similar to: Cluster GUI package worth publishing/enhancing?

Good morning! After I changed some permissions and owners of some files where Rowland told me, I have the next escenary: [root at proxy ~]# systemctl status named.service ● named.service - Berkeley Internet Name Domain (DNS)    Loaded: loaded (/usr/lib/systemd/system/named.service; enabled; vendor preset: disabled)    Active: active (running) since Mon 2018-06-11 08:54:10 AST; 12min ago  

I forgot to say that  I updated Centos from 7.4 to 7.5, and I updated samba4 to the new version. This Would be a problem of records of something like that. José Fermín Francisco Ferreras Registered User #579535 (LinuxCounter.net) El lunes, 11 de junio de 2018 9:45:03 a. m. GMT-4, Fermin Francisco <abcddo at yahoo.com> escribió: Sorry, the real e-mail is this: [root at pc ~]#

Sorry, the real e-mail is this: [root at pc ~]# systemctl status named.service ● named.service - Berkeley Internet Name Domain (DNS)    Loaded: loaded (/usr/lib/systemd/system/named.service; enabled; vendor preset: disabled)    Active: active (running) since Mon 2018-06-11 08:54:10 AST; 12min ago   Process: 1276 ExecStart=/usr/sbin/named -u named -c ${NAMEDCONF} $OPTIONS (code=exited,

----- Mensagem Original -----=20 Eu boicotaria estas empresas s=F3 pelo fato de financiarem os enjoativos e = rancentos Cassetas, mas me surpreendi ao descobrir que tantos produtos que = eu costumo comprar s=E3o transg=EAnicos. Fui conferir no site do greenpeace= e =E9 verdade!!!!!!!!!!!!! Boicote neles! ----- Original Message -----=20 C A M P A N H A=20 N A T A L S E M B A I X A R I A

This has got to be incredibly simple but I nevertheless can't figure it out as I am apparently brain dead. I just want to convert the elements of a character vector to variable names, so as to then assign formulas to them, e.g: z = c("model1","model2"); I want to assign formulas, such as lm(y~x[,1]) and lm(y~x[,2]), to the variables "model1" and

I admit that I've not done a thorough search on this topic, but from the several instructional manuals and/or tutorials I've looked at, I don't see any mention of relational database capabilities in R? Have I missed something, and if so, can someone point me in the right direction to get started? Thanks! Jim Bouldin, PhD Research Ecologist Department of Plant Sciences, UC Davis

I realize this is probably pretty basic but I can't figure it out. I'm looping through an array, doing various calculations and producing a resulting data frame in each loop iteration. I need to give each data frame a different name. Although I can easily create a new character string for writing each frame to an output file, I cannot figure out how to convert such strings to

Hi, I am trying to install R-0.64.0 version on SGI machine running IRIX 6.5. Machine details: 319 <mind.krasnow.gmu.edu:R-0.63.3> uname -a IRIX64 mind 6.5 07271714 IP27 Configure script runs fine and the output at the end of configure is: R is now configured for mips-sgi-irix6.5 Source directory: /usr/local/R-0.64.0 Installation directory: /usr/local C compiler:

I'm trying to run this simple random sample procedure and keep getting the error message shown. I don't understand this; I've designated x as a numeric vector, so what is going on here? Thanks. > x = as.vector(c(1:12));x [1] 1 2 3 4 5 6 7 8 9 10 11 12 > mode(x) [1] "numeric" > sample(x, 3) Error in sample(x, 3) : .Random.seed is not an integer vector

I'm trying to obtain the mean of the middle 95% of the values from each row of a matrix (that is, the highest and lowest 2.5% of values in each row are removed before calculating the mean). I am having all sorts of problems with this; for example the command: apply(matrix1,1,function(x) quantile(c(.05,.90),na.rm=T)) returns the exact same quantile values for each row, which is clearly

I can't seem to find info on how to unload packages that have been loaded. My goal in doing so is to gain access to functions that have been masked out by those packages. Or is there another way to do so? Thanks in advance. Jim Bouldin, PhD Research Ecologist Department of Plant Sciences, UC Davis Davis CA, 95616 530-554-1740

I realize this should be simple but I'm having trouble subsetting vectors and matrices, for example extracting all values meeting a certain criterion, from a vector. Cannot seem to figure out the correct syntax and help page not very helpful. Or should I be using some other function than subset. Thanks for any help. Jim Bouldin

When I try to run the following non-linear regression with variables index1 and prl3: > beta = 4 > nls(index1~beta*(1/prl3),start = list(beta = 4)) I get this error message: Error in nls(index1 ~ beta * (1/prl3), start = list(beta = 4)) : REAL() can only be applied to a 'numeric', not a 'logical' I've got no clue as to the REAL() to which this is referring. Any

Is there a command for partial matching of character strings? Specifically, I'd like to be able to calculate the mean of the values in any columns in a data frame or matrix that have identity in part of their column names. For example, columns labeled "mpw06a" and "mpw06b" match on the first five characters; their mean would be taken whereas any columns beginning with

I just upgraded from 2.8.1 to 2.10 on Windows Vista. BIG MISTAKE apparently because now when I type: > help(functionname) or ?functionname I get only a small text window giving some very basic info on the topic, e.g.: base-package package:base R Documentation The R Base Package Description: Base R functions Details: This package contains the

Mathew Brown Institute of Bioclimatology University of G?ttingen B?sgenweg 2 37077 G?ttingen, Germany t: +49 551 39 9359 mathew.brown at forst.uni-goettingen.de On 9/24/2011 6:00 PM, r-help-request at r-project.org wrote: > Send R-help mailing list submissions to > r-help at r-project.org > > To subscribe or unsubscribe via the World Wide Web, visit >

OK, what is the trick to extracting the overall p value from an lm object? It shows up in the summary(lm(model)) output but I can't seem to extract it: > test2 = apply(aa, 1, function(x) summary(lm(x[,1] ~ 0 + x[,3] + x[,6]))) > test2[[1]] Call: lm(formula = x[, 1] ~ 0 + x[, 3] + x[, 6]) [omitted summary output] F-statistic: 40.94 on 2 and 7 DF, p-value: 0.0001371 It does not seem

I continue to have great frustrations with NA values--in particular making summary calculations on rows or cols of a matrix containing them. For example, why does: > a = matrix(1:30,nrow=5) > is.na(a[c(1:2),c(3:4)]);a [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1 6 NA NA 21 26 [2,] 2 7 NA NA 22 27 [3,] 3 8 13 18 23 28 [4,] 4 9 14 19 24 29

Hi, keine ahnung. Das liegt jetzt bei Hr. Feld. Frag mal bei ihm nach. Gr??e Thushyanthan r-help-request at r-project.org wrote: > Send R-help mailing list submissions to > r-help at r-project.org > > To subscribe or unsubscribe via the World Wide Web, visit > https://stat.ethz.ch/mailman/listinfo/r-help > or, via email, send a message with subject or body 'help'

This seemingly should be quite simple but I can't solve it: I have a long character vector of geographic data (data frame column named "XY") whose elements vary in length (from 11 to 14 chars). Each element is structured as a set of digits, then an underscore, then more digits, e.g: > data.frame(head(as.character(XY))) head.as.character.XY.. 1 -448623_854854 2