similar to: Function Arguments

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

Hi, I noticed that by() returns an object of class 'by', regardless of what its argument 'simplify' is. ?by says that it always returns a list if simplify=FALSE, yet by.data.frame shows: ---<--------------------cut here---------------start------------------->--- function (data, INDICES, FUN, ..., simplify = TRUE) { if (!is.list(INDICES)) { IND <-

Suppose I have the following function myFun <- function(formula, data){ f <- formula(formula) dat <- model.frame(f, data) dat } Applying it with this sample data yields a new dataframe: qqq <- data.frame(grade = c(3, NA, 3,4,5,5,4,3), score = rnorm(8), idVar = c(1:8)) dat <- myFun(score ~ grade, qqq) However, what I would like is for the resulting dataframe (dat) to include

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:

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

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, 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 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

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

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

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

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 : > >

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

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! 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 everyone, I am fairly new to R, and I am aware that others have had this problem before, but I have failed to solve the problem from previous replies I found in the archives. As this is such a standard procedure in psychological science, there must be an elegant solution to this...I think. I would much appreciate a solution that even I could understand... ;-) Now, I want to calculate a

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

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,

Hi, Coming to R from SAS... I have a data.frame A with 2 long factors "x" and "y". I want to get a count of the number of rows with each level of "x" and "y" jointly. 'table' seemed like it would work, but as I have many levels, the matrix output is pretty useless to me (and I don't care about zero values). How can I get output that looks