similar to: loop? apply? I want to repeat a task through 208 itterations.

hey all, I'm just getting used to R and i'm having issues when it comes to reading my data in rows rather than columns. any good advice would be much appreciated ! here is the error: > data1 <- read.table(file.choose(),header=T) > x1 <- c(data1[1,1:5]) > shapiro.test(x1) Error in sort.int(x, na.last = na.last, decreasing = decreasing, ...) : 'x' must be atomic

I want to run Shapiro-Wilk test for each variable in my dataset, each grouped by variable groupFactor. I have these working commands: > data.n<-names(data) # put names into a vector called data.n > by(eval(parse(text=(paste("data",data.n[3],sep="$")))), data$factor, shapiro.test) #run shapiro.test but I must to change the variable number manualy. How to automate

Hello, I was wandering if it is possible to perform on a dataframe called 'all' a shapiro wilk normality test for COUNTS by variable Group ACTIVITY? Could it be done using plyer? I saw an eg that applies to an array but not to a dataframe: lapply(split(dataset1$Height,dataset1$Group),shapiro.test) Any thoughts would be much appreciated. My dataframe is in shape: dat ACTIVIT

Hi, does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can anybody tell me why the following sample doesn't give "W = 1" and "p-value = 1": R> x<-1:9/10;x [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 R> shapiro.test(qnorm(x)) Shapiro-Wilk normality test data: qnorm(x) W = 0.9925, p-value = 0.9986 I can't imagine a sample being

Hi everybody, somehow i dont get the shapiro wilk test for normality. i just can?t find what the H0 is . i tried : shapiro.test(rnorm(5000)) Shapiro-Wilk normality test data: rnorm(5000) W = 0.9997, p-value = 0.6205 If normality is the H0, the test says it?s probably not normal, doesn ?t it ? 5000 is the biggest n allowed by the test... are there any other test ? ( i know qqnorm

R Users: My question is probably more about elementary statistics than the mechanics of using R, but I've been dabbling in R (version 2.2.0) and used it recently to test some data . I have a relatively small set of observations (n = 12) of arsenic concentrations in background groundwater and wanted to test my assumption of normality. I used the Shapiro-Wilk test (by calling shapiro.test()

Hello, I have applied the Shapiro test to a matrix with 26.925 rows of data using the following F1.norm<-apply(F1.n.mat,1,shapiro.test) I would now like to view and export a table of the p and W values from the Shapiro test, but I am not sure how to approach this. I have tried the following with errors. > write.table(x=F1.norm,file="I:/R_Work/F1/Shapiro.csv",

Greetings! I want to have the coefficients that R uses in shapiro.test() for the Shapiro-Wilk test for a prticular sample size, i.e. the a[i] in W = Sum(a[i]*x[i])/(Sum(x[i] - mean(x))^2) (where the x[i] are sorted). Two questions: Q1: Is there a readymade R function from which I can extract these? Q2: I was wondering if I might be able to modify the code for the function shapiro.test() so

Hi, I am not so sure about an error note I got when using shapiro.test. I imported some data into R by wrinting it into a .txt file via > tab1<-read.table("etc....txt",header=T) > attach(tab1) The following object(s) are masked _by_ .GlobalEnv : ozon > ozon$V1 [1] 2.5 3.0 5.6 4.7 6.5 6.7 1.7 5.3 4.6 7.4 5.4 4.1 5.1 5.6 5.4 6.1 7.6[18] 6.2 6.0 5.5 5.8 8.2 3.1 5.8 2.6 Now

Hi The shapiro.test function outputs a value of the W statistic, which should be 1 if the distribution is normal, and a p-value for the test (as the documentation states). I'm a bit confused with some results. I'm getting a W=0.9977 and a p-value=0.1889. I was expecting that a W of 0.9977 would tell me that the distribution is normal so p-value should be small ... What am I missing ?

Hola, Estoy analizando unos datos para una tesis doctoral. Durante la investigación se han recogido distintas variables clínicas de dos grupos (n=30 y n=40). Me encuentro que las comparaciones de medias se han realizado mediante t-tests sin preocuparse de estudiar la distribución de las variables. Al revisar si las variables se ajustan a la distribución normal, aunque los QQplots no tienen

Why can't this function be used with the 'by' command? > x = array(runif(16), dim=c(8,2)) > x = data.frame(x) > x$group = rep(c('wt', 'app'), each=4) > shapiro.p = function(x) shapiro.test(x)[[2]] > apply(x[,1:2], 2, shapiro.p) X1 X2 0.4126345 0.2208781 > by(x[,1:2], x$group, shapiro.p) Error in `[.data.frame`(x, complete.cases(x)) :

I could not get wbinfo -g/u to work and was seeing a bunch of errors related to to not being able to enumerate groups. I saw somebody use idmap backend = ad and added this since I have been struggling to get ad working (still not working). Now, when I run wbinfo -g/-u, I am getting groups and users, but the domain it shows is different than what I expected. My domain I was using for workgroup

I have a large amount of data which I break down into a collection of vectors of 100-125 values each. I would like to test the normality of the vectors and compare them. In the interactive mode I can test any one vector using the Shapiro-Wilk test or the Kolmogorov-Smirnov test. My problem is that when I try to write out the results to a file, the term output is a mixture of alpha characters

whoops, forgot to copy the list on it. sorry. Well, an update. I can log in to the console using any domain profiles, but, I can not access the exposed home directory through NetBeui (My Network Places/Network Neighborhood). Also, how should I configure /etc/pam.d/sshd to allow domain users to authenticate and logon through an ssh client (PuTTY?, OpenSSH?) -----Original Message----- From:

Hi, In ?shapiro.test, there seems to be no option to pass mean and sd information of the Normal distribution which I want to compare my sample data to. For example in the code below, I want to test my sample to N(0, 10). shapiro.test(rnorm(100, mean = 5, sd = 3)) Is there any way to pass the information of the benchmark normal distribution?

Full_Name: Jason Polak Version: R version 2.5.0 (2007-04-23) OS: Xubuntu 7.04 Submission from: (NULL) (137.122.144.35) Dear R group, I have noticed a strange anomaly with the shapiro.test() function. Unfortunately I do not know how to calculate the shapiro test P values manually so I don't know if this is an actual bug. So, to produce the results, run the following code: pvalues = 0; for

Hello, I want to conduct normality test to a series of data and get the p-value for each subset. I am using the following codes, but it does not work. tapply(re, list(reg, ast), pvalue(shapiro.test)) Could anyone give me some advice? Many thanks. -- View this message in context: http://www.nabble.com/tapply-tf3851631.html#a10910748 Sent from the R help mailing list archive at Nabble.com.

This is probably a very simple question but I am completely stumped!I am trying to do shapiro.wilk(x) test on a relatively small dataset(75) and each time my variable and keeps coming out as 'NULL', and > shapiro.test(fcv) Error in complete.cases(x) : no input has determined the number of cases my text file looks like this: case 1.600972896 1.534026106 1.633468456 1.69019608

Hi, I use two similar scripts to put p-values of shapiro.test and, respectively, of wilcox.test in a table: a) d <- data.frame(dataset$GroupFactor, dataset[2:11]) # p-values for the shapiro test (by levels of GroupFactor) with(d, aggregate(d[,-1], list(d[,1]), FUN = function(x) shapiro.test(x)$p.value)) b) d <- data.frame(dataset$GroupFactor, dataset[2:11]) with(d, aggregate(d[,-1],