similar to: how to calculate this natural logarithm

Hello I have a time-comsuming program which need to simplify, I have tested the annotated program as follow: > #define function which will be call > calsta <- function(c, n=100000) + { + i <- seq(from=0, length=c) + logx <- lchoose(NT-n, CT-i) + lchoose(n, i) + logmax <- max(logx) + logmax + log(sum(exp(logx - logmax))) + } > CT=6000 #assignment to CT >

Hello T I want to order some calculation "result", there will be lots of "result" that need to calculate and order PS: the "result" is just a intermediate varible and ordering them is the very aim # problem: # For fixed NT and CT, and some pair (c,n). order the pair by corresponding result # c and n are both random variable CT<-6000 #assignment to CT

Hi there fellow R-users, Can anyone tell me how to build a prediction interval for a gaussian log-link model for the reponse variable?? I can find the standard error of the predictions but I cant seem to find the prediction interval. Is there a way I can calculate the prediction interval from the standard errors?? Here's the example: logX<-rnorm(100)

Hi all I was playing with termplot(), and came across what appears to be an inconsistency. It would appreciate if someone could enlighten me: > # First, generate some data: > y <- rnorm(100) > x <- runif(length(y),1,2) > # Now find the log of x: > logx <- log(x) > > # Now fit two models that are exactly the same, but specified differently: > m1 <-

lgamma(x) and lfactorial(x) are defined to return ln|Gamma(x)| {= log(abs(gamma(x)))} or ln|Gamma(x+1)| respectively. Unfortunately, we haven't chosen the analogous definition for lchoose(). So, currently > lchoose(1/2, 1:10) [1] -0.6931472 -2.0794415 NaN -3.2425924 NaN -3.8869494 [7] NaN -4.3357508 NaN -4.6805913 Warning message: In

Hello I have 4 1000*1000 matrix A,B,C,D. I want to use the corresponding element of the 4 matrices. Using the "for loop" as follow: E<-o for (i in 1:1000) {for (j in 1:1000) { E<-fisher.test(matrix(c(A[i][j],B[i][j],C[i][j],D[i][j]),2))#call fisher.test for every element } } It is so time-consuming Need vectorization Yours sincerely ZhaoXing Department of

Hi, I have been trying to use the function cochran.test from the Outliers package to test for homogeneity of variance. This works well except when I use transformed data. Would anyone have an idea why it doesn't work and how I could do the cochran test on transformed data? Thanks, Stephanie >library(outliers) > set.seed(1234) > x=rnorm(100) >

Dear R-users, I am using R version 2.10.1 under windows. In a barplot, I want to mark one of the bars with a special border color. For example: barplot(c(3, 7, 11), border = c(NA, "red", NA)) But how to do this when the bars are stacked? for example: barplot(matrix(1:6, ncol=3)) # border of second bar (i.e. the one with total height = 7) should be red again, I try: barplot(matrix(1:6,

>>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: >>>>> William Dunlap via R-devel >>>>> on Sun, 23 Jun 2019 10:34:47 -0700 writes: > include/Rmath.h declares a set of 'logspace' functions for use at the C > level. I don't think there are core R functions that call

I agree with many the sentiments about the wisdom of computing very small p-values (although the example below may win some kind of a prize: I've seen people talking about p-values of the order of 10^(-2000), but never 10^(-(10^8)) !). That said, there are a several tricks for getting more reasonable sums of very small probabilities. The first is to scale the p-values by dividing the

>From: Sundar Dorai-Raj <sundar.dorai-raj at PDF.COM> >Reply-To: sundar.dorai-raj at PDF.COM >To: Zhen Pang <nusbj at hotmail.com> >CC: r-help at stat.math.ethz.ch >Subject: Vectorising and loop (was Re: [R] optim "a log-likelihood >function") >Date: Wed, 29 Sep 2004 18:21:17 -0700 > > > >Zhen Pang wrote: > >> >>I also use

Aloha all, I know too little about what I'm about to write and hope I'm not wasting your time. For a class I'm teaching in archaeological data analysis, I'm trying to put together a routine that calculates the so-called Petersen index and, especially, confidence intervals for the index. This was introduced to archaeologists by N.R.J. Fieller and A. Turner in an article

>>>>> jing hua zhao >>>>> on Mon, 24 Jun 2019 08:51:43 +0000 writes: > Hi All, > Thanks for all your comments which allows me to appreciate more of these in Python and R. > I just came across the matrixStats package, > ## EXAMPLE #1 > lx <- c(1000.01, 1000.02) > y0 <- log(sum(exp(lx))) > print(y0) ## Inf

Hello Suppose I have a matrix mat=(1:16,2) [,1] [,2] [,3] [,4] [1,] 1 5 9 13 [2,] 2 6 10 14 [3,] 3 7 11 15 [4,] 4 8 12 16 I just want to use the "apply" function to the matrix partially there is a vector end=c(2,3,1,3) #sum the 1st 2 numbers of the 1st column #sum the 1st 3 numbers of the 2nd column #sum the 1st number of the 3rd column

Hello Suppose I have a matrix mat=(1:16,2) [,1] [,2] [,3] [,4] [1,] 1 5 9 13 [2,] 2 6 10 14 [3,] 3 7 11 15 [4,] 4 8 12 16 there is a vector end=c(2,3,1,3) #coerce the 1st 2 numbers of the 1st column to a vector [1] 1 2 #coerce the 1st 3 numbers of the 2nd column and append it to the previous vector [1] 1 2 5 6 7 #coerce the 1st number of

Hi, I was running samba-4.1.4 successfully on my fedora till the below issue occurs. I had put some simple DEBUG statements in "lib/util/debug.c" (reopen_logs_internal() and some other funcs) file to understand the logging mechanism. And the same is compiled fine without any issues. And I started my samba service(both nmbd and smbd), I got "(core dumped) segmentation fault"

Hi, I'm trying to calculate p-value to findout definitely expressed genes compare A to B situation. I got this data(this is a part of data) from whole organism , and each number means each expression values (that means, we could think 'a' gene is 13 in A situation, and it turns 30 in B situation) To findout probability, I'm going to use Audic - Claverie Method. ( The significance

R Version 2.2.0 Platform: Windows When I use barplot but select a ylim value greater than zero, the graph is distorted. The bars extend below the bottom of the graph. For instance the command produces a problematic graph. barplot(c(200,300,250,350),ylim=c(150,400)) Any help would be appreciated. Paul [[alternative HTML version deleted]]

OK, I see what you mean. But in those cases, we don't get the catastrophic failures from the if (k < 0) return 0.; if (k == 0) return 1.; /* else: k >= 1 */ part, because at that point k is sure to be integer, possibly after rounding. It is when n-k is approximately but not exactly zero and we should return 1, that we either return 0 (negative case) or n

> On 14 Jan 2020, at 16:21 , Duncan Murdoch <murdoch.duncan at gmail.com> wrote: > > On 14/01/2020 10:07 a.m., peter dalgaard wrote: >> Yep, that looks wrong (probably want to continue discussion over on R-devel) >> I think the culprit is here (in src/nmath/choose.c) >> if (k < k_small_max) { >> int j; >> if(n-k < k