search for: 0.18

Both of the approx functions work correctly individually, but they are not being distinguished in the for loop by the if statments. Any help would be appreciated. for loop of interest is below x <- (structure(list(Site = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,

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,

October 16, 2005 - Release 0.18 of Open Clip Art Library (http://www.openclipart.org), containing over 6900 SVG files created by over 500 artists, is now available for download and use. For the month of September, the Open Clip Art Library sought imagery related to pets including images of different breeds appropriate for use by animal shelters. For the upcoming 0.19 release, due November 1,

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 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

Dear R users: I've a list containing parameters (A , B & C), and I want create a data frame using data in these list. > coffee [[1]] A B C 1 1 0.5 7.78 2 1 1.0 6.06 3 1 2.0 3.67 4 1 4.0 1.35 5 1 6.0 0.49 6 1 8.0 0.18 [[2]] A B C 1 2 5 7.78 2 2 10 6.06 3 2 20 3.67

Hi all, I have 2 questions: 1)How do I calculate the mean on an imported txt file? I've imported the file below and that's what it looks like imported. How do I then calcuate the mean, median, or mode on the column LeafArea using the desktop R package? Any help would be greatly appreciated!! Thanks, Nat LeafType Leaflets LeafArea ShapeRatio LeafWeight LeafThickness 1 1

Hi, To plot a figure, I used lwd=0.18 in the function lines() to plot very fine lines. However, using png() or pdf() to save the figure, it appears that lwd has no effect on pdf() beyond some small number (e.g. 0.18) and the lines looks much thicker than png(). Setting lwd<0.18 does not make the lines any finer in pdf(). I also tried specifying the argument lwd in par() instead and still faced

The following errata for CentOS-2 have been built and uploaded to the centos mirror: RHSA-2008:0599-01 Critical: seamonkey security update Files available: seamonkey-1.0.9-0.18.el2.c2.1.i386.rpm seamonkey-chat-1.0.9-0.18.el2.c2.1.i386.rpm seamonkey-devel-1.0.9-0.18.el2.c2.1.i386.rpm seamonkey-dom-inspector-1.0.9-0.18.el2.c2.1.i386.rpm seamonkey-js-debugger-1.0.9-0.18.el2.c2.1.i386.rpm

Hallo, I am totally confused why Excel and R give me totally different regression results for the data below. If you know the solution, please enlighten me. In Excel I used LINEST() and Data>Data Analysis>Regression and both (fortunately) gave the same result. The coefficients were: /b0=1.16, b1=0.957, b2=0.024, R2=0.0027, adjusted R2=-0.017/ In R I used

New version of ffmpeg2theora, a command line tool to convert video files to Ogg Theora. new in version 0.19 - use libswscale api - add frontend mode and a simple PythonCard Frontend (only in svn right now or binary for os x *) - support for output larger than 2GB on 32bit systems - fix short option -k to output Ogg Skeleton - other smaller fixes binaries

Hello, A collegue of mine has compared the runtime of a linear model + anova in SAS and S+. He got the same results, but SAS took a bit more than a minute whereas S+ took 17 minutes. I've tried it in R (1.9.0) and it took 15 min. Neither machine run out of memory, and I assume that all machines have similar hardware, but the S+ and SAS machines are on windows whereas the R machine is Redhat

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 am new to R, and didn't know which list to send this to, since it is a bug report about a package, not about core R... I have created a regression tree using 4 predictors: 3 are unordered (binary) predictors, and the last is a date (integer), which I declare to be an ordered factor. However, the tree treats the date as if it were un-ordered, splitting into non-consecutive subsets. My code

Hi All, The version of "PKCS#11 support in OpenSSH" is ready for download. On download page http://alon.barlev.googlepages.com/openssh-pkcs11 you can find a patch for OpenSSH 4.5p1. Most of PKCS#11 code is now moved to a standalone library which I call pkcs11-helper, this library is used by all projects that I added PKCS#11 support into. The library can be downloaded from:

Hi listers, I have a simple doubt... I need to manipulate the data base as: var1 var2 7 0.1 7 0.1 8 0.12 10 0.15 12 0.18 20 0.31 I would like to aggregate the equal cases as var1 making the sum of var2... I would get... var1 var2 7 0.2 8 0.12 10 0.15 12 0.18 20 0.31 Thanks in advance, Marcio -- View this message in context:

Folks, I'm trying to get stats from a matrix for each transition from one state to another. I have a matrix x as below. structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0, 0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19, -0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.81, 0.39, -1.38, 0.06, 0.84, 0.51, -1, -1.29, 2.15, 0.39, 0.78, 0.85, 1.18, 1.66, 0.9, -0.94,

Dear all, I have a big file containing latitude points(-10 to 80) and corresponding values. Example data Lat=c(69.48134, 69.49439, 69.50736, 69.52026, 69.52438, 69.53308, 69.53746, 69.54365, 69.54582, 69.6884, 69.69272, 69.998, 70.00055, 70.00106, 70.00295, 70.00308, 70.00363, 70.00427, 70.00665, 70.00906, 70.01049, 70.01053, 70.01075, 70.01208, 70.01236, 70.01418, 70.01452, 70.01646,

I am having trouble mounting drives. We have a Windows 2008 (10.1.1.17, MRISRV02) server with folders I what access to. On my Fedora 8 client, in my fstab file, I have: //10.1.1.17/USERS/Jeff /home/mriuser/Desktop/jeff cifs rw,username=cluster,password=mrilab 0 0 and this works just fine. My samba is up to date: [mriuser at localhost ~]$ rpm -qa | grep "samba"

Hi all, I am trying to create images in R suitable for journal publication. I'd like to make an image that is 3.6" wide and 5.08" length, and 1200 dpi. When I create a tiff file that is 800 dpi and 4x4", it works (although doesn't look good). But when I create a tiff that is the size and dpi I want, it doesn't work. I get the following error: Error in tiff(filename =