similar to: dirattr.233808-rsynckbattr17 Zero K files

Hi, Can you help me since upgrading the latest version of rsync on my Xservers I have lost all resouce fork information and I am losing the file types on the backup server. The command I am using is sudo rsync -e ssh -avz /Times_QPS/ 192.168.22.72:/Times_QPS/ Am I missing something simple? -Jason Ferguson *********************************************************************************

I have the following as a line in a script kicked off from cron: sudo rsync -e ssh -avz /Times_QPS/ 192.168.22.72:/Times_QPS/ --eahfs It logs the output to a log file, however rsync caused the log file to go to an 80gb file and fill up my disk because it threw a mental when a file it went to sync was no longer there! Is there some way to get the sync to ignore errors? -Jason

Hi, I am trying to run rsync on OS X. I have it working but I get ?.-dirattr.095021 .-dirattr.095046? files in my destination directory. Every time I run rsync it creates another file. This is the string that I am using to run rsync ?rsync --progress --stats --rsh=/usr/bin/ssh --recursive --times -og --links user@Server:/Volumes/Production/test /Volumes/Production/test/? . Any ideas on how

I'm trying to use rsync to backup some database files on a Linux box to an OS X machine, so I can write them to a tape drive connected to the OS X machine. I'm aware that there are some "issues" about rsync and certain file forks found on HFS+ filesystems, but I'm not sure that should be stopping me from accomplishing what I want to do. I'm using rsync 2.5.5 on both

Hi, does anybody know, how a bash shell script looks, which automatically enters the ssh password? the rsync call should be: rsync -avz -e ssh /home/johndoe/data/repository billythegate@192.168.10.102:/home/johndoe/junk the call causes a password question. Is ist possible to automate that by a cron? If yes how should the script look to automatically enter the password? ThanX for an answer.

Hello, I've just spent several hours going over several Google searches trying to find a way to configure rsync to log into a file named "/var/log/rsync.log". So far, every instance where I've found someone asking about rsync logging remained unanswered (which is kind of weird in itself). As far as I can tell, the only way to do this is to setup rsync as a daemon process so

> > Message: 6 > Date: Wed, 15 Feb 2006 15:32:46 -0800 > From: Ezra Zygmuntowicz <ezra@yakima-herald.com> > Subject: [Rails] Rubuntu Live CD for Rails > To: Rails Mailing List <rails@lists.rubyonrails.org> > Message-ID: <EBB25945-FF95-4A90-9574-3AB85835B4B2@yakima-herald.com> > Content-Type: text/plain; charset=US-ASCII; delsp=yes; format=flowed > >

I asked: > In this discussion of seq(), can anyone explain to > me _why_ seq(to=n) and seq(length=3) have different > types? Martin Maechler <maechler@stat.math.ethz.ch> replied: well, the explantion isn't hard: look at seq.default :-) That's the "efficient cause", I was after the "final cause". That is, I wasn't asking "what is it

Hi, I wonder if the following (apparent) inconsistency is a bug or feature. Since scalars are simply vectors of length one I would think that a and a:a produce the same result. For example, > identical(4.01,4.01:4.01) [1] TRUE However, identical(4,4:4) [1] FALSE and > identical(4.0,4.0:4.0) [1] FALSE A closer look reveals that the colon operator produces objects

I am using openssh version 3.6p1 under AIX 5.1. I am trying to change the login herald. I have changed /etc/security/login.cfg, and when logging in via telnet, I see the new herald. When I log in through ssh, using public key authentication, I get the default herald. I am not sure if I am changing the correct file. If I have missed something in the documentation I apologize, but I don't see

The poly() function can create more variables than can be fitted when there are replicated values. In the example below, 'x' has only 5 distinct values, but I can apparently fit a 12th-degree polynomial with no error messages or even nonzero coefficients: R> x = rep(1:5,3) R> y = rnorm(15) R> lm(y ~ poly(x, 12)) Call: lm(formula = y ~ poly(x, 12)) Coefficients:

On 14/01/2020 10:07 a.m., peter dalgaard wrote: > Yep, that looks wrong (probably want to continue discussion over on R-devel) > > I think the culprit is here (in src/nmath/choose.c) > > if (k < k_small_max) { > int j; > if(n-k < k && n >= 0 && R_IS_INT(n)) k = n-k; /* <- Symmetry */ > if (k < 0) return 0.;

It took me by surprise to find that choose(4,5) delivers [1] NaN Warning message: NaNs produced in: choose(n, k) If we look at choose(4,5) as the number of ways of choosing 5 objects from 4 I would have expected 0 as result. Furthermore dhyper(5,4,6,5) does deliver 0 and this essentially equivalent to choose(4,5)*choose(6,0)/choose(10,5).

On 14/01/2020 10:50 a.m., peter dalgaard wrote: > > >> On 14 Jan 2020, at 16:21 , Duncan Murdoch <murdoch.duncan at gmail.com> wrote: >> >> On 14/01/2020 10:07 a.m., peter dalgaard wrote: >>> Yep, that looks wrong (probably want to continue discussion over on R-devel) >>> I think the culprit is here (in src/nmath/choose.c) >>> if (k

I sometimes buy myself a default value for a hash using something like this: h = Hash.new{|h,k| h[k] = ""}. I can now pass ''h'' any key and if a value has not yet been associated with that key I receive an empty string. I would love to do something similar in my partials so that I wouldn''t have to worry about passing in every variable that is referenced.

Yep, that looks wrong (probably want to continue discussion over on R-devel) I think the culprit is here (in src/nmath/choose.c) if (k < k_small_max) { int j; if(n-k < k && n >= 0 && R_IS_INT(n)) k = n-k; /* <- Symmetry */ if (k < 0) return 0.; if (k == 0) return 1.; /* else: k >= 1 */ if n is a near-integer, then k

At the risk of throwing oil on a fire. If we are talking about fractional values of choose() doesn't it make sense to look to the gamma function for the correct analytic continuation? In particular k<0 may not imply the function should evaluate to zero until we get k<=-1. Example: ``` r choose(5, 4) #> [1] 5 gchoose <- function(n, k) { gamma(n+1)/(gamma(n+1-k) * gamma(k+1))

Bringing up an old topic on a small technicality. In the help documentation on seq. Value: The result is of 'mode' '"integer"' if 'from' is (numerically equal to an) integer and 'by' is not specified. The arguement in seq specifying length also creates "double" which is not obvious in the wording, as "by" is not specified

Bringing up an old topic on a small technicality. In the help documentation on seq. Value: The result is of 'mode' '"integer"' if 'from' is (numerically equal to an) integer and 'by' is not specified. The arguement in seq specifying length also creates "double" which is not obvious in the wording, as "by" is not specified

This struck me as incorrect: > choose(3.999999, 4) [1] 0.9999979 > choose(3.9999999, 4) [1] 0 > choose(4, 4) [1] 1 > choose(4.0000001, 4) [1] 4 > choose(4.000001, 4) [1] 1.000002 Should base::choose(n, k) check whether n is within machine precision of k and return 1? Thanks, Erik *** sessionInfo() R version 3.6.0 beta (2019-04-15 r76395) Platform: x86_64-apple-darwin15.6.0