similar to: Bug in ftable?? (Was: Two-way tables of data, etc)

Dear all, I haven't received any feedback so far on my proposal to make "justify" argument available in stats:::format.ftable Is this list the appropriate place for this kind of proposal? I hope this follow-up to my message won't be taken as rude. Of course it's not meant to be, but I'm not used to the R mailing lists... Thank you in advance for your comments, Best,

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

Hello, I teach statistics and use R Commander for teaching. I have 2 students out of 169 that can't get scatterplots or plot to work. I have had them update packages, restart R/R Commander/their computers and even reinstall R/R Commander. One is using Windows 7 on a new pc and the other is a pc user (not sure the OS). They are both using R2.11.1 and R Commander 1.6-0. The data look like

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 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

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

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

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

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, 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

I would like to obtain a crosstabulation of means(var, quantiles...) i.e. I have a data frame with Var-i, Var-j, Var-k, Var-X, var-Y I like to have the mean of Var-X for each combination of Var-i,Var-j. One solution would be: by(var-i,Var-j,mean(Var-x)) but I would like it better formatted and with mean,S.Dev,n for each cell? Does anybody have some function to do this or some ideas how to go

Hello, I am performing a BACI analysis with ANOVA using the following glm: fit1<-glm(log(Cucs_m+1)~(BA*Otter)+BA+Otter+ID+Primary, data=b1) The summary(aov(fit1)) shows significance in the interaction; however, now I would like to determine what combinations of BA and Otter are significantly different (each factor has two levels). ID and PRIMARY substrates are categorical and included in

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36

Hi all: I have a quesion about ANOVA: Is SS(Sum of Square) of a specific factor constant with the number of factors changing? dat1 includes one factor g1,and g1's SS is called SS_g1_dat1. dat2 includes two factors g1,g2,and g1's SS is called SS_g1_dat2. My quesion is: Is SS_g1_dat1 equals to SS_g1_dat2? I have both "yes" and "no" reasons for the quesion,but

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