similar to: applying a function to data frame columns

Hi, We're running a cluster with two data nodes and one arbiter, and have sharding enabled. We had an issue a while back where one of the server's crashed, we got the server back up and running and ensured that all healing entries cleared, and also increased the server spec (CPU/Mem) as this seemed to be the potential cause. Since then however, we've seen some strange behaviour,

Hi, you mentioned that the arbiter bricks run out of inodes.Are you using XFS ?Can you provide the xfs_info of each brick ? Best Regards,Strahil Nikolov? On Sat, Jul 1, 2023 at 19:41, Liam Smith<liam.smith at ek.co> wrote: Hi, We're running a cluster with two data nodes and one arbiter, and have sharding enabled. We had an issue a while back where one of the server's

Hi, Thanks for your response, please find the xfs_info for each brick on the arbiter below: root at uk3-prod-gfs-arb-01:~# xfs_info /data/glusterfs/gv1/brick1 meta-data=/dev/sdc1 isize=512 agcount=31, agsize=131007 blks = sectsz=512 attr=2, projid32bit=1 = crc=1 finobt=1, sparse=1, rmapbt=0 =

Hi Liam, I saw that your XFS uses ?imaxpct=25? which for an arbiter brick is a little bit low. If you have free space on the bricks, increase the maxpct to a bigger value, like:xfs_growfs -m 80 /path/to/brickThat will set 80% of the Filesystem for inodes, which you can verify with df -i /brick/path (compare before and after).?This way?you won?t run out of inodes in the future. Of course, always

Hi Strahil, We're using gluster to act as a share for an application to temporarily process and store files, before they're then archived off over night. The issue we're seeing isn't with the inodes running out of space, but the actual disk space on the arb server running low. This is the df -h? output for the bricks on the arb server: /dev/sdd1 15G 12G 3.3G 79%

Thanks for the clarification. That behaviour is quite weird as arbiter bricks should hold?only metadata. What does the following show on host?uk3-prod-gfs-arb-01: du -h -x -d 1?/data/glusterfs/gv1/brick1/brickdu -h -x -d 1?/data/glusterfs/gv1/brick3/brickdu -h -x -d 1 /data/glusterfs/gv1/brick2/brick If indeed the shards are taking space -?that is a really strange situation.From which version

Hi Strahil, This is the output from the commands: root at uk3-prod-gfs-arb-01:~# du -h -x -d 1 /data/glusterfs/gv1/brick1/brick 2.2G /data/glusterfs/gv1/brick1/brick/.glusterfs 24M /data/glusterfs/gv1/brick1/brick/scalelite-recordings 16K /data/glusterfs/gv1/brick1/brick/mytute 18M /data/glusterfs/gv1/brick1/brick/.shard 0

Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1

Hi, let's see, if someone can help my with this one: I have the string as follows: > str<-("one","two","three") Now I want to concatenate the items to one string, seperateted by space or something else, >str >"one, two, three" If possible without a loop. My actual goal ist to create string like >str.names >"female = names1, male

Hi all, probably really simple to solve, but having no background in programming I haven't been able to figure this out: I have two dataframes like df1 <- data.frame(names1=c('aa','ab', 'ac', 'ad'), var1=c(1,5,7,12)) df2 <- data.frame(names2=c('aa', 'ab', 'ac', 'ad', 'ae'), var2=c(3,6,9,12,15)) Now I want merge

Hi, I am trying to install R on my Ultra-10 running Solaris 2.6.1. and got following massege when i configure. # ./configure loading cache ./config.cache checking for a BSD compatible install... /opt/apps/R-0.62.4/etc/install-sh -c checking whether ln -s works... yes checking for ranlib... : checking for bison... no checking for byacc... no checking for ar... no checking for ratfor... no checking

Hello gurus, I have some variables, and i am creating combinations for analysis in the end i need these variables to be displayed like "LEPTIN+SAA+PTH". currently i am using loop to perform this. I would appreciate any pointers to do it without the loop. > mols=c("LEPTIN","SAA","PTH","sEGFR") > samples=mols[1:3] > samples [1]

Dear R-Helpers, this is my first post ever to a mailing list, so please feel free to point out any missunderstandings on my side regarding the conventions of this mailing list. My problem: Assuming the following character vector is given: names <- c("filia Maria", "vidua Joh Dirck Kleve (oo 02.02.1732)", "Bernardus Engelb Franciscus Linde j.u.Doktor referendarius

Hi R users: I got this code to generate a graphic for each member of a lists. list1<-list(A=data.frame(x=c(1,2),y=c(5,6)),B=data.frame(x=c(8,9),y=c(12,6))) names1<-names(list1) sapply(1:length(list1),function(i) with(list1[[i]],plot(x,y,type="l",main=paste("Graphic of",names1[i])))) Is there a more elegant solution for not to use two separate lists? I would like to

Hi, If needed, lapply() tries to convert its first argument into a list before it starts doing something with it: > lapply function (X, FUN, ...) { FUN <- match.fun(FUN) if (!is.vector(X) || is.object(X)) X <- as.list(X) .Internal(lapply(X, FUN)) } But in practice, things don't always seem to "work" as suggested by this code (at least to the

useRs, I am looking to find the minimum positive value of some data I have. Currently, I am able to find the minimum of data after I apply some other functions to it: > x [1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 > sort(x) [1] 0 1 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 > diff(sort(x)) [1] 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 > min(diff(sort(x))) [1] 0

Hi R users, This seems like a simple problem but I have searched nabble for the answer and can't seem to find it. All I want to do is produce a boxplot where I have two boxes for one Individual but on the xaxis I only have one tick mark centred between the boxes so I can add the Individuals' name. I have 30 IDs and have shown the code I use below for a couple of IDs, I figure the data

useR's, I am trying trying to find out if there is a faster way to do a certain computation. I have successfully used FOR loops and the apply function to do this, but it can take some time to fully compute, but I was wondering if anyone may know of a different function or way to do this: > x [1] 1 2 3 4 5 > xk [1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 I want to do:

useR's, I want to randomly generate 500 numbers from the standard normal distribution, i.e. N(0,1), but I only want them to be generated in the range -1.5 to 1.5. Does anyone have a simple way to do this? Thanks, dxc13 -- View this message in context: http://www.nabble.com/simple-random-number-generation-tp18642611p18642611.html Sent from the R help mailing list archive at Nabble.com.

useR's I am trying to color the points on a scatter plot (code below) with two colors. Red for values 0.5 -1.0 and blue for 0.0 - .49. Does anyone know a easy way to do this? x <- runif(100, 0, 1) y <- runif(100, 0, 1) plot(y ~ x, pch=16) Thanks, dxc13 -- View this message in context: http://www.nabble.com/color-ranges-on-a-2D-plot-tp14893457p14893457.html Sent from the R help