similar to: Why can't this function be used with the 'by' command?

I am trying to do this, but it won't work: > library(lattice) > elemconc = data.frame(expand.grid(id=1:20, geno=c('exp', 'wt'), region=c('cor', 'cr', 'spine'), elem=c('fe', 'cu', 'zn')), conc=rnorm(360, 10)) > elemconc$geno = factor(elemconc$geno) > elemconc$region = factor(elemconc$region) > for (i in

Here is the code, assuming 8 cores in the cpu. library('modeest') library('snow') cl = makeCluster(rep('localhost', 8), 'SOCK') x = vector(length=50) x = sapply(x, function(i) i=sample(c(1,0), 1)) pastK = function(n, x, k) { if (n>k) { return(x[(n-k):(n-1)]) } else {return(NA)} } predR = function(x, k) { pastList = lapply(1:length(x), function(n)

Dear all, I found a strange thing with the snow package. This will work: y = matrix(1:4, 2) cl = makeCluster(rep('localhost', 8), type='SOCK') parMM(cl, y, y) This will not: y = matrix(1:4, 2) ncore = system('nproc') parMM(cl, y, y) Error in cut.default(i, breaks) : invalid number of intervals I also tried: cl = makeCluster(rep('localhost', ncore),

I have assayed the concentrations of various metal elements in different anatomic regions of two strains of mice. Now, for each element, in each region, I want to do a t test to find whether there is any difference between the two strains. Here is what I did (using simulated data as an example): # create the data frame > elemconc = data.frame(expand.grid(id=1:3, geno=c('exp',

> library(reshape2) > x = melt(airquality, id=c('month', 'day')) With reshape I can cast with multiple functions: > library(reshape) > cast(x, month+variable~., c(mean,sd)) month variable mean sd 1 5 ozone 23.615385 22.224449 2 5 solar.r 181.296296 115.075499 3 5 wind 11.622581 3.531450 4 5 temp 65.548387

> for (d in paste('df', 1:3, sep='')) { + assign(d, as.data.frame(replicate(3, rnorm(4)))) + } > dats = list(df1,df2,df3) > for (i in 1:length(dats)) { + names(dats[[i]]) = c('w', 'l', 'h') + } > dats [[1]] w l h 1 1.24319239 -0.05543649 0.05409178 2 0.05124331 -1.89346950 0.33896273 3 -1.69686777 -0.35963008

I found a strange bug in R recently (version 3.1.2): As you can see from the screenshots attached, when the cursor passes the right edge of the console, instead of start on a new line, it goes back to the beginning of the same line, and overwrites everything after it. This happens every time the size of the terminal is changed, for example, if you fit the terminal to the right half of the

> regions = c('cortex', 'hippocampus', 'brain_stem', 'mid_brain', 'cerebellum') > mice = paste('mouse', 1:5, sep='') > for (n in c('Cu', 'Fe', 'Zn', 'Ca', 'Enzyme')) { + assign(n, as.data.frame(replicate(5, rnorm(5)))) + } > names(Cu) = names(Zn) = names(Fe) = names(Ca) = names(Enzyme) =

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by

Dear list, I am trying to compile R on a 64-bit Ubuntu 13.04 machine and get the following error: make[2]: Entering directory `/home/kaiyin/opt/R-2.15.0/src/library/Recommended' begin installing recommended package MASS Error in untar2(tarfile, files, list, exdir) : incomplete block on file make[2]: *** [MASS.ts] Error 1 make[2]: Leaving directory

Buen día estimados Estoy tratando de hacer un tibble con los resultados de un apply que se muestran en la consola que me da R, no estoy seguro si eso se pueda hacer, pero me gustaría organizar los resultados de esa manera. mi código es: data("mtcars") Mtcars_matriz <- as.matrix(mtcars) apply(Mtcars_matriz, MARGIN =2, FUN = shapiro.test) DF2 <- tibble(Variable = NA, W = NA, Pvalue =

> d = data.frame(gender=rep(c('f','m'), 5), pos=rep(c('worker', 'manager', 'speaker', 'sales', 'investor'), 2), lot1=rnorm(10), lot2=rnorm(10)) > d gender pos lot1 lot2 1 f worker 1.1035316 0.8710510 2 m manager -0.4824027 -0.2595865 3 f speaker 0.8933589 -0.5966119 4 m sales

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

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",

In the course of applying Shapiro-Wilk to 100,000 samples of 60 items from 100,000 different distributions, I encountered a fatal error in apply(). This can be reconstructed as follows, using the attached data file distr.dat containing 2 lines of my original 100,000-line file: > version _ platform Windows arch x86 os Win32 system x86, Win32 status

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

Hi, I am trying to create a table with information from Shapiro Wilk's test of normality. However, it fails due to lack of sample size, it says, but the way I see it, this is not a problem. (See the table of sample sizes (almost) at the bottom). Applying a different function using a similar ftable call is not a problem (See the bottom table). This is R 2.1.0 on Linux (Gentoo). /Fredrik

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

Dear R-users Idea: Analysing tree height frequency with hist(), normal distribution (ks.test & shapiro.test) and skewness (package e1071 - thanks a lot for this useful package)as an indication of possible self-thinning in an experimental tree stand. Problem: Results from the ks.test and the shapiro.test are not comparable (see example of both tests). Tree height is a nice continuous