similar to: using dimnames as main when plotting with lapply

Dear R developers, I find the which.min function very handy. However, maybe you could consider a tiny addition to the example data in the the help text. By adding another zero to the vector x... x <- c(1:4, 0, 0:5,11) ...the example would more explicitly show that which.min actually 'giv[es] the index of the /first/ minimum or maximum respectively of x'. This will also more

Dear R developers, In the help text for ave() you can read in Arguments section: "... Grouping variables, typically factors, all of the same length as x" ...and in Value: "If ... is g1,g2,". I think a description in the Arguments section of how two or more grouping variables is specified together with an Example would be very helpful. In absence of a help text I just

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 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

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 R-Helpers, I was calling the TukeyHSD function and not getting confidence intervals or p-values. It turns out this was caused by missing data and the fact that I had previously turned on R Commander (Rcmdr). John Fox knew that Rcmdr sets na.action to na.exclude, which causes the problem. If you have this problem, you can either exit Rcmdr before calling TukeyHSD or you can set na.action to

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 : > >

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 All, I am currently doing the following to compute summary statistics of aggregated data: a = aggregate(warpbreaks$breaks, warpbreaks[,-1], mean) b = aggregate(warpbreaks$breaks, warpbreaks[,-1], sum) c = aggregate(warpbreaks$breaks, warpbreaks[,-1], length) ans = cbind(a, b[,3], c[,3]) This seems unnecessarily complex to me so I tried > aggregate(warpbreaks$breaks, warpbreaks[,-1],

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 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,

Dear all, I wish to create a POSIXct variable from date and time variables using the ymd_hm function in package lubridate. In some cases data for time is missing, which causes a problem for ymd_hm. I wish to find a smooth way to handle this. # Some example data: x <- data.frame(date = c("2011-09-22", "2011-07-28"), time = c("15:00", NA)) x # paste date and

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'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, Please, is there any way of controlling factors in row/columns when using ftable/xtabs? As far as I can see, the last cross-clasifing variable in the formula will appear in columns. The previous ones, in rows. For instance, is it possible to make tension and replicate appear in columns? ftable(xtabs(breaks ~ wool + tension + replicate, data = warpbreaks)) After some years using SAS

Dear R-Core, after switching to 2.3.0, all my trusted do.call constructs that worked in 2.2 and earlier fail. I noted that changes were introduced to do.call, but I could not find out how these relate to my problem. The following example works in 2.2 and earlier, but fails because rownames are partially NA. I can correct this by manually adding row names, but it's a bit of work to check this

Dear List: I found an error when I called the 'gee' function. I cannot solve and explain it. There are no errors when I used the 'geeglm' function. Both functions fit the gee model. The project supervisor recommends me to use the 'gee' function. But I cannot explain to him why this error happens. Would you help me solve this problem? I appreciate your help. In

Dear R-developeRs, Attached follows a patch against svn 34959 that adds the printing of p-values to the TukeyHSD.aov function in stats package. I also updated the corresponding documentation file and added a 'see also' reference to the simint function of the multcomp package. As it was already brought up in a previous thread [1] in R-help, one can obtain the adjusted