search for: tension

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 Behal...

I am still struggling to map a character string to an object name and vice versa in R. I thought the as.name() function might work, but observe the following behaviour ... > attach(warpbreaks) > levels(tension) [1] "L" "M" "H" > levels(as.name("tension")) NULL > objectname<-as.name("tension") > objectname tension > levels(objectname) NULL So even though it sets up a symbol, this symbol isn't recognised as an object name by functions s...

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 use the SpreadLevelPlot on th...

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 residuals from the...

...g 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 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...

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 spreadLeve...

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 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

...ilcoxon 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 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:...

...en 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 ) ftable( tp )...

...en 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 ) ftable( tp )...

Does anyone know if R has a function for splines under tension. I know there are numerous packages for spline interpolation within R i just can't find one that lets you determine the tension factor. Any help would be much appreciated! Sam -- View this message in context: http://r.789695.n4.nabble.com/Splines-under-tension-tp2173887p2173887.html Sent fro...

...es, 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 : > > 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 : > > > spreadLevelPl...

...nse 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]]

Hi, I am seeking help with an error when running the example from R Documentation for TukeyHSD. The error occurs with any example I run, from any text book or website. thank you... > plot(TukeyHSD(fm1, "tension")). Error in plot(confint(as.glht(x)), ylim = c(0.5, n.contrasts + 0.5), ...) : error in evaluating the argument 'x' in selecting a method for function 'plot': Error in UseMethod("vcov") : no applicable method for 'vcov' applied to an object of class &quot...

...oblem, you can either exit Rcmdr before calling TukeyHSD or you can set na.action to na.omit. The code below demonstrates the situation. Cheers, Bob data(warpbreaks) head(warpbreaks) # Introduce a missing value: warpbreaks$breaks[1] <- NA head(warpbreaks) # Do a model: fm1 <- aov(breaks ~ tension, data = warpbreaks) TukeyHSD(fm1, "tension", ordered = TRUE) # Setting na.exclude or starting Rcmdr will kill the confidence intervals: options(na.action = na.exclude) fm1 <- aov(breaks ~ tension, data = warpbreaks) TukeyHSD(fm1, "tension", ordered = TRUE) # Setting na.omit...

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 Ye

...t possible to build together a model and term component myself, so that glht will work for repeated measurements ANOVAs? If so, how would I do that? Best regards, Michael Zehetleitner For example (senseless, but it makes the point): --------- snip ------------- > amod <- aov(breaks ~ tension, data = warpbreaks) > amod.err <- aov(breaks ~ tension + Error(wool/tension), data = warpbreaks) > > class(amod) [1] "aov" "lm" > class(amod.err[["wool:tension"]]) [1] "aov" "lm" > > names(amod) [1] "coefficients&qu...

...-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" coefficient to the intercept value I get the mean value of breaks when wool equals "B". However, if I also consider the tension variable in the model, I'm unable to figure out the meaning of the intercept value: > lm(breaks ~ wool + tension, data=warpbreaks) Call: lm(formula = breaks ~ wool + tension, data = warpbreaks) Coefficients: (Intercept) woolB tensionM tensionH 39.278 -5.778...

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