Displaying 20 results from an estimated 7000 matches similar to: "variable names in lm formula ~."
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
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
--
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
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
2008 May 02
1
error in using by + median
dear all,
Could anyone explain me the behaviour of median() within by()?
(I am running R.2.7.0)
thanks,
vito
> H<-cbind(rep(0:1,l=20),matrix(rnorm(20*2),20,2))
> by(H[,-1],H[,1],mean)
INDICES: 0
V1 V2
-0.2101069 0.2954377
---------------------------------------------------------------------------------------------------------------------
INDICES: 1
V1
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
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)
2024 Jul 26
1
Automatic Knot selection in Piecewise linear splines
dear all,
I apologize for my delay in replying you. Here my contribution, maybe
just for completeness:
Similar to "earth", "segmented" also fits piecewise linear relationships
with the number of breakpoints being selected by the AIC or BIC
(recommended).
#code (example and code from Martin Maechler previous email)
library(segmented)
o<-selgmented(y, ~x, Kmax=20,
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,
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:
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)
>
2018 Jan 30
0
variable names in lm formula ~.
Functions are first class objects, so some kind of collision is bound to happen if you do this... so don't.
--
Sent from my phone. Please excuse my brevity.
On January 30, 2018 3:11:56 AM PST, "Vito M. R. Muggeo" <vito.muggeo at unipa.it> wrote:
>dear all,
>Is the following intentional? Am I missing anything in documentation?
>
2024 May 16
0
segmented 2.1-0 is released
dear R users,
I am pleased to announce that segmented 2.1-0 is now available on CRAN.
segmented focuses on estimation of breakpoints/changepoints of
segmented, i.e. piecewise linear, relationships in (generalized) linear
models. Starting with version 2.0-0, it is also possible to model
stepmented, i.e. piecewise constant, effects.
In the last release both models may be fitted via a formula
2024 May 16
0
segmented 2.1-0 is released
dear R users,
I am pleased to announce that segmented 2.1-0 is now available on CRAN.
segmented focuses on estimation of breakpoints/changepoints of
segmented, i.e. piecewise linear, relationships in (generalized) linear
models. Starting with version 2.0-0, it is also possible to model
stepmented, i.e. piecewise constant, effects.
In the last release both models may be fitted via a formula
2018 Jan 30
1
variable names in lm formula ~.
Well...
?terms.formula says:
"data: a data frame from which the meaning of the special symbol . can
be inferred. It is unused if there is no . in the formula."
So this seems to me to be an obscure bug, as I have found no warning
against this admittedly confusing but still, I think, legal syntax.
Note:
> d <- data.frame(log = runif(10), x = 1:10)
> y <- rnorm(10,5)
>
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
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
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Prof.