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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 A...

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. boxplot(breaks ~ tension @ each level of wool) Now should we not be making a-e normal and homoscedastic? Should we not make a giant collection of boxplots from a-e and...

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 there are only 9 cases per cell. If you examine all of these deviations simultaneously, that?s equivalent to examining the residual...

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 : > spreadLevelPlot(breaks ~tension * wool , warpbreaks) Error in spreadLevelPlot.f...

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

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 intercept is the mean value of breaks when wool equals "A", and that adding up the "woolB" coeff...

...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 <- tapply( breaks, list( wool, tension, variant ), mean ) tp # and hardly: ftable( xtabs( breaks ~ wool + tension + variant ) ) # It would be nice to flatten the tabulations, so I tried: ftable( tb ) tp <- as.table( tp ) fta...

...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 <- tapply( breaks, list( wool, tension, variant ), mean ) tp # and hardly: ftable( xtabs( breaks ~ wool + tension + variant ) ) # It would be nice to flatten the tabulations, so I tried: ftable( tb ) tp <- as.table( tp ) fta...

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 A #> 23 A #> 29 B #> 50 B but this doesn't: fun2 <-...

...1 equals to SS_g1_dat2: The formula for computing SS is:sum(sample size of level(i)*(mean of level(i)-TotalMean)^2),with i refers to each level in SS_g1_dat1 and SS_g1_dat2. Every element of the formula is constant,so SS is constant. Using the dataset "warpbreaks" from R: anova(lm(breaks~wool)) Analysis of Variance Table Response: breaks Df Sum Sq Mean Sq F value Pr(>F) wool 1 450.7 450.67 2.6684 0.1084 Residuals 52 8782.1 168.89 anova(lm(breaks~wool+tension)) Analysis of Variance Table Response: breaks Df Sum Sq Mean Sq F value Pr(>F) wool...

...nction 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 avoid loops? Thanks Daniel Here is the R help example for "by" require(stats) attach(warpbreaks) by(warpbreaks, tension, function(x) lm(breaks ~ wool, data = x)) *->works great by(warpbreaks,tension,function(x)t.test(breaks ~ wool,data=warpbreaks,paired = TRUE)) *Same output for each level of tension: tension: L Paired t-test data: breaks by wool t = 1.9956, df = 26, p-value = 0.05656 alternative hypothesis: true difference in means is n...

...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 : > > 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 : > > > spreadLevelPlot(brea...

...he 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 returns the list with maximal values for breaks per wool/tension subset maxValues = aggregate(warpbreaks$breaks, list(wool = wool, tension = tension), max) # now i'd like the subset of the rows in warpbreaks that are associated with these maximal values Thank you in advance! Tim. [[alternative HTML version deleted]]

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 Y...

...ogrammer. Ultimately want to have code that I can turn into a function. Will then use the output produced to make tables using odfWeave and Sweave/knitr. Thanks, Paul breaks <- as.data.frame(lapply(warpbreaks, function(x) rep(x, warpbreaks$breaks))) Freq <- with(breaks, addmargins(table(wool, tension), 2)) Prop <- round( prop.table(Freq, 2) * 100, 2 ) Freq <- addmargins(Freq, 1) Prop <- addmargins(Prop, 1) require(odfWeave) class(Freq) <- "character" class(Prop) <- "character" FreqProp <- matrixPaste(Freq, "(", Prop, "%)", sep...

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

...em 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 the best format ##Summary of warpbreaks suggests three tension levels (H, M, L) > summary(warpbreaks) breaks wool tension Min. :10.00 A:27 L:18 1st Qu.:18.25 B:27 M:18 Median :26.00 H:18 Mean :28.15 3rd Qu.:34.00 Max. :70.00 ## Subset the dataset and keep only those observations with "L" > wb.subset <- warpbreaks[...

...ing the pairwise comparisons of interaction effects in a multi-way factorial ANOVA, because I find comparisons of interactions between all different variables different to interpret. For example (supposing a p<0.10 cutoff just to be able to use this example): > summary(fm1 <- aov(breaks ~ wool*tension, data = warpbreaks)) Df Sum Sq Mean Sq F value Pr(>F) wool 1 450.7 450.7 3.7653 0.0582130 . tension 2 2034.3 1017.1 8.4980 0.0006926 *** wool:tension 2 1002.8 501.4 4.1891 0.0210442 * Residuals 48 5745.1 119.7 --- Signif. codes: 0 '***...

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

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