similar to: 2.0, hourly performance stats

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

Hi, I am profiling some workloads for the voluntary and involuntary context switches. I am interested in finding out the reasons causing these two types of context switches. As far as I understand, involuntary context switch happens on expiration of time slice or when a higher priority process comes in. While the voluntary switch generally happens when a process is waiting for I/O etc. So to

Hi all, I have 2 questions: 1)How do I calculate the mean on an imported txt file? I've imported the file below and that's what it looks like imported. How do I then calcuate the mean, median, or mode on the column LeafArea using the desktop R package? Any help would be greatly appreciated!! Thanks, Nat LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness 1 1

Dear list, I have observed a weird behaviour from R --- apologies if I am missing something obvious! df3f826f28 df3f826f28 Say you type in R: >c.preec <- 10074 >c.gd <- 2200 >p1 <- .2 >c.neo <- p1*9451 + (1-p1)*3883 >n.preec <- 3710 >n.gd <- 2650 >n.neo <- 2120 >n.pcos <- 53000 >unit.met <- 94 >cost.met <- 94*n.pcos >effect <-

Hi, User issues should generally go to samba at lists.samba.org which I've cc'd. How large is your database/how many users do you have? Is it still slow if you reverse the search expression (&(sAMAccountName=XXXX)(objectClass=person))? There have been a large number of performance improvements made since 4.3, so upgrading to the latest versions might be a good idea before any

Hi, 25,000 is definitely higher than what I would reasonably expect Samba 4.3 to handle. It should be significantly better in more recent versions (preferably to 4.7.5 by joining a new DC). When you have slow responses, is that with a number of concurrent users (or is it possible that there are specific users that are consistently slow)? Cheers. On 07/03/18 01:49, Kitsda Yonpladyod wrote:

Hi there, Here is a minimum working example: ---------------------------------------------------------------- lower = 0 upper = 1 n_bins = 50 interval = (upper - lower) / n_bins bins = vector(mode="numeric", length=n_bins) breaks = seq(from=lower + interval, to=upper, by=interval) for(idx in breaks) { bins[idx / interval] = idx } print(bins)

Commit-ID: d4ab7343978bbec7141f8462236ba6a47574205f Gitweb: http://git.kernel.org/?p=libs/klibc/klibc.git;a=commit;h=d4ab7343978bbec7141f8462236ba6a47574205f Author: Ben Hutchings <ben at decadent.org.uk> AuthorDate: Sat, 14 Jan 2023 03:15:20 +0100 Committer: Ben Hutchings <ben at decadent.org.uk> CommitDate: Sat, 14 Jan 2023 18:09:28 +0100 [klibc] resource: Avoid using

Short: get rid of the loops I use and optimize runtime Dear all, I want to calculate for each row the amount of the month ago. I use a matrix with 2100 rows and 22 colums (which is still a very small matrix. nrows of other matrixes can easily be more then 100000) Table before Year month quarter yearmonth Service ... Amount 2009 9 Q3 092009 A ...

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")

The following is mainly a capture of parts of multiple off-line discussions within members of the Crossbow team (Gopi, Thiru, Roamer, May-Lin, Thirumailai, Nitin, KB, ...), I thought I''d open it up to other participants. Crossbow''s core scheduling involves switching a NIC (or individual Rx rings on the NIC) to polling mode. The receive interrupt will become not only rarer,

Hi, LLVM 2.1-pre1 test results: Linux (SUSE) on x86 (P4) Release mode, but with assertions enabled LLVM srcdir == objdir # of expected passes 2250 # of expected failures 5 I ran the llvm-test suite on my desktop while I was also working on that PC, so don't put too much trust in the timing info. Especially during the "spiff" test the machine was swapping

I've been playing around with snapshots lately but I've got a problem on one of my servers running 7-STABLE amd64: FreeBSD paladin 7.1-PRERELEASE FreeBSD 7.1-PRERELEASE #8: Mon Nov 10 20:49:51 GMT 2008 tdb@paladin:/usr/obj/usr/src/sys/PALADIN amd64 I run the mksnap_ffs command to take the snapshot and some time later the system completely freezes up: paladin# cd /u2/.snap/ paladin#

HI, Is it possible to specify CONFIG_PREEMPT_RT=y in the Dom0 kernel? It seems that it is forbidden as it is not a menu option in make menuconfig and gets removed if I manually add it to the .config. I want to run a sound server on Dom0 and am having problems with xruns at present, so need to be able to run a preemptive kernel. Thanks, Chris _______________________________________________

Hi, I am running ClamAV process which is invoked by Cron scheduler around 3:00 AM ( Midnight) on remote server running CentOS Linux release 7.6.1810 (Core). Is there a way to measure what was the system resources consumptions like CPU, Network, I/O, Memory and Computing load is being used and is it possible to find out how much time it took to complete the scan? Any help will be highly

Hi all, I computed a Wilks Lambda Test with manova: > df.man <- manova(df.mul ~ 1, na.rm=TRUE) > summary(df.man, intercep=T, test="Wilks") Df Wilks approx F num Df den Df Pr(>F) (Intercept) 1 0.0002824 393.3 9 1 0.03911 * Residuals 9 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*'

Btw., another question that made me wonder already when doing the original patch: why is it that x86-64 properly uses locking for mm_pin_all(), yet i386 doesn''t need to? Jan _______________________________________________ Xen-devel mailing list Xen-devel@lists.xensource.com http://lists.xensource.com/xen-devel

Hi List, I have a raster and a polygon with attribute ID and Class. I want to have the fraction of each class present in each pixel of raster. I have use the raster::extract to get the value and weights as below. ex<-extract(raster,polygon,weighted=TRUE) this gives me [[1]] value weight 13943 0.24 13958 0.02 13959 0.84 13960 0.19 13987 0.03 13988 0.31 13990 0.30 [[2]]

Thank you Jim Holtman and Mark Leeds for your help. Original question: >How do I do the following more concisely? > Bout[is.na(Bout$bd.n), 'bd.n'] <- 0 > Bout[is.na(Bout$ht.n), 'ht.n'] <- 0 > Bout[is.na(Bout$dbh.n), 'dbh.n'] <- 0 >. . . Solution: for (i in c('bd.n', 'ht.n', 'dbh.n')) Bout[is.na(Bout[[i]]), i] <- 0 Toy

Dear list, I made a logistic regression model (MyModel) using lrm and penalization by pentrace for data of 104 patients, which consists of 5 explanatory variables and one binary outcome (poor/good). Then, I found bootcov and robcov function in rms package for calculation of confidence range of coefficients and odds ratio by bootstrap covariance matrix and Huber-White sandwich method,