misplaced. One misplaced panel is on top of the subcomponent describing ranges of the conditioning variable. This is what I did: X <- cut(rnorm(200), 2) Y <- runif(200) Z <- rnorm(200) given.Z <- co.intervals(Z, number = 4, overlap = .25) coplot(Y ~ X | Z,

I have a specific question and a general question. Specific Question: I want to do an analysis on a data frame by 2 or more class variables (i.e., use 2 or more columns in a dataframe to do statistical classing). Coming from SAS, I'm used to being able to take a data set and have the output of the analysis in a dataset for further manipulation. I have a data set with vote totals, with one

Dear all, model.frame behaves in a way I don't expect when both its formula and subset argument are passed through a function call. This works as expected: model.frame(~wool, warpbreaks, breaks < 15) #> wool #> 14 A #> 23 A #> 29 B #> 50 B fun1 <- function(y) model.frame(~wool, warpbreaks, y) fun1(with(warpbreaks, breaks < 15)) #> wool #> 14

Dear list members, I have been using R to create a heatmap where my data has continous variables from 0 to 100. When I create the heatmap, although the branches are correct, they do not order themselves so that the row with the most zeros is at one end and the row with the most 100s is at the other, which is what I would like them to do, so as to create a colour gradient down the graphic. I have

Hello, I would like to do a large number of e.g. 1000 paired ttest using the by-function. But instead of using only the data within the 1000 groups, R caclulates 1000 times the ttest for the full data set(The same happens with Wilcoxon test). However, the by-function works fine with the lme function. Did I just miss something or is it really not working? If not, is there any other possibility to

I'm manipulating a large dataset and need to eliminate some observations based on specific identifiers. This isn't a problem in and of itself (using which.. or subset..) but an imprint of the deleted observations seem to remain, even though they have 0 observations. This is causing me problems later on. I'll use the dataset warpbreaks to illustrate, I apologize if this isn't in

Dear all, When I use aggregate function as: attach(warpbreaks) aggregate(warpbreaks[, 1], list(wool = wool, tension = tension), sum) The results are right but I get a warning message: "number of items to replace is not a multiple of replacement length." BTW: I use R version 2.4.1 in Ubuntu 7.04. Your kind solutions will be great appreciated. Best wishes Yours, sincerely, Xingwang

Hi! I'm failing to understand the value of the intercept value in a multiple linear regression with categorical values. Taking the "warpbreaks" data set as an example, when I do: > lm(breaks ~ wool, data=warpbreaks) Call: lm(formula = breaks ~ wool, data = warpbreaks) Coefficients: (Intercept) woolB 31.037 -5.778 I'm able to understand that the value of

hi list, i have to ask you again, having tried and searched for several days... i want to do a TukeyHSD after an Anova, and want to get the adjusted p-values after the Tukey Correction. i found the p.adjust function, but it can only correct for "holm", "hochberg", bonferroni", but not "Tukey". Is it not possbile to get adjusted p-values after

Hi, I'd like to select one row in a data frame per subset which is maximal for a particular value. I'm pretty close to the solution in the sense that I can easily select the maximal values per subset using "aggregate", but I can't really figure out how to select the rows in the original data frame that are associated with these maximal values. library(stats) # this

Hi all, I'm demonstrating a block entry regression using R for my regression class. For each block, I get the R**2 and the associated F. I do this with separate regressions adding the next block in and then get the results by writing separate summary() statements for each regression. Is there a more convenient way to do this and also to get the change in R**2 and associated F test for

I am attempting to produce crosstabulations between two variables for subgroups defined by a third factor variable. I'm using by() and CrossTable() in package gmodels. I get the printing of the tables first and then a printing of each level of the INDICES. For example: library(gmodels) by(warpbreaks, warpbreaks$tension, function(x){CrossTable(x$wool, x$breaks > 30,