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CentOS Errata and Security Advisory 2009:0398 https://rhn.redhat.com/errata/RHSA-2009-0398.html The following updated files have been uploaded and are currently syncing to the mirrors: s390: updates/s390/RPMS/seamonkey-1.0.9-0.36.el3.centos3.s390.rpm updates/s390/RPMS/seamonkey-chat-1.0.9-0.36.el3.centos3.s390.rpm updates/s390/RPMS/seamonkey-devel-1.0.9-0.36.el3.centos3.s390.rpm

Send CentOS-announce mailing list submissions to centos-announce at centos.org To subscribe or unsubscribe via the World Wide Web, visit http://lists.centos.org/mailman/listinfo/centos-announce or, via email, send a message with subject or body 'help' to centos-announce-request at centos.org You can reach the person managing the list at centos-announce-owner at centos.org When

CentOS Errata and Security Advisory CESA-2009:0398 seamonkey security update for CentOS 3 x86_64: https://rhn.redhat.com/errata/RHSA-2009-0398.html The following updated file has been uploaded and is currently syncing to the mirrors: x86_64: updates/x86_64/RPMS/seamonkey-1.0.9-0.36.el3.centos3.i386.rpm updates/x86_64/RPMS/seamonkey-1.0.9-0.36.el3.centos3.x86_64.rpm

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

CentOS Errata and Security Advisory CESA-2009:0398 seamonkey security update for CentOS 3 i386: https://rhn.redhat.com/errata/RHSA-2009-0398.html The following updated file has been uploaded and is currently syncing to the mirrors: i386: updates/i386/RPMS/seamonkey-1.0.9-0.36.el3.centos3.i386.rpm updates/i386/RPMS/seamonkey-chat-1.0.9-0.36.el3.centos3.i386.rpm

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

CentOS Errata and Security Advisory 2009:0398 https://rhn.redhat.com/errata/RHSA-2009-0398.html The following updated files have been uploaded and are currently syncing to the mirrors: ia64: updates/ia64/RPMS/seamonkey-1.0.9-0.36.el3.centos3.ia64.rpm updates/ia64/RPMS/seamonkey-chat-1.0.9-0.36.el3.centos3.ia64.rpm updates/ia64/RPMS/seamonkey-devel-1.0.9-0.36.el3.centos3.ia64.rpm

Hello, I would like to run a series of regressions on my data (response variable over time): 1) regression from T1 to T2 2) regressions from T1 through T3 3) regression from T1 through T4, etc. I have been struggling to find a way to do this through commands, as opposed to cutting up the data manually (my dataset has over 6000 rows/observations). An illustrative dataset can be created thusly:

I have a data frame called e, dim is 27,3, the first 5 lines look like this: V1 V2 V3 V4 1 1673 0.36 0.08 Smith 2 167 0.36 0.08 Allen 3 99 0.37 0.06 Allen 4 116 0.38 0.07 Allen 5 95 0.41 0.08 Allen I am trying to calculate the proportion/percentage of V1 which would have values >0.42 if V2 was the mean of a normal distribution with V1

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")

I tried sending this a day or two ago but I didn''t see it appear. I''ve just tried the wxRuby 0.36 release but appear to get the following error (I got it from bigdemo.rb too). C:\opt\ruby\lib\ruby\gems\1.8\gems\wxruby2-preview-0.0.36-i386-mswin32\samples\minimal>minimal Our Initialize was called Their Initialize returned 1

I have a dataframe say: date price_g price_s 0.34 0.56 0.36 0.76 . . . . . . and so on. say, 1000 rows. Is it possible to add two columns to this dataframe, by computing say diff(log(price_g) and diff(log(price_s)) ? The elements in the first row of these columns cannot be computed, but

I have the clinical study data. Year 0 Year 3 Retinol (nmol/L) N Mean +-sd Mean +-sd Vitamin A group 73 1.89+-0.36 2.06+-0.53 Trace group 57 1.83+-0.31 1.78+-0.30 where N is the number of male for the clinical study. I want to test if the mean serum retinol has increased over 3 years among subjects in the vitamin A group. > 1.89+0.36

Dear list, Im trying to do the following operation but im not able to do it This is my table: 1 2 3 1 0 7 4 2 0 2 0 3 0 1 3 4 0 3 4 what i would like to do is divide each row values with the corresponding column' sum,namely: 1 2 3 1 0 0.54 0.36 2 0 0.15 0 3 0 0.08

Processing commands for control at bugs.debian.org: > user debian-qa at lists.debian.org Setting user to debian-qa at lists.debian.org (was anbe at debian.org). > tags 714794 - moreinfo Bug #714794 {Done: Gilles Filippini <pini at debian.org>} [release.debian.org] pu: package sikuli/1.0~x~rc3.tesseract3-dfsg1-5+deb7u1 Removed tag(s) moreinfo. > tags 710035 - moreinfo Bug #710035

$ R --version R 1.6.1 (2002-11-01). So I would like to perform principal components analysis on a 16X16 correlation matrix, [princomp(cov.mat=x) where x is correlation matrix], the problem is princomp complains that it is not non-negative definite. I called eigen() on the correlation matrix and found that one of the eigenvectors is close to zero & negative (-0.001832311). Is there any way

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

Hello tftp-hpa community, I am trying to locate a tftp server that I can integrate with an external program/database application, and was wondering if tftp-hpa might have the capability I am looking for. Upon the tftp server receiving a read request, and before it sending a read request reply, I would like to communicate to an external program (and/or database). In real-time, the external

Short: get rid of the loops I use and optimize runtime Dear all, I want to calculate for each row the amount of the month ago. I use a matrix with 2100 rows and 22 colums (which is still a very small matrix. nrows of other matrixes can easily be more then 100000) Table before Year month quarter yearmonth Service ... Amount 2009 9 Q3 092009 A ...