similar to: How to generate a pairwise non-parametric comparison table?

Hi! I have been working several years with R but it's my first public question. I hope I'll be clear :) . This question is related to obtaining letter-based representation of non-parametric pairwise comparisons. I have a dataframe with this structure (but with quite more rows and cols): A B C factor 1 2 2 one 2 1 2 one 2 2 3 two 2 3 2 two 1 4 2 three 9 8 1 three I have no normality,

how do I check the rank of a matrix ? say A= 1 0 0 0 1 0 then rank(A)=2 what is this function? thanks I did try help.search("rank"), but all the returned help information seem irrelevant to what I want. I would like to know how people search for help information like this. rank(base) Sample Ranks SignRank(stats) Distribution of the

Hi I get a matrix with the columns named like "X13 X22 X1 X14 ...", i.e. not successively. That has it's good reasons, but now how can I rename the columns to "X1 X2 X3 ..."? Thanks a lot, David

Hi I am trying to replicate a vector in n rows for comparison purposes with another matrix. foo <- c(1,2,3) bar <- rbind(foo,foo) # does the trick for 2 rows bar <- rbind(rep(foo,2)) # does something else How do I generate a matrix with all rows=foo without writing 'foo' n times as arg? Thanks, David

Hi all I am very astonished that R generates a "syntax error" when I want to split up a line with a backslash, which usually works in any shell script. R itself generates the "+" symbols at the beginning of following lines in a splitted line so I've tried with them as well, but also without success. Searching in the h-help archive did not reveal any answer either.

Can anyone suggest a good non-parametric test, and an R implementation of the test, that allows for two or more factors? Wilcoxon signed rank allows for only one. Thanks, John John Sorkin M.D., Ph.D. Chief, Biostatistics and Informatics Baltimore VA Medical Center GRECC and University of Maryland School of Medicine Claude Pepper OAIC University of Maryland School of Medicine Division of

Hi cor(x,apply(x,1,sum)) gives me the correlations of each column with the sums of each row (correct me if I'm wrong, please). What I need are the correlations of each column with the sums of each row except the entry in the given column. It seems that for any one column i I get it by doing: cor(x[,i],apply(x[,-i],1,sum)) But I struggle to get it for all the columns. I was trying things

Hi, A client has a consumption measure on each of four products. The sample size is 75. The consumption distributions are highly skewed for each product. He would like a pairwise comparison test of the products, much like Tukey's HSD but using medians rather than means. Is there such a median comparison test in R? Thanks, Walt ________________________ Walter R. Paczkowski, Ph.D.

Hello, I think I am right in saying that a 2 sample wilcox.test is equal to a 2 sample kruskal.test and a 2 sample t.test is equal to a 2 sample anova. This is also stated in the ?kruskal.test man page: The Wilcoxon rank sum test (wilcox.test) as the special case for two samples; lm together with anova for performing one-way location analysis under normality assumptions; with Student's t

I am writing a little abstraction for a series of tests. For example, I am running an anova and kruskal.test on a one-factor model. That isn't a particular problem, I have an interface like: my.function <- function(model,data) { print(deparse(substitute(data))) a <- anova(lm(formula,data)) print(a) if(a$"Pr(>F)"[1] < 0.05) { pairwise.t.test(???) }

Well fellow R users, I throw myself on your mercy. Help me, the unworthy, satisfy my employer, the ungrateful. My feeble ramblings follow... I've searched R-Help, the R Website and done a GOOGLE without success for a one way ANOVA procedure to analyse data that are both non-normal in nature and which exhibit unequal variances and unequal sample sizes across the 4 treatment levels. My

Dear R users, I would like to know if there is a way in R to execute a post-hoc test (factor levels comparison, like Tukey for ANOVA) of a non-parametric analysis of variance with kruskal.test() function. I am comparing three different groups. The preliminary analysis using the kruskal-wallis-test show significance, but I still don''t know the relationship and the significance level

Recently, I ran a series of Kruskal-Wallace tests [kruskal.test()] using by() to group by site Output is a list: >Herb.KW Herb.df$ID: 10-1 Kruskal-Wallis rank sum test data: Indicator_Rating by Year Kruskal-Wallis chi-squared = 15.24, df = 7, p-value = 0.03302 ----------------------------------------------------------------------------------------------------- Herb.df$ID: 18-1

This is only partly an R question, but this seems a good place to ask. Like many, if not most, statistics programs, R has no post hoc tests for Kruskal Wallis or Friedman, and as far as I can make out they are not available in any of the R packages. I would be grateful if someone could point me in the right direction if I am wrong. As I said this is not uncommon. Parametric Anova procedures

Dear all, I run a kruskal wallis test and found significant results. Then, I conducted all pairwise comparisons and found no significant results. Could anyone please give me a hint as to why this happens or redirect me towards a specific web page where I can find more info? (I used alpha=5% and made no bonferroni or other correction for the pairwise comparisons) Thank you   Dr. Iasonas

Hi there, I am doing multiple comparisons for data that is not normally distributed. For this purpose I tried both functions kruskal{agricolae} and kruskalmc{pgirmess}. It confuses me that these functions do not yield the same results although they are doing the same thing, don't they? Can anyone tell my why this happens and which function I can trust? kruskalmc() tells me that there are no

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Dear R users, I have a dataset with two variables: $esan - a grouping factor with 8 levels and $reus. I'd like to do wilcox.test on this dataset as sugested Weiwei here: https://stat.ethz.ch/pipermail/r-help/2007-July/136627.html. I tried to adapt his recommendation but no succes. Can anyone help me? Regards, Iurie Malai, Senior Lecturer Department of Psychology Faculty of Psychology and

Hi all, The Kruskal-Wallis test is a generalization of the two-sample Mann-Whitney test to *k* samples. That being the case, the Kruskal-Wallis test with *k*=2 should give an identical p-value to the Mann-Whitney test, should it not? x1<-c(1:5) x2<-c(6,8,9,11) a<-wilcox.test(x1,x2,paired=FALSE) b<-kruskal.test(list(x1,x2),paired=FALSE) a$p.value [1] 0.01587302 b$p.value [1]

Hi I have a matrix of type "character" (strangly) and am trying to apply the function "aggregate" to it, but unfortunately without success. It seems to me that aggregate would work, if I only could get rid of the quotation marks around each item. So for a very simple example which looks as my actual data look (of course here I put the quotation marks on purpose, but I have