search for: 0.33

Dear R gurus, I have data for which I want to estimate the markov transition matrix that generated the sequence, and preferably obtain some measure of confidence for that estimation. e.g., for a series such as 1 3 4 1 2 3 1 2 1 3 4 3 2 4 2 1 4 1 2 4 1 2 4 1 2 1 2 1 3 1 I would want to get an estimate of the matrix that generated it [[originally: [,1] [,2] [,3] [,4] [1,] 0.00 0.33 0.33

Hi there, I ran a test on an Apache server, the workload is a helloworld.c compiled cgi, very simple. OS is SLES 10. The stress tool is ab (apache bench). The performance looks big degration from native to Dom0: Running in prefork mode: Native Dom0 Performance(request/s) 3700 . 650 CPU%

Hi there, I ran a test on an Apache server, the workload is a helloworld.c compiled cgi, very simple. OS is SLES 10. The stress tool is ab (apache bench). The performance looks big degration from native to Dom0: Running in prefork mode: Native Dom0 Performance(request/s) 3700 . 650 CPU%

Hi, I have a DF like this: DF = data.frame(read.table(textConnection(" A B C 1 b1 1999 0.25 2 c1 1999 0.25 3 d1 1999 0.25 4 a2 1999 0.25 5 c2 1999 0.25 6 d2 1999 0.25 7 a3 1999 0.25 8 b3 1999 0.25 9 d3 1999 0.25 10 a4 1999 0.25 11 b4 1999 0.25 12 c4 1999 0.25 13 b1 2001 0.5 14 a2 2001 0.5 15 b1 2004 0.33 16 c1 2004 0.33 17 a2 2004 0.33 18 c2 2004

Full_Name: Woolton Lee Version: 2.1 OS: windows Submission from: (NULL) (128.118.224.46) I did the following ('g' and 'h' are both numeric vectors) > i <- abs(g-h) creating a vector 'i' with values, > i [1] 0.08 0.00 0.33 0.00 0.00 0.00 0.00 0.33 0.00 0.00 0.08 0.08 0.20 0.00 0.13 Now, I want to create a new vector =1 whenever 'i' = 0.33 and =0

Hello, I am trying to create multiple matrices (to run a PVA) but can't import all of them from a .csv without the numbers treated as labels and not factors. I can enter the matrix slowly: Site05_96 <- matrix(c(0.07,0,0.03,0.00,NA,0.00, 0.09,0.166666667,0.31,0.42,NA,0.00, 0.00,0,0.00,0.00,NA,0.00, 0.00,0,0.00,0.00,NA,0.00,

I've been attempting to export a graphics file of my lattice plots and no matter what color I use for the bg= argument in the bmp/jpeg/png function, the background is grey. bmp(filename = "c:/my_paper/resids.bmp", width=1024, height=1024, pointsize=10, bg="transparent") # these are lattice plots. plot them in a 3x2 matrix print( hgqqplot, position=c(0.0, 0.66, 0.5, 1.0 ),

CentOS Errata and Bugfix Advisory 2018:0167 Upstream details at : https://access.redhat.com/errata/RHBA-2018:0167 The following updated files have been uploaded and are currently syncing to the mirrors: ( sha256sum Filename ) x86_64: 62c118ef7947f9b6afff67882c85a927454b9e982026a92f5e7ad30d777d42c6 spice-glib-0.33-6.el7_4.1.i686.rpm

The following errata for CentOS-2 have been built and uploaded to the centos mirror: RHSA-2009:0437-02 Critical: seamonkey security update Files available: seamonkey-1.0.9-0.33.el2.c2.1.i386.rpm seamonkey-chat-1.0.9-0.33.el2.c2.1.i386.rpm seamonkey-devel-1.0.9-0.33.el2.c2.1.i386.rpm seamonkey-dom-inspector-1.0.9-0.33.el2.c2.1.i386.rpm seamonkey-js-debugger-1.0.9-0.33.el2.c2.1.i386.rpm

Hi R users, I have a data frame that contains 10K obs and 200 variables where I am trying to format the numeric columns to look like the output table below (format to 2 decimal places) but I am having no luck.. Can someone tell me the best way to accomplist this? Thanks in advance for any help! str(ad.test) 'data.frame': 10,000 obs. of 200 variables: $ ID : Factor

Dear Rhelp, I would like to get alternate tick labels for the xyplot: library(lattice) library(grid) xy <-xyplot(decrease ~ treatment, OrchardSprays, ??? main= "Some plot", ??????? groups = rowpos, type = "a", ??? ??????????? page???? = function(n){ ???????????????????????? grid.text(LETTERS[j], ???????????????????????? y = 0.95, ???????????????????????? x = 0.15,

Dear All, we have: t<-seq(0,24,1) a<-10*exp(-0.05*t) b<-10*exp(-0.07*t) c<-10*exp(-0.1*t) d<-10*exp(-0.03*t) z<-data.frame(a,b,c,d) res<-t(apply(z, 1, quantile, probs=c(0.3))) my goal is to do a 'reverse" of the function here that produces "res" on a data frame, ie: to get the answer 0.3 back for the percentile location when I have

Hi: I compiled and run tftp-hpa-0.33 on my Solaris 2.7 but it does not support file size larger than 16MB. The tftpd time out when 16MB of file is transfered. Can anyone offer a hint for how to enable tftp-hpa to support large file size ? Very appreciate Geefu Wang

Dear R-List, I am sure there must be a very simple way to do this - I just do not know how... This is what I want to do: #my dataframe df<-data.frame(id=c("x01","x02","x03","x04","x05","x06"),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA),sum=c(3,9,15,24,15,6))    id    a     b     c   sum 1 x01  1     2    NA   3 2

$ 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

In this Release I changed all the stuff i was annoyed of. Here the changes: 0.33 We are now also able to load a config-file... - Optional Config-File Support! The Internal Config-Area still exists. I created an example-config and a colorless one for all who don't want a colorful output - Better handling of multiple process protection for the script - Better Checks for set variables

Hi, I have some problems with my repeated measures analysis. When I compute it with SPSS I get different results than with R. Probably I am doing something wrong in R. I have two groups (1,2) both having to solve a task under two conditions (1,2). That is one between subject factor (group) and one within subject factor (task). I tried the following: aov(Score

> On Jun 15, 2017, at 12:37 PM, Andras Farkas via R-help <r-help at r-project.org> wrote: > > Dear All, > > we have: > > t<-seq(0,24,1) > a<-10*exp(-0.05*t) > b<-10*exp(-0.07*t) > c<-10*exp(-0.1*t) > d<-10*exp(-0.03*t) > z<-data.frame(a,b,c,d) > > res<-t(apply(z, 1, quantile, probs=c(0.3))) > > > > my

I'll appreciate if some one can help me with the following loop. This is the logic of the loop, if we have the following data; > x.df x.dif . . . . 102 0.00 103 0.42 104 0.08 105 0.00 106 0.00 107 0.00 108 -0.16 109 -0.34 110 0.00 111 -0.17 112 -0.33 113 0.00 114 0.00 115 0.00 116 0.33 117 0.17 118 0.00 . . . . I'm trying to find i's where for (i

Dear all I am trying nlme together with Pinheiro/Bates book. I constructed grouped data object with suitable plotting layout (according to some common factor, panels from bottom to top are in increasing order). When I do nlme(... some stuff...) I get fitted object which I can plot with plot(augPred(fit.nlme6, level=0:1)) but it results in completely different ordering. Is there any way