similar to: crosstab means

Hi All, I am currently doing the following to compute summary statistics of aggregated data: a = aggregate(warpbreaks$breaks, warpbreaks[,-1], mean) b = aggregate(warpbreaks$breaks, warpbreaks[,-1], sum) c = aggregate(warpbreaks$breaks, warpbreaks[,-1], length) ans = cbind(a, b[,3], c[,3]) This seems unnecessarily complex to me so I tried > aggregate(warpbreaks$breaks, warpbreaks[,-1],

Hello all,, Im looking for a simple function to produce a crosstab from a dumped sql query result. Its very hard to produce crosstabs with most databases (Access being the exception), so with the vast array of R packages, Im sure this has to have already been implemented somewhere. Examples are always good: Take a csv dump like name code user1 100 user2 100 user1 200 user2 210 user1 300 user2

Hello, I am wondering if it is possible, or what the correct way to code a three-way crosstab in R using the survey package? I have been using the following code to complete two way crosstabs, but have not seen any three-way code. Two-Way: svyby(~factor(a), ~factor(b), data, svymean) Thanks! Rachel [[alternative HTML version deleted]]

Dear list. New to R, I'm looking for a way of using crosstab to output low-dimensional (higher than 2) contingency tables (frequencies, per-cents by rows, % by columns, mean, quantiles....) I'm looking for something of the following sort dataframe: singers, categorical variates: voice category (soprano,mezzo-soprano, ...) , voice type( drammatic, spinto, lirico-spinto, lirico,

I have the following data: x <- as.factor(c(1,1,1,2,2,2,3,3,3)) y <- as.factor(c(10,10,10,20,20,20,30,30,30)) z <- c(100,100,NA,200,200,200,300,300,300) I could create the cross tab of x and y with Sum of z as its elements using the xtabs function as follows: # X Vs. Y with Sum Z xtabs(z ~ x + y) y x 10 20 30 1 200 0 0 2 0 600 0 3 0 0 900 How do I replace

Perhaps this is a common question but I haven't been able to find the answer. I have data with many factors, each taking many values. However, only relatively few combinations appear in the data, ie have nonzero counts, in other words the resulting table is sparse. Say we have 10 factors each with 10 levels. The result of table() would exceed the memory space (on a 32bit machine). Is there

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

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

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

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

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 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, 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 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 : > >