similar to: What method I should to use for these data?

hai, I want to perform a simulation like this仯伜 Suppose that I have one population ,it's size is 500, is composed of x,y and z. The probability of x, y and z is respectively is 0.3, 0.5, 0.2. I wan to simulate a new same size population based ratio of x, y and z, how can I get and assess the number of x, y and z. ? luan Key Laboratory for Sustainable Utilization of Marine Fisheries

hello,everyone. I have one question: example 1 > x=numeric(0) > y=5 > print(x+y) numeric(0) example 2 > x=numeric(1) > y=5 > print(x+y) [1] 5 why the print(x+y) is numeric(0) at the first example, but the result is 0 at the second example? __________________________________________________ 佈伵伝仮伱佲伔佈G佊伿佅佷仯伃佒佇伖侜伒佢佉伝伨侙佄佫伬伂伝侙佊伿伡侢伾仹伻伵伋伂伌侒佊伿佅佷

Dear R user: Now I am studying "matrix population models" and Rmetasim package,but I find it's very difficult to understand this model fully and can't find a good teacher.My question is that:Now I do some research about one marine shrimp: Penaeus chinensis, it's life cycle is one year. after it breed it's offsprings, it will died. This shrimp's life cycle is dicrete

I've noticed that glm and lrm give extremely different results if you attempt to fit a saturated model to a dataset with zero cells. Consider, for instance the data from, Agresti's Death Penalty example [0]. The crosstab table is: , , PENALTY = NO VIC DEF BLACK WHITE BLACK 97 52 WHITE 9 132 , , PENALTY = YES VIC DEF BLACK WHITE BLACK 6 11

I want to th current folder is the specific folder when R start.It is very boring to modify the current directory manually. How can I do it?

A big compile time regression. Any ideas? Ciao, Duncan. On 22/07/11 19:13, llvm-testresults at cs.uiuc.edu wrote: > > bwilson__llvm-gcc_PROD__i386 nightly tester results > > URL http://llvm.org/perf/db_default/simple/nts/253/ > Nickname bwilson__llvm-gcc_PROD__i386:4 > Name curlew.apple.com > > Run ID Order Start Time End Time > Current 253 0 2011-07-22 16:22:04

Dear R-tisans, I am trying to calculate the 12th root of a transition (square) matrix, but can't seem to obtain an accurate result. I realize that this post is laced with intimations of quantitative finance, but the question is both R-related and broadly mathematical. That said, I'm happy to post this to R-SIG-Finance if I've erred in posting this to the general list. I've

LLVMers, The 2.2 prerelease is now available for testing: http://llvm.org/prereleases/2.2/ If anyone can help test this release, I ask that you do the following: 1) Build llvm and llvm-gcc (or use a binary). You may build release (default) or debug. You may pick llvm-gcc-4.0, llvm-gcc-4.2, or both. 2) Run 'make check'. 3) In llvm-test, run 'make TEST=nightly report'. 4) When

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Dear R listers, I have a data file looks like the following: Testing marker: s_1 --------------------------------------------- Allele df(0) -LnLk(0) df(T) -LnLk(T) ChiSq p 3 7995 29320.30 7994 29311.85 16.90 4e-05 (2229/8000 probands) Testing marker: s_2 --------------------------------------------- Allele df(0)

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

On 23.11.2012, at 15:12, john skaller <skaller at users.sourceforge.net> wrote: > > On 23/11/2012, at 5:46 PM, Sean Silva wrote: > >> Adding LLVMdev, since this is intimately related to the optimization passes. >> >>> I think this is roughly because some function level optimisations are >>> worse than O(N) in the number of instructions. >>

I'm getting an error message when using the new version of sqldf, > library(sqldf) > str(kdv) 'data.frame': 71 obs. of 3 variables: $ dpss: num 0.117 0.144 0.164 0.166 0.165 ... $ npdp: num 0.1264 0.0325 0.0109 0.0033 0.0055 ... $ logk: num 1.12 1.29 1.41 1.41 1.42 ... > test=sqldf("select * from kdv") Error in get("fun", env = this, inherits =

To whom it may concern: I am trying a factorial design a system of mine that has two factors. Each factor was set at four different levels, with one replication for each of the combinations. My data is as follows: A B Response 1 600 2.5 0.0257 2 600 2.5 0.0254 3 600 5

Hello, I have a question regarding estimation of the dispersion parameter (theta) for generalized linear models with the negative binomial error structure. As I understand, there are two main methods to fit glm's using the nb error structure in R: glm.nb() or glm() with the negative.binomial(theta) family. Both functions are implemented through the MASS library. Fitting the model using these

Dear List, My data consist of nine columns and about 50,000 rows. It looks like this. -9.0225 3.46464 2.80926 -0.3847 3.73735 1.1058 -2.98936 1.38901 -8.1846 -2.4315 -5.1189 1.8225 3.3798 1.7874 4.693 -3.9286 1.4266 5.7849 -3.4894 -4.0305 3.7879 3.5195 2.9186 2.8685 -6.126 4.978 4.9381 4.5282 3.62558 -3.0455 4.6518 1.39746 0.68652 3.5708 -3.6404 -4.2963 -1.3183 0.6752 -4.0382 -2.5386

Hi everyone, I need help. I want to have a "uniform" kind distribution. When I used sample function I got almost twice many zeros compared to other numbers. What's wrong with my command ? temp <-sample(0:12, 2000, replace=T,prob=(rep(1/13,13))) hist(temp) Thanks in advance, Taka,

Hi all: Can someone advice me on how to hold the residuals Mean sq value on a string so it can be used in other calculations. I was trying something like this: Msquare<-dfr$Mean sq but fails..Thanks dfr <- read.table(textConnection("percentQ Efficiency 1.565 0.0125 1.94 0.0213 0.876 0.003736 1.027 0.006 1.536 0.0148 1.536 0.0162 2.607 0.02 1.456 0.0157 2.16 0.0103

Hello, How does one specify a formula to lm inside a function (with variable names not known in advance) and have the formula appear explicitly in the output? For example, f <- function(d) { in.model <- sample(c(0,1), ncol(d)-1, replace=T) current.model <- lm(paste(names(d)[1], "~", paste(names(d[2:ncol(d)])[which(in.model == 1)], collapse= "+")), data=d) #***

Hi, R-users, data is a matrix like this AMR BS GE HR MO UK SP500 1974 -0.3505 -0.1154 -0.4246 -0.2107 -0.0758 0.2331 -0.2647 1975 0.7083 0.2472 0.3719 0.2227 0.0213 0.3569 0.3720 1976 0.7329 0.3665 0.2550 0.5815 0.1276 0.0781 0.2384 1977 -0.2034 -0.4271 -0.0490 -0.0938 0.0712 -0.2721 -0.0718 1978 0.1663 -0.0452 -0.0573 0.2751 0.1372 -0.1346