similar to: (PR#9811) sequence(c(2, 0, 3)) produces surprising results,

Hello, Based on yesterday's R-help thread (help: program efficiency), and following Bill's suggestions, it appeared that sequence: > sequence function (nvec) unlist(lapply(nvec, seq_len)) <environment: namespace:base> could benefit from being written in C to avoid unnecessary memory allocations. I made this version using inline: require( inline ) sequence_c <- local( {

> -----Original Message----- > From: linux-pci-owner at vger.kernel.org [mailto:linux-pci-owner at vger.kernel.org] > On Behalf Of Yijing Wang > Sent: Saturday, July 26, 2014 8:39 AM > To: linux-kernel at vger.kernel.org > Cc: Xinwei Hu; Wuyun; Bjorn Helgaas; linux-pci at vger.kernel.org; > Paul.Mundt at huawei.com; James E.J. Bottomley; Marc Zyngier; linux-arm- > kernel at

> -----Original Message----- > From: linux-pci-owner at vger.kernel.org [mailto:linux-pci-owner at vger.kernel.org] > On Behalf Of Yijing Wang > Sent: Saturday, July 26, 2014 8:39 AM > To: linux-kernel at vger.kernel.org > Cc: Xinwei Hu; Wuyun; Bjorn Helgaas; linux-pci at vger.kernel.org; > Paul.Mundt at huawei.com; James E.J. Bottomley; Marc Zyngier; linux-arm- > kernel at

I evaluated the Bayes factor in the k=2 allele case with a "triangular" prior under the null as in the example in the help file: HWETriangBF2(nvec=c(88,10,2)) [1] 0.4580336 When I swap the n11 entry and n22 entry of nvec, I received totally different Bayes factor: > > HWETriangBF2(nvec=c(2,10,88)) [1] 5.710153 > In my understanding, defining the genotype frequency as

hey guys, I am working on a function to make a duplicated value unique. For example, the original vector would be like : a = c(2,1,1,3,3,3,4) I'll like to transform it into: a.nodup = 2, 1.01, 1.02, 3.01, 3.02, 3.03, 4 basically, find the duplicates and assign a unique value by adding a small amount and keep it in order. I come up with the following codes, but it runs slow if t is large. Is

Hello: If anyone could guide me with this I would greatly appreciate it. Thanking you in advance for your assistance. Using a 3-level input factor alternative so that a function(below) can compute both a two-sided and one-sided p-values. Making the two-sided test the default. And produce output information about which alternative was tested. Where would I place the ifelse statement?

> -----Original Message----- > From: linux-pci-owner at vger.kernel.org [mailto:linux-pci-owner at vger.kernel.org] > On Behalf Of Yijing Wang > Sent: Saturday, July 26, 2014 8:39 AM > To: linux-kernel at vger.kernel.org > Cc: Xinwei Hu; Wuyun; Bjorn Helgaas; linux-pci at vger.kernel.org; > Paul.Mundt at huawei.com; James E.J. Bottomley; Marc Zyngier; linux-arm- > kernel at

> -----Original Message----- > From: linux-pci-owner at vger.kernel.org [mailto:linux-pci-owner at vger.kernel.org] > On Behalf Of Yijing Wang > Sent: Saturday, July 26, 2014 8:39 AM > To: linux-kernel at vger.kernel.org > Cc: Xinwei Hu; Wuyun; Bjorn Helgaas; linux-pci at vger.kernel.org; > Paul.Mundt at huawei.com; James E.J. Bottomley; Marc Zyngier; linux-arm- > kernel at

This was sent to me by someone on the R-list ( I don't know her ) but I don't have time to look at this right now so I told her I would send it to the R-list because she said it keeps getting bounced when she sends it. #======================================================================= ==================================================================== I am a bit confused with

> -----Original Message----- > From: linux-pci-owner at vger.kernel.org [mailto:linux-pci-owner at vger.kernel.org] > On Behalf Of Yijing Wang > Sent: Saturday, July 26, 2014 8:39 AM > To: linux-kernel at vger.kernel.org > Cc: Xinwei Hu; Wuyun; Bjorn Helgaas; linux-pci at vger.kernel.org; > Paul.Mundt at huawei.com; James E.J. Bottomley; Marc Zyngier; linux-arm- > kernel at

> -----Original Message----- > From: linux-pci-owner at vger.kernel.org [mailto:linux-pci-owner at vger.kernel.org] > On Behalf Of Yijing Wang > Sent: Saturday, July 26, 2014 8:39 AM > To: linux-kernel at vger.kernel.org > Cc: Xinwei Hu; Wuyun; Bjorn Helgaas; linux-pci at vger.kernel.org; > Paul.Mundt at huawei.com; James E.J. Bottomley; Marc Zyngier; linux-arm- > kernel at

pci_enable_msix currently returns -EINVAL if you ask for more vectors than supported by the device, which would typically cause fallback to regular interrupts. It's better to return the table size, making the driver retry MSI-X with less vectors. Signed-off-by: Michael S. Tsirkin <mst at redhat.com> --- Hi Jesse, This came up when I was adding MSI-X support to virtio pci driver, which

pci_enable_msix currently returns -EINVAL if you ask for more vectors than supported by the device, which would typically cause fallback to regular interrupts. It's better to return the table size, making the driver retry MSI-X with less vectors. Signed-off-by: Michael S. Tsirkin <mst at redhat.com> --- Hi Jesse, This came up when I was adding MSI-X support to virtio pci driver, which

Hi All, I would like to compute the simple finite-difference approximation to the gradient of a scalar function of a large number of variables (on the order of 1000). Although a one-time computation using the following function grad() is fast and simple enough, the overhead for repeated evaluation of gradient in iterative schemes is quite significant. I was wondering whether there are

Me: Win XP 4 gig ram R 2.9.2 library(foreign) # to read/write SPSS files library(doBy) # for summaryBy library(RODBC) setwd("C:\\Documents and Settings\\............00909BR") gc() memory.limit(size=4000) ## PROBLEM: I have memory limit problems. R and otherwise. My dataframes for merging or subsetting are about 300k to 900k records. I've had errors such as vector size too large.

Hi, I have the following dataset (simplified for example). __DATA__ 300.35 200.25 104.30 22.00 31.12 89.99 444.50 22.10 43.00 22.10 200.55 66.77 Now from that I wish to do the following: 1. Compute variance of each row 2. Pick top-2 row with highest variance 3. Store those selected rows for further processing To achieve this, I tried to: a) read the table and compute variance for each row,

Dear list members, I am trying to use the maptools library to display geographical data. At the moment I have some trouble understanding how the " auxvar " variable is supposed to be used in the plot.Map function. I am using R Version 1.8.1 (2003-11-21) on Linux Looking at the plot.Map function itself, I see that it calls a color.ramp function (I am reporting only the relevant

Hi there I'm trying to find the mle fo a multinomial model ->*L(N,h,S?x)*. There is only *N* I want to estimate, which is used in the number of successes for the last cell probability. These successes are given by: p^(N-x1-x2-...xi) All the other parameters (i.e. h and S) I know from somewhere else. Here is what I've tried to do so far for a imaginary data set:

Thanks. This is exactly the doubt I had. Rui Barradas ?s 05:26 de 05/08/2018, Kenny Bell escreveu: > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)]

El dom., 5 ago. 2018 a las 6:27, Kenny Bell (<kmbell56 at gmail.com>) escribi?: > > This should more clearly illustrate the issue: > > c(1, 2, 3, 4)[-seq_len(4)] > #> numeric(0) > c(1, 2, 3, 4)[-seq_len(3)] > #> [1] 4 > c(1, 2, 3, 4)[-seq_len(2)] > #> [1] 3 4 > c(1, 2, 3, 4)[-seq_len(1)] > #> [1] 2 3 4 > c(1, 2, 3, 4)[-seq_len(0)] > #>