Displaying 20 results from an estimated 5000 matches similar to: "R-help Digest, Vol 67, Issue 23"
2008 Sep 22
2
Why isn't R recognising integers as numbers?
I have a number of files containing anywhere from a few dozen to a few
thousand integers, one per record.
The statement "refdata18 =
read.csv("K:\\MerchantData\\RiskModel\\Capture.Week.18.csv", header =
TRUE,na.strings="")" works fine, and if I type refdata18, I get the integers
displayed, one value per record (along with a record number). However, when
I try "
2008 Oct 08
0
Applying an R script to data within MySQL? How to?
I am trying something I haven't attempted before and the available
documentation doesn't quite answer my questions (at least in a way I can
understand). My usual course of action would be to extract my data from my
DB, do whatever manipulation is necessary, either manually or using a C++
program, and then import the data into R. Now I need to try to do it all
within R+RMySQL+MySQL.
I
2008 Sep 24
0
Trouble understanding the behaviour of stableFit(fBasics)
Can anyone explain such different output:
> stableFit(s,alpha = 1.75, beta = 0, gamma = 1, delta = 0,
+ type = c("q", "mle"), doplot = TRUE, trace = FALSE, title = NULL,
+ description = NULL)
Title:
Stable Parameter Estimation
Call:
.qStableFit(x = x, doplot = doplot, title = title, description =
description)
Model:
Student-t Distribution
Estimated
2008 Sep 21
1
Multiple plots per window
Hi all,
I'm currently working through "The Analysis of Time Series" by Chris
Chatfield. In order to also get a better understanding of R, I play
around with the examples and Exercises (no homework or assignement,
just selfstudy!!).
Exercise 2.1 gives the following dataset (sales figures for 4 week
intervals):
> sales2.1.dataframe
1995 1996 1997 1998
1 153 133 145 111
2
2008 Sep 22
0
Warnings in fitdistr() from MASS.
For a lark, I experimented a bit with the data from Ted Byers' recent
postings. The result of fitdistr() seemed sensible, but I was bothered
by the warnings about NaNs that arose. Warnings always make me nervous.
Explicitly this is what I did:
TXT <- "0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2008 Sep 21
0
Unexpected behaviour when testing for independence, with multiple factors
>Ben Bolker <bolker <at> ufl.edu> writes:
>
>I would try
>
>fligner.test(dT ~ Topology:Drift:lambda)
>
>in response to:
>
>Javier Acuna <javier.acuna.o <at> gmail.com> writes:
>
> Hi, I'm a new user of R. My background is Electrical Engineering, so
> please bear with me if this is a silly question.
>
> I'm trying to assess
2004 Apr 05
1
fligner.test (ctest) (PR#6739)
Full_Name: Karel Zvara
Version: 1.8.1
OS: MS Winows 2000
Submission from: (NULL) (195.113.30.163)
The test statistics of the fligner.test (ctest package) depends on the order of
cases:
> fligner.test(count~spray,data=InsectSprays)
Fligner-Killeen test for homogeneity of variances
data: count by spray
Fligner-Killeen:med chi-squared = 14.4828, df = 5, p-value =
0.01282
>
2008 Sep 21
1
Calculating interval for conditional/unconditional correlation matrix
Hi there,
Could anyone please help me to understand what should be done in order not to get this error message: Error: evaluation nested too deeply: infinite recursion / options(expressions=)?
Here is my code:
determinant<-
function(x){det(matrix(c(1.0,0.2,0.5,0.8,0.2,1.0,0.5,0.6,0.5,0.5,0.5,1.0,x,0.8,0.6,x,1.0),ncol=4,byrow=T))}
matrix<-
2012 Jan 10
1
different results from fligner.test
I've made fligner test with the same data, changing the orders of the
variables, and this what i get
> fligner.test(rojos~edadysexo*zona*ano*estacion)
Fligner-Killeen test of homogeneity of variances
data: rojos by edadysexo by zona by ano by estacion
Fligner-Killeen:med chi-squared = 15.7651, df = 2, p-value = 0.0003773
> fligner.test(rojos~ano*edadysexo*zona*estacion)
2008 Sep 17
2
Unexpected behaviour when testing for independence with multiple factors
Hi, I'm a new user of R. My background is Electrical Engineering, so
please bear with me if this is a silly question.
I'm trying to assess whether the results of an experiment satisfy the
hypothesis of homoscedasticity (my ultimate goal is to use ANOVA).
The result of the experiment is mean delay (dT), which depends on
three factors, topology, drift, and lambda. The first two factors are
2011 Oct 18
1
Repeat a loop until...
Dear all,
I know there have been various questions posted over the years about loops but I'm afraid that I'm still stuck. I am using Windows XP and R 2.9.2.
I am generating some data using the multivariate normal distribution (within the 'mnormt' package). [The numerical values of sanad and covmat are not important.]
> datamat <-
2019 Jun 21
1
[Suggested patch] to fligner.test - constant values can produce significant results
In specific cases fligner.test() can produce a small p-value even when both
groups have constant variance.
Here is an illustration:
fligner.test(c(1,1,2,2), c("a","a","b","b"))
# p-value = NA
But:
fligner.test(c(1,1,1,2,2,2), c("a","a","a","b","b","b"))
# p-value < 2.2e-16
2009 Aug 20
1
Understanding R code
What is
1. par.ests <- optimfit$par
2. fisher <- hessb(negloglik, par.ests, maxvalue=maxima);
3. varcov <- solve(fisher);
4. par.ses <- sqrt(diag(varcov));
Thanks a lot,
fit.GEV <- function(maxima)
{
sigma0 <- sqrt((6. * var(maxima))/pi)
mu0 <- mean(maxima) - 0.57722 * sigma0
xi0 <- 0.1
theta <- c(xi0, mu0, sigma0)
#10/5/2007: removed assign() for maxima.nl
2008 Aug 22
1
Test of Homogeneity of Variances
I am testing the homogeneity of variances via bartlett.test and fligner.test. Using the following example, how should I interpret the p-value in order to accept or reject the null hypothesis ?
set.seed(5)
x <- rnorm(20)
bartlett.test(x, rep(1:5, each=4))
Bartlett test of homogeneity of variances
data: x and rep(1:5, each = 4)
Bartlett's K-squared = 1.7709, df = 4, p-value =
2019 Jun 18
0
Small bug in fligner.test - constant values can produce significant results (patch attached)
In specific cases fligner.test() can produce a small p-value even when both groups have constant variance.
Here is an illustration:
fligner.test(c(1,1,2,2), c("a","a","b","b"))
# p-value = NA
But:
fligner.test(c(1,1,1,2,2,2), c("a","a","a","b","b","b"))
# p-value < 2.2e-16
2009 Sep 14
2
Function "Varcov" in Design (Ver. 2.2-0) package
Hi,
I'm running into an error message for the "anova"-function I never got
before with the Design (Version 2.2-0) package.
There seems to be a missing function "Varcov", please check my
function calls (it's in german but I think you get the error):
> library(Design) ## attaching Design package and dependent packages
Lade n?tiges Paket: Hmisc ## loading
2003 Jul 27
1
multiple imputation with fit.mult.impute in Hmisc
I have always avoided missing data by keeping my distance from
the real world. But I have a student who is doing a study of
real patients. We're trying to test regression models using
multiple imputation. We did the following (roughly):
f <- aregImpute(~ [list of 32 variables, separated by + signs],
n.impute=20, defaultLinear=T, data=t1)
# I read that 20 is better than the default of
2008 Sep 21
1
How to plot "greater than" symbol on the x-axis
Hello everyone,
I want to plot a "greater than" symbol (the "_" under ">") on the x-axis in the labels. Is it possible to do it?
Thanks.
Bingshan
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2005 Feb 02
0
Not reproducing GLS estimates
Dear List:
I am having some trouble reproducing some GLS estimates using matrix
operations that I am not having with other R procedures. Here are some
sample data to see what I am doing along with all code:
mu<-c(100,150,200,250)
Sigma<-matrix(c(400,80,16,3.2,80,400,80,16,16,80,400,80,3.2,16,80,400),n
c=4)
sample.size<-100
temp <-
2008 Oct 16
1
Two last questions: about output
Here is my little scriptlet:
optdata =
read.csv("K:\\MerchantData\\RiskModel\\AutomatedRiskModel\\soptions.dat",
header = FALSE, na.strings="")
attach(optdata)
library(MASS)
setwd("K:\\MerchantData\\RiskModel\\AutomatedRiskModel")
for (i in 1:length(V4) ) {
x = read.csv(as.character(V4[[i]]), header = FALSE, na.strings="");
y = x[,1];
fp =