similar to: merge performace degradation in 2.9.1

Hello there, I am not quite sure how to interpret the output of Rprof (in the following the output I was staring at). I was poking around the web a little bit for documentation but without much success. I guess if I want to figure out what takes so long in my code the 2nd table $by.total and the total.pct column (pct = percent) is the most helpful. What does it mean that [ or [.data.frame is

Dear R-devel: I could use some advice about matrix calculations and steps that might make for faster computation of generalized inverses. It appears in some projects there is a bottleneck at the use of svd in calculation of generalized inverses. Here's some Rprof output I need to understand. > summaryRprof("Amelia.out") $by.self self.time self.pct

Hi, I am trying to use a CGI service (Pubchem PUG) via RCurl and am running into a problem where the data must be supplied via POST - but I don't know the keyword for the argument. The data to be sent is an XML fragment. I can do this via the command line using curl: I save the XML string to a file called query.xml and then do curl -d @query.xml

On 02/13/2018 at 08:41 AM Floimair Florian wrote: > No you're reading it wrong. > > There are 188K received with no loss, and 16441K transmitted. This doesn't make any sense to me, either. There can't be more packages transmitted than received. It's the same codec in and out and it's been running exactly the same time. > ...........Receive.........

Hi Carsten, On 02/11/2018 at 07:46 PM Carsten Bock wrote: > Hi, > > Lost percent (%).... Are you sure? I'm seeing here: ...........Receive......... .........Transmit.......... Count Lost Pct Jitter Count Lost Pct Jitter RTT.... 188K 0 0 0.000 188K 16641K 8809 0.000 0.026 => This doesn't sound reliable to me: there are 188K packets and 16641K

That's very helpful and instructive, thank you! Jason Stout, MD, MHS Box 102359-DUMC Durham, NC 27710 FAX 919-681-7494 ________________________________ From: John Fox <jfox at mcmaster.ca> Sent: Saturday, October 14, 2023 10:13 AM To: Jason Stout, M.D. <jason.stout at duke.edu> Cc: r-help at r-project.org <r-help at r-project.org> Subject: Re: [R] Create new data frame with

Could this gap in sequence numbers caused by a codec change generate errors like the one below? [2018-02-13 12:57:43] WARNING[4917][C-0004c2cb] codec_sangoma.c: [526559][g722toulaw] Got Seq 15944 but expecting 10106 (time since last read = 0ms), dropped 5838 packets On 02/13/2018 01:24 PM, Andres wrote: > On 2/13/18 11:55 AM, Michael Maier wrote: >> On 02/13/2018 at 08:41 AM Floimair

Under the hood, sapply() is also a loop (at the interpreted level). As is lapply(), etc. -- Bert On Sun, Oct 15, 2023 at 2:34?AM Jason Stout, M.D. <jason.stout at duke.edu> wrote: > > That's very helpful and instructive, thank you! > > Jason Stout, MD, MHS > Box 102359-DUMC > Durham, NC 27710 > FAX 919-681-7494 > ________________________________ > From: John

Dear Rs: Hi, I am trying to read a big text file (nrows=243440, ncols=144). It seems the computational time of all the read methods (scan,readtable,read.delim) is not linear to the number of rows I want to read in: things became really slow once I tried to read in 100000 lines compare to 10000 lines). If I am reading the profiling result right, I guess scan wouldn't help either. My

Hello, (This is follow up from this thread: http://www.nabble.com/execution-time-of-.packages-td22304833.html but with a different focus) I am often confused by the result of the profiler, when a loop is involved. Consider these two scripts: script1: Rprof( ) x <- numeric( ) for( i in 1:10000){ x <- c( x, rnorm(10) ) } Rprof( NULL ) print( summaryRprof( ) ) script2:

I have a query which returns a data set like so: > salaries yearID POS pct 1 2009 RF 203 2 2009 DH 200 3 2009 1B 198 4 2009 3B 180 5 2009 LF 169 6 2009 SS 156 7 2009 CF 148 8 2009 2B 97 9 2009 C 86 10 2008 DH 234 11 2008 1B 199 12 2008 RF 197 13 2008 3B 191 14 2008 SS 180 15 2008 CF 164 16 2008 LF 156 17 2008 2B 104 18 2008

Hello, could somebody please tell me the meaning of "Pct" as seen in asterisk cli: ...........Receive......... .........Transmit.......... Count Lost Pct Jitter Count Lost Pct Jitter RTT.... Thanks, Michael

Well, here's one way to do it: (dat is your example data frame) Cutoff <- seq(0, .15, .01) Pop <- with(dat, sapply(Cutoff, \(p)sum(Totpop[Pct >= p]))) I think there must be a more efficient way to do it with cumsum(), though. Cheers, Bert On Sat, Oct 14, 2023 at 12:53?AM Jason Stout, M.D. <jason.stout at duke.edu> wrote: > > This seems like it should be simple but I

If one makes the reasonable assumption that Pct is much larger than Cutoff, sorting Cutoff is the expensive part e.g O(nlog2(n) for Quicksort (n = length Cutoff). I believe looping is O(n^2). Jeff's approach using findInterval may be faster. Of course implementation details matter. -- Bert On Mon, Oct 16, 2023 at 4:41?AM Leonard Mada <leo.mada at syonic.eu> wrote: > > Dear

Hello, Is the following a bug? I always thought that df$varname <- does the same as df["varname"] <- > df <- data.frame(weight=round(runif(10, 10, 100)), sex=round(runif(100, 0, 1))) > df$pct <- df["weight"] / ave(df["weight"], df["sex"], FUN=sum)*100 > names(df) [1] "weight" "sex" "pct" ###

Any votes for/against? ----- Forwarded message from Rik Faith <faith@alephnull.com> ----- Date: Wed, 28 Nov 2001 12:55:29 -0500 From: Rik Faith <faith@alephnull.com> To: mbp@samba.org Subject: rsync patch X-Mailer: VM 6.96; XEmacs 21.1; Linux 2.4.16 (light) Here is a patch that adds rate information (e.g., kB/s) to the --progress display. I just noticed that 2.4.7pre4 is coming

I have an R function (about 1000 lines long) that takes more than 20 times as long to run under R Windows 1.9.0 and 1.8.1 than it does under 1.7.1. Profile results indicate that the $<-.data.frame operation is the culprit, but I don't understand exactly what that is (assignment of data frame elements to another variable?), or why it's only a problem under 1.8.1 and 1.9.0. Any advice?

This seems like it should be simple but I can't get it to work properly. I'm starting with a data frame like this: Tract Pct Totpop 1 0.05 4000 2 0.03 3500 3 0.01 4500 4 0.12 4100 5 0.21 3900 6 0.04 4250 7 0.07 5100 8 0.09

Hi all, thank you for your answers; i've tried both cca from vegan library, and dudi.coa from ade4 library; one last question: my deal is mainly with contingency tables, like the one i'm posting here acciaieria<-c(.41,.02,.44,.04,.09) laminatoio<-c(.34,.28,.26,.01,.11) fonderia<-c(.48,.05,.34,.08,.05) leghe<-c(.45,.19,.25,.03,.08)

I generate a groupedData object library(nlme) obj <- groupedData(mg10 ~ time | gp, data = common, outer = ~pct) gp has 101 levels, and pct has 3. There are 38, 25, 38 gps in each of the levels of pct respectively. I fit my model fit.rtg <- lme(mg10 ~ time * group, data = obj, random = ~time * group | gp) Now I try to plot the results. I would like to print 40 panels on each