similar to: getting elements out of list automatically

>>>>> Karolis Koncevi?ius >>>>> on Mon, 9 Dec 2019 23:43:36 +0200 writes: > So I tried adding Infinity support for all cases. > And it is (as could be expected) more complicated than I thought. "Of course !" Thank you, Karolis, in any case! > It is easy to add Inf support for the test. The problems start with conf.int=TRUE.

I would like to preform a t.test to each of the measured variables (sand.silt etc.) with a mean and sd for each of the treatments (up or down), and out put this as a table.... I am having a hard time starting- maybe it is to close to lunch. Any suggestions would be greatly appreciated. Stephen Sefick x <- (structure(list(sample. = structure(c(1L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 2L, 3L,

Hi all, I have a distribution, and take a sample of it. Then I compare that sample with the mean of the population like here in "Wilcoxon signed rank test with continuity correction": > wilcox.test(Sample,mu=mean(All), alt="two.sided") Wilcoxon signed rank test with continuity correction data: AlphaNoteOnsetDists V = 63855, p-value = 0.0002093 alternative hypothesis:

Dear everyone, I am trying to do a Wilcoxon one-sided test on my gene expression data. These are the data i have in R: data.matrix (matrix, numeric) containing all gene expression data (42 rows=genes, 42 columns=tumors), no column header or row names data.cl (vector, numeric) consisting of 42 0's and 1's to indicate class 0 or class 1 for each column in data.matrix I want to do a

Dear R-ers, I am currently running some Wilcoxon tests in R-64. How do I find the degrees of freedom in the output I am receiving? > wilcox.test(good$TRUE, good$x4a, paired=FALSE) Wilcoxon rank sum test with continuity correction data: good$TRUE and good$x4a W = 2455, p-value < 2.2e-16 alternative hypothesis: true location shift is not equal to 0 Thank you, Stephen.

Hi, I have a data set: > Dataset X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 1 user1 m 22 19 28 24 12 18 9 7 4 5 4 7 5 7 9 2 user2 f 25 19 23 18 18 15 6 8 6 6 7 10 7 7 7 3 user3 f 28 21 24 18 15 12 10 6 7 9 5 10 5 9 5 4 user4 f 26 19 26 21 12 18 6 6 5 1 3 8 6 5 6 5 user5 m 21 22 26 18 9 6 4 6 1

Hello everyone, I'm having some trouble interpreting the results of a Wilcoxon (Mann-Whitney U) test. Hope you can help. This is the R script that I am running: a <- c(1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 5, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 3, 1, 1) b <- c(1, 2, 1, 1, 2, 3, 2, 2, 1, 2, 1, 1, 1, 2) wilcox.test(a, b, alternative="t", mu=0, exact=FALSE, paired=FALSE) #1st

Thank you for a fast response. Nice to see this mailing list being so alive. Regarding Inf issue: I agree with your assessment that Inf should not be removed. The code gave me an impression that Inf values were intentionally removed (since is.finite() was used everywhere, except for paired case). I will try to adjust my patch according to your feedback. One more thing: it seems like you

This is my first post to the mailing list and I guess it's a pretty stupid question but I can't figure it out. I hope this is the right forum for these kind of questions. Before I started using R I was using STATA to run a Wilcoxon signed-rank test on two variables. See data below:

I have a question concerning the Wilcoxon signed-rank test, and specifically, which R subroutine I should use for my particular dataset. There are three different commands in R (that I'm aware of) that calculate the Wilcoxon signed-rank test; wilcox.test, wilcox.exact, and wilcoxsign_test. When I run the three commands on the same dataset, I get different p-values. I'm hoping that

Dear, I'm using R software to evaluate Wilcoxon Rank Sum Test and I' getting one Warning message as this: > C1dea_com [1] 1.000 0.345 0.200 0.208 0.508 0.480 0.545 0.563 0.451 0.683 0.380 0.913 1.000 0.506 > C1dea_sem [1] 1.000 0.665 0.284 0.394 0.509 0.721 0.545 0.898 0.744 0.683 0.382 0.913 1.000 0.970 > wilcox.test(C1dea_sem,C1dea_com, paired = TRUE, alternative =

Por si alguno pudiera ayudarme. Al realizar el t.test para una muestra, junto con el valor de t y el p-valor, la función proporciona la estimación de la media y su INTERVALO DE CONFIANZA. Desde el punto de vista de la estadística de rangos esto se puede hacer mediante: > iris$MEDIANA <- with(iris, 2.95) > median(iris$Sepal.Width - iris$MEDIANA, na.rm=TRUE) # median difference [1]

I have run a wilcox test on a dataframe and I have returned to me data which I am wanting to sort and/or pick the max. I have tried the following code to pick the max: v<-wilcox.test.run w<-max(v,na.rm=FALSE)   I have also tried w<-pmax(p-value,na.rm=FALSE) for the second line   and this returns the following error:invalid 'type'  (list) of argument This error is understandable

Dear List, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately

This was filed as a bug report on the Debian r-base package. It is more properly a bug report on the ctest package in R. The default method for wilcox.test manipulates x and y without checking the class or data.class of these objects. Possible solutions are - create wilcox.test.factor (if appropriate) - check the class and/or data.class of x and y in wilcox.test.default and produce error

When I use wilcox.test, I get vastly different p-values than the problems from Statistics textbooks. For example: The following problem comes from "Applied Statistics and Probability for Engineers", 2nd Edition, by D. C. Montgomery. Page736, problem 14.7. The problem is to compare the sample data with a population median of 8.5. The book answer is p = 0.25, wilcox.test answer is p =

>>>>> Martin Maechler >>>>> on Thu, 12 Dec 2019 17:20:47 +0100 writes: >>>>> Karolis Koncevi?ius >>>>> on Mon, 9 Dec 2019 23:43:36 +0200 writes: >> So I tried adding Infinity support for all cases. And it >> is (as could be expected) more complicated than I >> thought. > "Of course

Buena tarde, Estoy llevando a cabo un trabajo y no encuentro la forma de llenar una matriz con el p_value y un estadístico calculado. Un poco mas detallado, tengo muestras aleatorias, calculo por ejemplo la prueba t, wilcoxon, etc y requiero llevar por una parte los p_value de k muestras a una matriz creada anteriormente y por otro lado el valor del estadístico de las mismas k muestras a otra

Hi, I am using wilcox.test function to test the difference between the means of two samples. The data points are paired, so I am using a paired test. There is one strange case. Sample A has a higher mean than a sample B. However, wilcox.test function says that sample B has a significantly higher "mean rank" than sample A. How is it possible? Here is the code (data file is attached):

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