similar to: How to manipulate an abitrary dimensioned array.

The following snippet suggests that there is either a bug in qr(,LAPACK=T), or some bug in my understanding. Note that the detected rank is correct (= 2) using the default LINPACK qr, but incorrect (=3) using LAPACK. This is running on Linux Redhat 9.0, using the lapack library that comes with the Redhat distribution. I'm running R 1.7.1 compiled from the source. If the bug is in my

Dear R-experts! Running a meta-analysis (using the magnificent metafor-package), I use cohen's d as a main outcome measure in a random-effects model. For most of the samples cohen's d is derived form a comparison of two groups (A & B). However some studies report results from an ANOVA (one-factor with three levels: C,D,E) whereas two groups (C,D) correspond to one group in the other

Hi, how can I create an array of lists of three components? This approach does not work: n1 <- 2 n2 <- 4 n3 <- 5 res <- array(rep(vector("list",3), n1*n2*n3), dim = c(n1,n2,n3)) res[1,1,1] # is not a list with three components... The goal is that res[1,1,1] is a list with three components. Also, appending the components didn't work. For example, I tried: component

Hi there, I am a newbie of llvm and here is my question situation. Assume that there is a function F which contains a loop named L, a array b[100]. I want to collect the statistical information of array index operation op(i) (take add and mul simply) of i in the loop L. Pseudocode lists below. void F(arg1, arg2){ int b[100]; for(int i=0; i<n; i++){ op1(i); op2(i);

Dear R People: I'm having trouble with something that should be very simple. I'm setting up a matrix outside of a loop and writing items into it during the loop. Here is the output: > glob3b("sites.info") dim 27 3 [1] "/raid1/osg-app" Error in xy[i, ] : incorrect number of dimensions Here is the function: > glob3b function(xx) { x.df <-

I''m using this gem https://github.com/justintv/APNS to send push notifications to an iPhone app and amazingly it works for single notifications, such that I can do... APNS.send_notification(Device.first.device_token,''test message'') Works great... Now it should be a simple task (as per the docs to send multiple notifications) Gem says like this... device_token =

(On Unix/Linux) Is it possible to set the --no-save command line option as an option in the .Rprofile file. I have looked in the sources and do not see any obvious (user initiated) ways of changing the command line defaults, but I am hoping I have missed something. I *know* I can write my own shell script wrapper that calls R with whatever arguments I want (and I do that). However there

It's also very common in Fortran. -David Michael Ferguson via llvm-dev <llvm-dev at lists.llvm.org> writes: >> It seems that the main advantage of your proposal is that it would >> allow for non-constant strides (i.e. variable length arrays) in >> dimensions other than the first one. Do these appear frequently >> enough in the programs

Dear all, I have data like 3 coulmns and many rows. Each entry is less than 10. Example x y z 1 5 3 2 2 3 7 8 3 8 9 5 4 5 4 6 -------------------------- --------------------------- I have to sum entries of each coulmn (seperately) till it be 10. This i have to start for each row. And I want to assign no. of rows needed including that row too(it to be 10 or 10+, the moment it exceeds 10, i

> It seems that the main advantage of your proposal is that it would allow for non-constant strides (i.e. variable length arrays) in dimensions other than the first one. Do these appear frequently enough in the programs that you're interested in to be worth optimizing for? Yes - at least in Chapel (which is one of the motivating languages) these are very common. In other words, typical

We could also simply extend the existing inrange mechanism to non-constantexpr GEPs.  It would remove an inconsistency in the semantics, be relatively straight forward, and solve the motivating example. (I didn't read the proposal in full, so there may be other examples it doesn't solve.) Philip On 7/22/19 10:01 AM, Peter Collingbourne via llvm-dev wrote: > The restrictions of

Estimados Usuarios-R: Estoy convirtiendo un programa en Matlab a R. El original lo saqu de: Lai, M., & Poltera, Y. (2009). Lecture with computer exercises: Modelling and simulating social systems with matlab. Tech. rep., Swiss Federal Institute of Technology (December 2009). 27. Ahora estoy convirtiendo la siguiente funcin: function sizes = gridsizes(N,varargin) % gridsizes(N) calculates

Hi, Algorithm question: I have two sets of "intervals", where an interval is an ordered pair [a,b] of two numbers. Is there an efficient way in R to generate the intersection of two lists of same? For concreteness: I'm representing a set of intervals with a data.frame: > list1 = as.data.frame(list(open=c(1,5), close=c(2,10))) > list1 open close 1 1 2 2 5

I'm solving 4 complex equations simultaneously. Code is below. The code returns only zero's for the solution though there should also be a non-zero result. I'm pretty confident that the equations are correct because they are straight from a published paper and I checked them pretty thoroughly. The parameter values I used are from the published paper as well. Any suggestions for how

Hello r helpers! Below is the whole coding for my programme. Before proceed more further, let me explain for you. First of all, I need to compute trimmed mean. Till that step is ok. Then I need to compute ssdw which is sum of square deviation. If I do equal trimming at both tail of distribution that I chose, I will use the first ssd formulae which is "a". But if I am doing unequal

Hi, final<-data.frame() ?? for (m1 in 4:10) { ?????? for (n1 in 4:10){? ?????????? for (x1 in 0: m1) { ????????????? for (y1 in 0: n1) { final<- rbind(final,c(m1,n1,x1,y1)) res}}}} ?final1<-within(final,{flag<-ifelse(x1/m1>y1/n1, 1,0)}) ?head(final1) #? m1 n1 x1 y1 flag #1? 4? 4? 0? 0??? 0 #2? 4? 4? 0? 1??? 0 #3? 4? 4? 0? 2??? 0 #4? 4? 4? 0? 3??? 0 #5? 4? 4? 0? 4??? 0 #6? 4? 4?

Hi, I'm doing a function that describe two populations in competition. that's the function that i wrote: exclusao<-function(n10, n20, k1, k2, alfa, beta, t){ n1<-k1-(alfa*n20) n2<-k2-(beta*n10) if(t==0){plot(t, n10, type='b', xlim=range(c(1:t),c (1:t)), ylim=range(n10, n20), xlab='tempo', ylab='tamanho populacional') points(t, n20, type='b',

Windows 7, R 2.12.1 Colleagues, I am trying to understand the n.for.2means function. The code below is a copy of the function (renamed to n.for.2means.js). I have inserted a single line of code towards the bottom of the function which uses the cat function to print the value of n1. You will note the value (preceded by stars) is printed as 1. The function (1) prints a lot of output without any

Hi, Here are some new library functions for klibc. Some of them are required for udev, which currently has a klibc_fixups.c file that implements these functions. mh -- Martin Hicks Wild Open Source Inc. mort@wildopensource.com 613-266-2296 # This is a BitKeeper generated patch for the following project: # Project Name: The kernel C library # This patch format is intended

Hi I need to find the Hessian matrix for a complicated function from a certain kind of data but i keep getting this error Error in f1 - f2 : non-numeric argument to binary operator the data is given by U<-runif(n) Us<-sort(U) tau1<- 2 F1tau<- pgamma((tau1/theta1),shape,1) N1<-sum(Us<F1tau) X1<- Us[1:N1]