Displaying 20 results from an estimated 1000 matches similar to: "error in using by + median"
2009 Nov 02
3
partial matching with grep()
dear all,
This is a probably a silly question.
If I type
> grep("x",c("a.x" ,"b.x","a.xx"),value=TRUE)
[1] "a.x" "b.x" "a.xx"
Instead, I would like to obtain only
"a.x" "b.x"
How is it possible to get this result with grep()?
many thanks for your attention,
best,
vito
--
2008 Jun 30
2
difference between MASS::polr() and Design::lrm()
Dear all,
It appears that MASS::polr() and Design::lrm() return the same point
estimates but different st.errs when fitting proportional odds models,
grade<-c(4,4,2,4,3,2,3,1,3,3,2,2,3,3,2,4,2,4,5,2,1,4,1,2,5,3,4,2,2,1)
score<-c(525,533,545,582,581,576,572,609,559,543,576,525,574,582,574,471,595,
557,557,584,599,517,649,584,463,591,488,563,553,549)
library(MASS)
library(Design)
2006 Nov 03
1
difference in using with() and the "data" argument in glm call
Dear all,
I am dealing with the following (apparently simple problem):
For some reasons I am interested in passing variables from a dataframe
to a specific environment, and in fitting a standard glm:
dati<-data.frame(y=rnorm(10),x1=runif(10),x2=runif(10))
KK<-new.env()
for(i in 1:ncol(dati)) assign(names(dati[i]),dati[[i]],envir=KK)
#Now the following two lines work correctly:
2010 Mar 04
1
only actual variable names in all.names()
dear all,
When I use all.vars(), I am interest in extracting only the variable names..
Here a simple example
all.vars(as.formula(y~poly(x,k)+z))
returns
[1] "y" "x" "k" "z"
and I would like to obtain
"y" "x" "z"
Where is the trick?
many thanks
vito
--
====================================
Vito M.R. Muggeo
Dip.to Sc
2010 Oct 25
1
building lme call via call()
dear all,
I would like to get the lme call without fitting the relevant model.
library(nlme)
data(Orthodont)
fm1 <- lme(distance ~ age, random=list(Subject=~age),data = Orthodont)
To get fm1$call without fitting the model I use call():
my.cc<-call("lme.formula", fixed= distance ~ age, random = list(Subject
= ~age))
However the two calls are not the same (apart from the data
2007 Nov 28
3
using names with functions..
Dear all,
I have the following (rather) strange problem..
For some reasons, I finally work with a variable whose name includes an
R function, "a.log(z)", say. And that is a problem when I call it in a
formula, for instance:
> myname<-"a.log(z)"
> dd<-data.frame("a.log(z)"=1:10,y=rnorm(10))
> o<-lm(y~1,data=dd)
>
2008 Dec 17
0
OFF topic testing for positive coeffs
Dear all,
This is off-topic,
however I hope someone can give me useful suggestion..
Given the regression model
y = b0 + b1*x + e
I am interested in testing for positive coeffs, namely
H0: b0>0 AND b1>0
H1: b0,b1 unconstrained
It is simple to estimate the model under H0 and H1 (there are several
suggestions on the Rlist about estimation but nothing about testing..)
perform a likelihood
2005 Mar 11
0
Negative binomial regression for count data,
Dear list,
I would like to know:
1. After I have used the R code (http://pscl.stanford.edu/zeroinfl.r) to fit a zero-inflated negative binomial model, what criteria I should follow to compare and select the best model (models with different predictors)?
2. How can I compare the model I get from question 1 (zero-inflated negative binomial) to other models like glm family models or a logistic
2007 Dec 06
1
differences in using source() or console
Dear all,
Is there *any* reason explaining what I describe below?
I have the following line
myfun(x)
If I type them directly in R (or copy/past), it works..
However if I type in R 2.6.1
> source("code.R") ##code.R includes the above line
Error in inherits(x, "data.frame") : object "d" not found
namely myfun() does not work correctly.
In particular the
2013 Feb 14
0
IWSM 2013: LAST call for papers
dear all,
apologizes for this OT
===========================
28th International Workshop on Statistical Modelling (IWSM), Palermo
(Italy) 8-12 July 2013. http://iwsm2013.unipa.it
Dear friend,
For your information, I would like to bring to your attention that
deadline for submission of abstracts is
FEBRUARY 18
If you are still interested to visit Palermo (and taste its specialities
:-))
2013 Jan 18
0
OT: IWSM 2013
dear all,
apologizes for this off topic.
I would like to inform you that registration and paper submission for
the 28th International Workshop on Statistical Modelling (IWSM)
to be held in Palermo (Italy) 8-12 July 2013 is now open at
http://iwsm2013.unipa.it
Register at http://iwsm2013.unipa.it/?cmd=registration and then submit
your abstract. Deadlines for Abstract submission is February 4,
2008 Jan 16
1
strange behaviour of is.factor()
Dear all,
It appears that the function is.factor() returns different results when
used inside the apply() function: that is, is.factor() fails to
recognize a factor..
Where is the trick?
many thanks,
vito
> df1<-data.frame(y=1:10,x=rnorm(10),g=factor(c(rep("A",6),rep("B",4))))
> is.factor(df1[,1])
[1] FALSE
> is.factor(df1[,2])
[1] FALSE
>
2006 Feb 27
3
how to use the basis matrix of "ns" in R? really confused by multi-dim spline filtering?
Hi all,
Could anybody recommend some easy-to-understand and example based
notes/tutorials on how to use cubic splines to do filtering on
multi-dimension data?
I am confused by the 1-dimensional case, and more confused by
multi-dimensional case.
I found all the books suddenly become very abstract when it comes to this
subject.
They don't provide examples in R or Splus at all.
Specifically,
2012 Jun 01
1
getting the name of the working .Rdata file
dear all,
I do not if it is a nonsense question..
Is it possible in the R session to get the name of the current .Rdata
file that I ran?
I mean: suppose I double click the file myfile.Rdata. ls() returns the
names of the objects in the current workspace (that is saved in
myfile.Rdata). In the current R session, I would like to obtain
"myfile.Rdata". Is it possible?
Thanks in
2006 Mar 01
1
a strange problem with integrate()
Dear all,
I am stuck on the following problem with integrate(). I have been out of
luck using RSiteSearch()..
My function is
g2<-function(b,theta,xi,yi,sigma2){
xi<-cbind(1,xi)
eta<-drop(xi%*%theta)
num<-exp((eta + rep(b,length(eta)))*yi)
den<- 1 + exp(eta + rep(b,length(eta)))
result=(num/den)*exp((-b^2)/sigma2)/sqrt(2*pi*sigma2)
2012 Mar 21
1
glmnet() vs. lars()
dear all,
It appears that glmnet(), when "selecting" the covariates entering the
model, skips from K covariates, say, to K+2 or K+3. Thus 2 or 3
variables are "added" at the same time and it is not possible to obtain
a ranking of the covariates according to their importance in the model.
On the other hand lars() "adds" the covariates one at a time.
My question
2004 May 11
2
bilinear and non linear
Dear all,
there are R packages able to simulate or estimate bilinear model for time
series?
I know it is an open problem, but do exist something for very simplified
bilinear models?
Alternatively, what kinfd of non linear time series models are performed
in R?
If R is not able, could someone suggest me for some commercial softwares
to deal with bilinear models?
i'm afraid of a negative
2004 Dec 14
1
correlation in lme4
Dear all,
I have tried to consider a correlation structure in lme (package lme4), but
without success.
I have used something like:
> risul<-lme(y~x+ z , data=mydata, random=~ x | g, correlation = corAR1())
but the result is the same as:
> risul<-lme(y~x+ z , data=mydata, random=~ x | g).
Can anybody help me?
Antonella
**************************************************
Prof.
2018 May 21
2
removing part of a string
dear all,
I am stuck on the following problem. Give a string like
ss1<- "z:f(5, a=3, b=4, c='1:4', d=2)"
or
ss2<- "f(5, a=3, b=4, c=\"1:4\", d=2)*z"
I would like to remove all entries within parentheses.. Namely, I aim to
obtain respectively
"z:f()" or "f()*z"
I played with sub() and gsub() but without success..
Thank you very
2009 Jun 15
1
Linear Models: Explanatory variables with uncertainties
One of the assumptions, on which the (General) Linear Modelling is
based is that the response variable is measured with some
uncertainties (or weighted), but the explanatory variables are fixed.
Is it possible to extend the model by assigning the weights to the
explanatory variables as well? Is there a package for doing such a
model fit?
Thanks