Hi
The package VGAM checks the assumption by the argument parallel and
comparing the 2 models with and without.
I am not sure if package rms does so as well
Below are some other packages that do ordinal regression, some have
marginal capability
ordinal
repolr
multgee
geepack
nnet
I am not
Yes; see clm and clmm2 (mixed effects) in the ordinal package for
fitting proportional odds models. See section 3 of
http://cran.r-project.org/web/packages/ordinal/vignettes/clm_tutorial.pdf
to see how to test the proportional odds assumption with clm - it is
equivalent for clmm2 models. For an
On 14-02-28 4:47 PM, Christian De Santis wrote:
Dear R List,
today i have installed Ubuntu as i decided to give this a try after running R
in windows was always running out of memory in making my figures. I have copied
my working folders into Ubuntu and changed the WD and the \\ into /. I
Among various possibilities, you might consider a logistic or probit regression
model with ARMA errors specified via Gaussian copula. This approach is
implemented in the package gcmr (Gaussian Copula Marginal Regression).
Example: logistic model with covariates S1 and S2 and AR(1,2) errors
Thanks for your reply Ista and Duncan,
i solved the issue and you were both right. Ista, my file was spelled with a
capital L (as in Light) and it wasn't recognized. However, this was never a
problem in windows as Duncan pointed out. You live and learn. :-)
Thanks very much,
Christian
Dear Users,
I have a problem with cvm.test using with for loop. It makes R crash while
calculating. In the function, x and y are lists, also for loops controls the
columns that will be tested:
function (x,y)
{
cval=matrix(ncol=dim(x)[2],nrow=dim(x)[2])
for(i in
Hi,
You could try:
#If mat1 is the matrix
dimnames(mat1) - list(1:nrow(mat1),1:ncol(mat1))
setNames(as.data.frame.table(mat1),c(m,n,value))
A.K.
On Friday, February 28, 2014 11:40 PM, Chirag Gupta cxg...@email.uark.edu
wrote:
Hi list
I have a matrix of size m x n (m and n are different,
R-helpers:
I am getting an error when trying to fit a GEE model. Below is code
reproducing the error.
###
library(foreign)
muscatine -
read.dta('http://www.hsph.harvard.edu/fitzmaur/ala2e/muscatine.dta')
muscatine$gender - as.factor(muscatine$gender)
muscatine$y -
Hello,
First of all, x and y don't seem to be lists, they seem to be matrices.
Second, the instrucions a-x[,i] and b-y[,j] change the variables a
and b that exist outside the function. Why?
Third, exactly what do you mean by crash? What is the error message?
Finally, your example, as is, is
Hi All
I am totally new to R so this question may sound basic to many of you. I am
trying to use R for time series analysis of some financial instruments.
Currently i have hourly data of a stock which has OPEN/HIGH/LOW/CLOSE in a
CSV file. I used read.table to import the data in R in to a
Hi,
Let say I Have following data:
Info - structure(list(Person = structure(1:6, .Label = c(A, B, C,
+ D, E, F), class = factor), Attr1 = c(0.52, 0.14, 0.63,
+ 0.43, 0.89, 18.46), Attr2 = c(0.06, 3.35, 0.62, 1.42, 1.96, 8.38
+ )), .Names = c(Person, Attr1, Attr2), row.names = c(NA,
+ -6L),
On 02/03/14 05:04, Christian De Santis wrote:
Thanks for your reply Ista and Duncan,
i solved the issue and you were both right. Ista, my file was spelled
with a capital L (as in Light) and it wasn't recognized. However,
this was never a problem in windows as Duncan pointed out. You live
and
Hi everybody
I guess my problem is pretty easy to solve, but I didn't manage to get it
working...
Me, as a Swiss, I'm trying to substitute German Umlauts (ä, ü, ö) in a
string variable in a dataset; meaning ä to ae, ü, to ue and ö to
oe.
Here's a reproducible example:
# defining
One possibility would be to use the absolute value of the
differences. Then they would be all positive:
abs(Info$Attr1 - Info$Attr2)
[1] 0.46 3.21 0.01 0.99 1.07 10.08
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX
Hello,
Maybe with some simple modifications of swiss2. (Note that your 2nd
function wasn't returning a value.)
swiss2 - function(m, data){
for (i in seq_len(ncol(m))){
data - gsub(paste(m[,i], sep=,)[1],paste(m[,i], sep=,)[2], data,
ignore.case=F, perl = F, fixed = F,
Hi All,
My sample code looks like
options(stringsAsFactors = FALSE);
clean = function(x)
{
loc = agrep(ABC, x$name);
x[loc,]$new_name - NEW;
x;
}
name = c(12, dad, dfd);
y = data.frame(name = as.character(name), idx = 1:3);
y$new_name = y$name;
z - clean(y)
The snippet does not
There is a website that populates a table with athlete scores during a
competition. I would like to be able to extract those scores from the website
and place them into a data frame if this is possible. The website is at the
link below:
http://games.crossfit.com/leaderboard
One complication
On 03/02/2014 05:51 AM, Ron Michael wrote:
Hi,
Let say I Have following data:
Info- structure(list(Person = structure(1:6, .Label = c(A, B, C,
+ D, E, F), class = factor), Attr1 = c(0.52, 0.14, 0.63,
+ 0.43, 0.89, 18.46), Attr2 = c(0.06, 3.35, 0.62, 1.42, 1.96, 8.38
+ )), .Names = c(Person,
Hi Brant
I have not got Fitzmaurice etal but from their web site it seems that you
are trying to do ordinal GEE
With GEE models particularly ordinal models you MUST get your data structure
correct otherwise it can fail or even R can crash
try
f1 =
ordgee(ordered(y) ~ factor(gender) + cage +
Duncan,
Thank you for your reply. The example is in fact not ordinal (the response
variable Y is an indicator of the presence or absence of obesity). I too saw
their code snippet online where they use an ordinal GEE, but the outcome
variable is binary as can be seen from the imported data
Hi,
I am trying to predict the time to certain type of failure given the
following data on Certain Factory Equipments. The data I have are
readings collected every day for sensor installed on those equipments
.On Same day, an equipment can have different Repaires performed,and on
some days
Hi,
Not sure about your expected result. If you need a logical index as the new
variable:
##check str(Food)
Food[,-1] - lapply(Food[,-1],function(x) as.numeric(as.character(x)))
str(Food)
Food1 - within(Food,Healthy.food - GRASSI 20)
HFood - Food1[Food1$Healthy.food,-7]
dim(HFood)
#[1]
Hi,
Please use ?dput() to show the data
e.g. dput(head(data,20))
Also, your heading and description is a bit confusing.
##Assuming that this is how your data looks like
dat - read.table(text=1 0.1
2 0.5
3 0.7
8 0.01
3 0.2
4 0.5
6 0.3
8 0.1,sep=,header=FALSE)
# dput(dat)
vec1 - c(1=0.1,
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