search for: 0.72

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 ________________________________________________________________________ Linux-Mandrake Security Update Advisory ________________________________________________________________________ Package name: pam Date: August 1st, 2000 Advisory ID: MDKSA-2000:029 Affected versions: 6.0, 6.1, 7.0, 7.1

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

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

--------------------------------------------------------------------- Red Hat, Inc. Security Advisory Synopsis: Updated PAM packages are available. Advisory ID: RHSA-2000:044-02 Issue date: 2000-07-21 Updated on: 2000-07-21 Product: Red Hat Linux Keywords: N/A Cross references: N/A

> It's the configuration part that I was looking for some help with, but > as noted I can install neither winesetuptk nor wineinstall You don't need winesetuptk or wineinstall anymore, because wine automatically sets up a .wine directory with the appropriate contents. Also, the wine config file isn't needed to successfully run wine anymore. If you don't have a .wine

Hello, I am using Redhat 6.1 on pentium. I have a problem with sshd in openssh, when I try to connect by ssh to sshd even on the local machine, the pam module will not authenticate my password. Here is a transcript of sshd -ddd: debug1: sshd version OpenSSH_2.3.0p1 debug1: Seeding random number generator debug1: read DSA private key done debug1: Seeding random number generator debug1: Bind to port

Hello R User, I was trying to display r.squared and p value in table from regression, but I could not display these parameters in the table (matrix) for example individual <- c(1,1,6,8,8,9,9,9,12,12) day <- c(4,17,12,12,17,3,9,22,13,20) condition <- c(0.72, 0.72, 0.67, 0.73, 0.76, 0.65, 0.68, 0.78, 0.73, 0.71) test <- data.frame(individual, day, condition) ind.id <-

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 ______________________________________________________________________________ Caldera Systems, Inc. Security Advisory Subject: DoS on gpm Advisory number: CSSA-2000-024.0 Issue date: 2000 July, 6 Cross reference: ______________________________________________________________________________ 1. Problem Description There are security

Dear R list As suggested by Prof Brian Ripley, I have tried to read acf literature. The main problem is I am not the statistician and hence have some problem in understanding the concepts immediately. I came across one literature (http://www.stat.nus.edu.sg/~staxyc/REG32.pdf) on auto-correlation giving the methodology. As per that literature, the auto-correlation is arrived at as per following.

I've a matrix like this: 1985 1.38 1.27 1.84 2.10 0.59 3.47 1986 1.05 1.13 1.21 1.54 0.21 2.14 1987 1.33 1.21 1.77 1.44 0.27 2.85 1988 1.86 1.06 2.33 2.14 0.55 1.40 1989 2.10 0.65 2.74 2.43 1.19 1.45 1990 1.55 0.00 1.59 1.94 0.99 2.14 1991 0.92

Hello, I?m wondering if anyone can offer advice on the out-of-memory error I?m getting. I?m using R2.12.2 on Windows XP, Platform: i386-pc-mingw32/i386 (32-bit). I am using the quantreg package, trying to perform a quantile regression on a dataframe that has 11,254 rows and 5 columns. > object.size(subsetAudit.dat) 450832 bytes > str(subsetAudit.dat) 'data.frame': 11253 obs.

I have two alternative Cox models with C-statistics 0.72 and 0.78. My question is if 0.78 is significantly greater than 0.72. I'm using rcorrp.cens. I cannot find the U statistics in the output of the function. This is the output of the help example: > x1 <- rnorm(400) > x2 <- x1 + rnorm(400) > d.time <- rexp(400) + (x1 - min(x1)) > cens <- runif(400,.5,2) > death

--------------------------------------------------------------------- Red Hat, Inc. Red Hat Security Advisory Synopsis: Updated webalizer package available Advisory ID: RHSA-2001:140-05 Issue date: 2001-10-24 Updated on: 2001-10-30 Product: Red Hat Linux Keywords: webalizer cross-site scripting malicious html security update Cross

To whom it may concern, I don't know whether this is really a bug with the Pspline package or only a problem with my installation. Things work fine in Linux but not in Mac OS X (Darwin). Both system run the latest public versions of R and Pspline. predict.smooth.Pspline produces only NaN instead of predicted values when norder>2: > library (Pspline) > tt <- seq

I am looking for simple introduction to cluster analysis using R, that would be understandable to a novice in statistics. Or, could someone perhaps help me understand how to proceed in my analysis? I am very new to both statistics and R, but am trying hard to avoid having to use SPSS as everyone around me... I have dataset on people presenting their opinions on different religious

This adds [f]getc() and fgets() for parsing config files. Probably hard to avoid. Still trying to decide if I actually want to add system() or not. -hpa

--------------------------------------------------------------------- Red Hat, Inc. Security Advisory Synopsis: Zope update Advisory ID: RHSA-2000:052-04 Issue date: 2000-08-11 Updated on: 2000-08-18 Product: Red Hat Powertools Keywords: Zope Cross references: N/A

Hello R-helpers: I have data like this: sample replicate height weight age A 1.00 12.0 0.64 6.00 A 2.00 12.2 0.38 6.00 A 3.00 12.4 0.49 6.00 B 1.00 12.7 0.65 4.00 B 2.00 12.8 0.78 5.00 C 1.00 11.9 0.45 6.00 C 2.00 11.84 0.44 2.00 C 3.00 11.43 0.32 3.00 C 4.00 10.24 0.84 4.00 D

I'm trying to replace NA with 0 value... I've write a loop, but don't work... Where's the problem? > cimfasy_rwl 1991 0.92 0.72 0.50 1.29 0.54 1.22 1992 2.15 1.28 1.23 2.26 1.22 3.17 1993 1.50 0.87 1.68 1.97 0.83 2.55 1994 0.69 0.00 0.76 1.89 0.60 0.87 1995 1.13 1.04

Dear R People: Suppose we have a regression model that we will call y.lm We run the Durbin Watson test for autocorrelation and we find that there is positive autocorrelation, and phi = 0.72, say. What is our next step, please? Do we calculate the following yprime_t = y_t - 0.72y_t-1, x1prime_t = x1_t - 0.72x1_t-1, and so on, and re-fit the linear mode? I haven't done this in a while.