similar to: justify hard coded in format.ftable

One can use as.data.frame(as.matrix(tab)) to avoid calling as.data.frame.matrix directly (although I find I do use as.data.frame.matrix anyways sometimes even though it is generally better to call the generic.). Also note that the various as.data.frame methods do not address the examples in the SO links I posted which is why I mentioned it. On Thu, May 14, 2020 at 9:22 AM SOEIRO Thomas

Thanks for the links. I agree that such a feature would be a nice addition, and could make ftable even more useful. In the same spirit, I think it could be useful to mention the undocumented base::as.data.frame.matrix function in documentation of table and xtabs (in addition to the already mentioned base::as.data.frame.table). The conversion from ftable/table/xtabs to data.frame is a common task

>>>>> Gabor Grothendieck >>>>> on Thu, 14 May 2020 06:56:06 -0400 writes: > If you are looking at ftable could you also consider adding > a way to convert an ftable into a usable data.frame such as > the ftable2df function defined here: > https://stackoverflow.com/questions/11141406/reshaping-an-array-to-data-frame/11143126#11143126 > and there

Further to the discussion between Murray Jorgensen and Brian Ripley, it seems to me better to choose tabulations that will not come and bite you. Suppose your data are sligtly irregular, e.g. (for the sake of the argument): data( warpbreaks ) warpbreaks$variant <- rep( 1:5, len=54 ) attach( warpbreaks ) tb <- table( wool, tension, variant ) tb # in this case you would like to see: tp

Further to the discussion between Murray Jorgensen and Brian Ripley, it seems to me better to choose tabulations that will not come and bite you. Suppose your data are sligtly irregular, e.g. (for the sake of the argument): data( warpbreaks ) warpbreaks$variant <- rep( 1:5, len=54 ) attach( warpbreaks ) tb <- table( wool, tension, variant ) tb # in this case you would like to see: tp

>>>>> SOEIRO Thomas >>>>> on Wed, 13 May 2020 20:27:15 +0000 writes: > 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? Yes, it is.. Actually such a post is even a "role

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 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 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 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 list, I cannot make the latex command to output a ftable objet the way I want it. Is it posible? I found a post in the archives saying that one should use the rgroup and n.rgroup arguments to supply the row names, but so far I have been unsuccessful. This is what 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

I am quite new to R, so please bear over with me if I have problems with the R terminology. I want to (try to) use R for some analyses within vegetation ecology, using the vegan package. I have my data in a postgresql database, and I manage to get them into R as a dataframe with columns for respectively: Name of the analysed m2, Name of the species, coverage of species in the square in %. I

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