similar to: calculating p-values of columns in a dataframe

Hi, this is a made-up example. Function "myfun" returns two arguments. Can "apply" be used so that "myfun" is called only once? Thanks Serguei mat<-matrix(runif(50),nrow=10,ncol=5) myfun<-function(x) { mymean<-mean(x) mysd<-sd(x) return(mymean,mysd) } out1<-t(apply(mat,1,function(x) myfun(x)$mymean)) out2<-t(apply(mat,1,function(x)

Hi all, I have two very large samples of data (10000+ data points) and would like to perform normality tests on it. I know that p < .05 means that a data set is considered as not normal with any of the two tests. I am also aware that large samples tend to lead more likely to normal results (Andy Field, 2005). I have a few questions to ensure that I am using them right. 1) The Shapiro-Wilk

Dear R-users, please consider the following minimal example: \documentclass[a4paper,titlepage,onecolumn,12pt]{article} \usepackage[italian]{babel} \usepackage{amssymb} \usepackage[utf8x]{inputenc} \usepackage[pdftex]{graphicx} \begin{document} <<label=test, echo=FALSE, results=tex>>= df.data1 <- cbind.data.frame(A = rnorm(18), B =factor(rep(LETTERS[1:6],

Hello, everybody! I'm back to ask the obvious on behalf of the silent majority :) Today a student asked me "what standard error was used in this t.test output". I looked into it and was a little surprised that a t.test output object does not have a "slot" for the standard error. Of course, we can reconstruct the se=mu-hat/t, but I was surprised. Do you think it would

When building R-devel with gcc 5.2.0 (mingw-w64 v4) on Windows, make check fails reg-tests-1b.R at the following check: x <- c(1:2, NA) sx <- sd(x) !is.nan(sx) Here 'sx' should be 'NA' but it is 'NaN'. It turns out this problem only appears when the function is byte compiled with optimization level 3: mysd <- function (x, na.rm = FALSE) sqrt(var(if

Hi R-helpers, Can some one tell me how to train 'mynn' of this type?: mynn <- nnet(y ~ x1 + ..+ x8, data = lgist, size = 2, rang = 0.1, decay = 5e-4, maxit = 200) I assume that this nn is untrained, and to train I have to split the original data into train:test data set, do leave-one-out refitting to refine the weights (please straighten this up if I was wrong). I just don't know

I'm using R-2.6.2 on Fedora Linux 9. I've been experimenting with plotmath. I wish it were easier to combine expressions in plotmath with values from the R program itself. There are two parameters in the following example, the mean "mymean" and standard deviation "mystd". I am able to use bquote to write elements into the graph title like mu = mymean and R will

Dear All, Need some help as I was trying to follow the guide below. https://wiki.samba.org/index.php/Samba4/HOWTO/Join_a_domain_as_a_DC Until the steps of ldbsearch -H /usr/local/samba/private/sam.ldb '(invocationid=*)' --cross-ncs objectguid and my ldbsearch reply with such a result. ldb: unable to dlopen /usr/lib64/samba/ldb/acl.so : /usr/lib64/ldb/libreplace.so: version

Hello everybody When one is working with complex matrices, "transpose" very nearly always means *Hermitian* transpose, that is, A[i,j] <- Conj(A[j,i]). One often writes A^* for the Hermitian transpose. I have only once seen a "real-life" case where transposition does not occur simultaneously with complex conjugation. And I'm not 100% sure that that wasn't a

Dear R users, My set-up: OS=Windows XP, R-2.9.2, Rtools210 I faced the follwing problem with the package compilation: There is no "/inst" or "/etc" subdirectory in the package-zip-file. And the content of the "/etc" subdirectory is lost, too. I tried a simplified "test" package. The "test" package has the following structure (see also attachement:

Dear All, I have a database of 200 observations named myD. In the dataframe there are a column named code (with codes varying from 1 to 77), a column named "prevalence" with some quantitative measurements are given and an column named Pr_mean, with no values. I would like to set a cycle to compute the average of prevalence values for each different code and store the averages under the

Dear List, I am having problems converting a file with fractional seconds to class POSIXct. I have set my options to include digits.secs and my format to just time, but my output is the current date with my time lacking the fractions of a second. For example: options(digits.secs=3) t<-c("06:00:00.100","06:00:01.231") myt<-as.POSIXct(t,format="%H:%M:%S")

Hi all, I am currently using the code snippet below to minimize a function which I wasn't able to minimize properly with optim (cliffy surface, constraints, poles). Obviously this iteration is probably the poorest way of implementing such a function and it is expectedly slow. I've already written some vectorized functions, but I'm afraid that I'm missing the "big

I believe the issue is that on Windows the sqrt function when called with a NaN does not return the same NaN, as it does on other platforms. We have #if (defined(_WIN32) || defined(_WIN64)) && defined(__GNUC__) && \ __GNUC__ <= 4 # define R_sqrt(x) (ISNAN(x) ? x : sqrt(x)) #else # define R_sqrt sqrt #endif for implementing the SQRT opcode. I suspect this came from Duncan

On 14/09/2015 9:36 AM, luke-tierney at uiowa.edu wrote: > I believe the issue is that on Windows the sqrt function when called > with a NaN does not return the same NaN, as it does on other platforms. > We have > > #if (defined(_WIN32) || defined(_WIN64)) && defined(__GNUC__) && \ > __GNUC__ <= 4 > # define R_sqrt(x) (ISNAN(x) ? x : sqrt(x)) > #else

I would like to take the standard deviation of a column, but only for a subset of the rows in that column with a given index. The following loop worked fine when I wanted the mean, but is not working for the standard deviation: for (i in 1:length(x[1,])){ a<-tapply(x[,i],x[,2],sd, na.rm=TRUE) xnew<-cbind(xnew,a)} I have tried re-defining the sd as follows (as suggested on this board),

Hello, I am work with a linear regression model: y=ax+b with the function of lm. y= observed migration distance of butterflies x= predicted migration distance of butterflies Usually the result will show if the linear term a is significantly different from zero based on the p-value. Now I would like to test if the linear term is significantly different from one. (because I want to know

R programming question, not machine learning, although that's the content. Apologies to all for whom the following code is eye-burning. I am using foreach() to run a simulation on a randomForest model (actually conditional randomForest ... "party" package). The simulation is in two dimensions. examining how "mtry" and "ntrees" are related in terms of predictive

> On Jul 11, 2017, at 6:25 AM, Cristina Silva <csilva at ipma.pt> wrote: > > Dear Jeff, > > I am sorry, I didn't notice that it was not plain text. I hope that it is now in the correct format. I explain the problem again, now with more detais. > I am collecting the track positions of our research vessel from www.marinetraffic.com. In the page, the data appear in a

What in the world am I missing?? > x<-rnorm(20) > mean(x) [1] -0.2272851 > results<-boot(x,mean,R=5) > results[2] $t [,1] [1,] -0.2294562 [2,] -0.2294562 [3,] -0.2294562 [4,] -0.2294562 [5,] -0.2294562 Jeff Morris Ortho-Clinical Diagnostics A Johnson & Johnson Co. Rochester, NY Tel: (585) 453-5794 [[alternative HTML version deleted]]