similar to: beginner's questions about lme, fixed and random effects

Will the various client packages (centos in my case) be able to automatically handle the upgrade vs new install decision, or will we be required to do something manually to determine that? It?s a little unclear that things will continue without interruption because of the way you describe the change from GD1 to GD2, since it sounds like it stops GD1. Early days, obviously, but if you could

On Thu, Nov 2, 2017 at 7:53 PM, Darrell Budic <budic at onholyground.com> wrote: > Will the various client packages (centos in my case) be able to > automatically handle the upgrade vs new install decision, or will we be > required to do something manually to determine that? We should be able to do this with CentOS (and other RPM based distros) which have well split glusterfs

Just so I am clear the upgrade process will be as follows: upgrade all clients to 4.0 rolling upgrade all servers to 4.0 (with GD1) kill all GD1 daemons on all servers and run upgrade script (new clients unable to connect at this point) start GD2 ( necessary or does the upgrade script do this?) I assume that once the cluster had been migrated to GD2 the glusterd startup script will be smart

On Fri, Nov 3, 2017 at 8:50 PM, Alastair Neil <ajneil.tech at gmail.com> wrote: > Just so I am clear the upgrade process will be as follows: > > upgrade all clients to 4.0 > > rolling upgrade all servers to 4.0 (with GD1) > > kill all GD1 daemons on all servers and run upgrade script (new clients > unable to connect at this point) > > start GD2 ( necessary or

On Thu, Nov 2, 2017 at 4:00 PM, Amudhan P <amudhan83 at gmail.com> wrote: > if doing an upgrade from 3.10.1 to 4.0 or 4.1, will I be able to access > volume without any challenge? > > I am asking this because 4.0 comes with DHT2? Very short answer, yes. Your volumes will remain the same. And you will continue to access them the same way. RIO (as DHT2 is now known as) developers

if doing an upgrade from 3.10.1 to 4.0 or 4.1, will I be able to access volume without any challenge? I am asking this because 4.0 comes with DHT2? On Thu, Nov 2, 2017 at 2:26 PM, Kaushal M <kshlmster at gmail.com> wrote: > We're fast approaching the time for Gluster-4.0. And we would like to > set out the expected upgrade strategy and try to polish it to be as > user

does RIO improves folder listing and rebalance, when compared to 3.x? if yes, do you have any performance data comparing RIO and DHT? On Thu, Nov 2, 2017 at 4:12 PM, Kaushal M <kshlmster at gmail.com> wrote: > On Thu, Nov 2, 2017 at 4:00 PM, Amudhan P <amudhan83 at gmail.com> wrote: > > if doing an upgrade from 3.10.1 to 4.0 or 4.1, will I be able to access > > volume

We're fast approaching the time for Gluster-4.0. And we would like to set out the expected upgrade strategy and try to polish it to be as user friendly as possible. We're getting this out here now, because there was quite a bit of concern and confusion regarding the upgrades between 3.x and 4.0+. --- ## Background Gluster-4.0 will bring a newer management daemon, GlusterD-2.0 (GD2),

Dear R experts, I am having difficulty using loops productively and would like to please ask for advice. I have a dataframe of ids and groups. I would like to break down the dataframe into groups, find the unique sets of ids, then reassemble. My thought was to use a loop, but I have been unable to finish this loop in a logical way. I would like to find the unique ids for group 1, group 2,

ls is a list of character vectors created by strsplit() I want to concatenate the 1st 4 character elements of each list item as a new vector called file. I admit to being confused about list syntax even after numerous readings. Here's what I tried: ls <- list(c("Focused", "10k", "A12", "t04.tif", "+", "µm"),

Hi Bill, I put stringsAsFactors = FALSE still did not work. tdat <- read.table(textConnection("A B C Y A12 B03 C04 0.70 A23 B05 C06 0.05 A14 B06 C07 1.20 A25 A23 A12 3.51 A16 A25 A14 2,16"),header = TRUE ,stringsAsFactors = FALSE) tdat$D <- 0 tdat$E <- 0 tdat$D <- (ifelse(tdat$B %in% tdat$A, tdat$A[tdat$B], 0)) tdat$E <- (ifelse(tdat$B %in% tdat$A, tdat$A[tdat$C], 0))

Dear R helpers, please ignore my earlier mail. Here is the corrected mail. Please forgive me for the lapses on my part. Extremely sorry.   Here is the corrected mail.     Dear R helpers   I am having following data   Name           Numbers A11                  12 A12                  17  A13                   0 A11                  11  A12                   6 A13                   0

Thank you Rui, I did not get the desired result. Here is the output from your script A B C Y D E 1 A12 B03 C04 0.70 0 0 2 A23 B05 C06 0.05 0 0 3 A14 B06 C07 1.20 0 0 4 A25 A23 A12 3.51 1 1 5 A16 A25 A14 2,16 4 4 On Wed, Dec 13, 2017 at 4:36 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote: > Hello, > > Here is one way. > > tdat$D <- ifelse(tdat$B %in% tdat$A,

Hi all, I have a data frame tdat <- read.table(textConnection("A B C Y A12 B03 C04 0.70 A23 B05 C06 0.05 A14 B06 C07 1.20 A25 A23 A12 3.51 A16 A25 A14 2,16"),header = TRUE) I want match tdat$B with tdat$A and populate the column values of tdat$A ( col A and Col B) in the newly created columns (col D and col E). please find my attempt and the desired output below Desired output

Use the stringsAsFactors=FALSE argument to read.table when making your data.frame - factors are getting in your way here. Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, Dec 13, 2017 at 3:02 PM, Val <valkremk at gmail.com> wrote: > Thank you Rui, > I did not get the desired result. Here is the output from your script > > A B C Y D E > 1 A12 B03 C04 0.70 0 0

All, I'd like to plot the main relationship of a grouped data object for all levels of a factor in a single panel. The sample code below creates a separate panel for each level of the factor. I realize that this could be done in other ways, but I'd like to do it via plotting the grouped data object. thanks! dave z = rnorm(18, mean=0, sd=1) x = rep(1:6, 3) y =

Hello, Here is one way. tdat$D <- ifelse(tdat$B %in% tdat$A, tdat$A[tdat$B], 0) tdat$E <- ifelse(tdat$B %in% tdat$A, tdat$A[tdat$C], 0) Hope this helps, Rui Barradas On 12/13/2017 9:36 PM, Val wrote: > Hi all, > > I have a data frame > tdat <- read.table(textConnection("A B C Y > A12 B03 C04 0.70 > A23 B05 C06 0.05 > A14 B06 C07 1.20 > A25 A23 A12 3.51

Hi, I understand the idea behind compare_numeric() is to compare strings containing digits in a special way: Do a normal string-compare up to the point where both string elemnts are numerical. Find then an outcome based on the number of consecutive digits in the strings while disregarding the value of the digits, eg a12b < a123. I guess then this order should hold: a12 == a22 < a1b, for

Good morning! Please, could you help me with the problem? I have a matrix *144x73.* An example of a matrix: [,1] [,2] [,3] [,4] [,5] [1,] 277.4 276.24 275.62 276.55 278.05 [2,] 277.4 276.24 275.55 276.42 277.72 [3,] 277.4 276.24 275.50 276.22 277.39 [4,] 277.4 276.24 275.42 276.02 277.02 [5,] 277.4 276.22 275.37 275.82 276.64 And I want to *compare*its cells

??? > y <- sort( c("a1","a2","a10","a12","a100")) > y [1] "a1" "a10" "a100" "a12" "a2" > mixedsort(y) [1] "a1" "a2" "a10" "a12" "a100" **Please read the docs!** They say that mixedsort() and mixedorder() both take a **single