similar to: pairwise.wilcox.test for paired samples

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Dear Friends: I like so much to work with R program. Congratulations for your work. I need R for work with multivariate data. My question is: With the pairs(X) command my output is a pairwise scatterplot symmetric matrix. Like: | X1 |X1 vs X2|X1 vs X3|X1 vs X4| |X2 vs X1| X2 |X2 vs X3|X2 vs X4| |X3 vs X1|X3 vs X2| X3 |X3 vs X4| |X4 vs X1|X4 vs X2|X4 vs X3| X4 | It is

Dear Developers, I'm updating my R installation via svn (currently: Revision 44626). For some time now (sorry don't know the exact revision number) make for R devel yields the following error make[4]: *** Keine Regel vorhanden, um das Target ?../../../src/include/R_ext/GraphicsBase.h?, ben?tigt von ?g_fontdb.o?, zu erstellen. Schluss. (translation: no rule to make target ...

Hi there, I am trying to do multiple pairwise Wilcoxon signed rank tests in a manner similar to: a <- c(runif(1000, min=1,max=50), rnorm(1000, 50), rnorm(1000, 49.9, 0.5), rgeom(1000, 0.5)) b <- c(rep("group_a", 1000), rep("group_b", 1000), rep("group_c", 1000), rep("group_d", 1000)) pairwise.wilcox.test(a, b, alternative="two.sided",

Dear developers, we develop our packages via r-forge and svn. Under R version 2.8.0 Under development (unstable) (2008-06-02 r45826) I now observe the following warning for our package "distrEx" * checking for executable files ... WARNING Found the following executable file(s): src/.svn/text-base/distrEx.dll.svn-base Source packages should not contain undeclared executable files.

hI I have to calculate V statistic for each row of a large dataframe (28000). I can not use multtest package for paired wilcox test. I have been using for loop which are. Is there a way to speed the computation with another method like using apply or tapply? My data set looks like this: 11573_MB 11911_MB 11966_MB 12091_MB 12168_MB 12420_MB................ cg00000292

Dear all, I'm a newbie to R and I would really apperciate any help with the following: I have two lists, l1 and l2: l1: $"A*0101" [1] 0.076 0.109 0.155 0.077 0.09 0 0 0.073 [9] 0.33 0.0034 0.0053 $"A*0247" [1] 0 0 0.5 .004 0 0 0 $"A*0248" [1] 0 0 0.3 0 0.06 .... l2: $"A*1101" [1] 0.17 0.24 0.097 0.075 0.067 $"A*0247" numeric(0)

Hello, I would like to use mapply to avoid using a loop but for some reason, I can't seem to get it to work. I've included copies of my code below. The first set of code uses a loop (and it works fine), and the second set of code attempts to use mapply but I get a "subscript out of bounds" error. Any guidance would be greatly appreciated. Xj, Yj, and Wj are also lists, and s2,

Dear readers of the list, I have a problem a comparison of two data from a vector. The comparison yields FALSE but should be TRUE. I have checked for mode(), length() and attributes(). See the following code (R2.10.0): ----------------------------------------------- # data vector of 66 double data X =

Below is a version of [.data.frame that is faster for subscripting rows of large data frames; it avoids calling duplicated(rows) if there is no need to check for duplicate row names, when: i is logical attr(x, "dup.row.names") is not NULL (S+ compatibility) i is numeric and negative i is strictly increasing "[.data.frame" <- function (x, i, j,

Conversion of a data frame to a matrix using as.matrix() when a column of the data frame is POSIXt and all other columns are numeric has changed in R 1.9.0 from R 1.8.1. The new behavior issues a warning message and converts to a character matrix. In R 1.8.1, such an object was converted to a numeric matrix. Here is an example. #### R 1.9.0 #### > foo <- data.frame(

Converting a data.frame with a nested data.frame to a matrix fails: x <- structure(list(a = data.frame(letters)), class = "data.frame", row.names = .set_row_names(26)) as.matrix(x) #> Error in ncol(xj) : object 'xj' not found The offending code is here, in the definition of as.matrix.data.frame (source/base/all.R): for (j in pseq) {

I got the error. I haven't been able to get a stand along case so that I can show it here. But could somebody give some clue on what could cause this error? Since I never defined xj[i], I don't understand where this error come from. Error in xj[i] : invalid subscript type 'list'

I would like to suggest a modification to the make.names() function. The current implementation has two problems: 1. It doesn't check if a name matches an R keyword (like "function"). 2. The uniqueness algorithm is not invariant to concatenation. In other words, make.names(c("a","a","a"),unique=T) !=

I would like to suggest a modification to the make.names() function. The current implementation has two problems: 1. It doesn't check if a name matches an R keyword (like "function"). 2. The uniqueness algorithm is not invariant to concatenation. In other words, make.names(c("a","a","a"),unique=T) !=

does anybody know why there are the two warnings in the example above? Regards Knut > day_4 [1] 540 1 1 1 1 1 1 300 720 480 > day_1 [1] 438 343 1 475 1 562 500 435 1045 890 > is.vector (day_1) [1] TRUE > is.vector (day_4) [1] TRUE > wilcox.test(day_4 ,day_1,paired=TRUE,alternative="two.sided",exact=TRUE,conf.int=TRUE) Wilcoxon

The point estimates produced by wilcox.test are perverse (not wrong, just brain damaged). The Hodges-Lehmann estimator that goes with the signed rank test is the median of the Walsh averages. The Hodges-Lehmann estimator that goes with the rank sum test is the median of the pairwise differences. wilcox.test agrees except that it uses the following very peculiar definition of "sample

>>>>> Jan van der Laan <rhelp at eoos.dds.nl> >>>>> on Fri, 6 Oct 2017 12:13:39 +0200 writes: > It is actually model.matrix that crashes, not glm. Same > crash occurs with e.g. lm. > model.matrix(dob_mon ~ dob_day*dob_mon, data = tab) > also crashes R. Yes, segmentation fault. It only happens when these are *logical*

Dear R-ians, I'm looking for a computational simplified formula to calculate a measure for heterogeneity (let's say H ): H = sqrt [ (Si (Sj (Xi - Xj)?? ) ) /n ] where: sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Xi = element of X with index i Xj = element of X with index j I can simplify the formula to: H = sqrt [ ( 2 * n * Si (Xi) - 2 Si (Sj

Your second issue seems like a more or less unavoidable floating-point computation issue. The paired test operates by computing differences between corresponding values of x and y. It's not impossible to try to detect "almost-ties" (by testing for differences less than, say, sqrt(.Machine$double.eps)), but it's delicate and somewhat subjective/problem-dependent. Example: