I see a rich set of graphic device functions to redirect that output. Are
there commands to redirect text as well. I have a set of functions that
execute many linear regression tests serially and I want to capture this in a
file for printing.
Thanks,
Stan Hopkins
[[alternative HTML
James MacDonald jmacdon at med.umich.edu writes:
Hi all,
Most of the analyses I do are short little once-and-done type things that are
easily encapsulated in a .Rnw
file. However, I sometimes end up with projects that take an extended amount
of time. Usually these
projects are not
Dear useRs,
I have written a function that implements a Bayesian method to
compare a patient's score on two tasks with that of a small control
group, as described in Crawford, J. and Garthwaite, P. (2007).
Comparison of a single case to a control or normative sample in
neuropsychology:
The AMMI senso strictu part starts from the corrected data for genotype and
environment. In most cases where AMMI is applied (maybe also in the
agricolae package, I haven't checked), starts from the interaction effects
obtained through a general linear model anova.
It should be possible to
Stan Hopkins wrote:
I see a rich set of graphic device functions to redirect that output. Are
there commands to redirect text as well. I have a set of functions that
execute many linear regression tests serially and I want to capture this
in a file for printing.
Thanks,
Stan
Here are two simple ways:
=== method1 ===
cat(line1,\n,file=output.txt)
cat(line2,\n,file=output.txt,append=TRUE)
=== method2 ===
sink(output.txt)
cat(line1,\n)
cat(line2,\n)
out - lm(y~x,data=data.frame(x=1:10,y=(1:10+rnorm(10,0,0.1
print(out)
sink()
And then there is 'Sweave'. Check out,
At 22:36 23.07.2007 +, D L McArthur wrote:
James MacDonald jmacdon at med.umich.edu writes:
Hi all,
Most of the analyses I do are short little once-and-done type things
that are
easily encapsulated in a .Rnw
file. However, I sometimes end up with projects that take an extended
amount
I have been modelling spatial data using function likfit in package geoR. Now
that I have the spatial regression models, I would like to calculate the amount
of variance explained by both the trend and the spatial parts of the model.
Any advice on how to do this would be much appreciated.
I don't know why it doesn't work but I think people generally recommend that
you use wireframe() in lattice rather than persp(), because wireframe is more
customizable (the pdf document referred to in this post is pretty good):
http://tolstoy.newcastle.edu.au/R/e2/help/07/03/12534.html
Here's an
You need to make use of the profiling methods described in 'Writing R
Exensions'. My machine is about 4x faster than yours: I get
Each sample represents 0.02 seconds.
Total run time: 62.08041 seconds.
Total seconds: time spent in function and callees.
Self seconds: time spent in
On 24-Jul-07 01:09:06, Andrew Clegg wrote:
Hi folks,
I've looked through the list archives and online resources, but I
haven't really found an answer to this -- it's pretty basic, but I'm
(very much) not a statistician, and I just want to check that my
solution is statistically sound.
07/24/2007 11:10:07 [GMT+0100]
For security reasons certain items found in an email with your address as the
sender have not been accepted.
File name: attachment.zip
Filtered by: Malformed messages
Sender: r-help@stat.math.ethz.ch
Recipients: [EMAIL PROTECTED]
CC:
I think your way is probably the easiest (shockingly). For instance, here are
some alternatives - I think in both cases you have to calculate the
coefficient of determination (R^2) manually. My understanding is that
multiple R^2 in your case is the usual R^2 because you only have one
predictor
Well spoken. And since log transformations are nonlinear and 'compresses' the
data, it's not surprising to find that the fit doesn't look so nice while the
fit metrics tell you that a model does a good job.
--- [EMAIL PROTECTED] wrote:
On 24-Jul-07 01:09:06, Andrew Clegg wrote:
Hi folks,
Dear R users,
how can I extrapolate values listed in the summary of an lm model but not
directly available between object values such as the the standard errors of
the calculated parameters?
for example I got a model:
mod - lm(Crd ~ 1 + Week, data=data)
and its summary:
summary(mod)
Call:
Thanks to Patrick Burns and Mark Lyman for their suggestions in
solving my problem.
Patrick suggested creating the list directly (the solution I opted for)
On 7/23/07, Patrick Burns [EMAIL PROTECTED] wrote:
Why not make the list directly:
list.to.convert - vector('list', n)
for(x in 1:n)
It's not very clear to me but I think summary(mod)$coef[, Std. Error]
is wat you need?
Cheers,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel
try this:
coef(summary(mod))[, Std. Error]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web:
Manuele Pesenti wrote:
Dear R users,
how can I extrapolate values listed in the summary of an lm model but not
directly available between object values such as the the standard errors
of
the calculated parameters?
for example I got a model:
mod - lm(Crd ~ 1 + Week, data=data)
Il giorno mar, 24/07/2007 alle 06.39 +0100, Prof Brian Ripley ha
scritto:
On Mon, 23 Jul 2007, Iwona Szyd?owska wrote:
Hello,
I would like to ask You, how to generate random numbers from an
exponential power family with a shape parameter p less than 1(p-0). I
found the rnormp
Hi all,
there is tune() in the e1071 package for doing this in general, and,
among others, a tune.nnet() wrapper (see ?tune):
tmodel = tune.nnet(Species ~ ., data = iris, size = 1:5)
summary(tmodel)
Parameter tuning of `nnet':
- sampling method: 10-fold cross validation
- best
Dear R-Help community,
I am trying to overlay a single contour line over a correlation plot using
levelplot in the lattice package. These are the two arrays:
1) a correlation plot over Africa - so each grid square is a different colour
dependent on correlation - this is in an array: result_cor
Stephen, Ted -- thanks for your input. I'm glad to know I was barking
up the right-ish tree at least.
On 7/24/07, Ted Harding [EMAIL PROTECTED] wrote:
There are not enough data to properly identify the non-linearity,
but the overall appearance of the data plot suggests to me that
you should
Hi everybody,
I am working with gams and I have found some questions when plotting gams
models.
I am using mgcv, and my model looks something like this:
model- gam(x ~ s(lat,long))
I can plot the output of the model using plot(model) or plot.gam(model) and I
get a surface plot.
That is
Hope these help for alternatives to lm()? I show the use of a 2nd order
polynomial as an example to generalize a bit.
Sometimes from the subject line two separate responses can appear as reposts
when in fact they are not... (though there are identical reposts too). I
should probably figure a way
see ?predict.gam
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 24/07/07, Lucia Zarauz [EMAIL PROTECTED] wrote:
Hi everybody,
I am working with gams and I have found some questions when plotting gams
models.
I am using mgcv, and my model looks something like
Hi Henrique,
Thank you for your suggestion.
Actually, I have already tried it, but I am confused because the plot I get is
not the same as the output of plot(model) or plot.gam(model). The yaxis is
different
On the other hand, if I build a more complex model, as for example:
model- gam(x ~
2007/7/24, Lucia Zarauz [EMAIL PROTECTED]:
Hi everybody,
I am working with gams and I have found some questions when plotting gams
models.
I am using mgcv, and my model looks something like this:
model- gam(x ~ s(lat,long))
I can plot the output of the model using plot(model) or
Thank you for your suggestion.
Actually, I have already tried it, but I am confused because the plot I get
is not the same as the output of plot(model) or plot.gam(model). The yaxis
is different
-- `plot.gam' will always plot the `centered smooth', i.e. the smooth
constrained to sum to
On Tue, 24 Jul 2007, Stephen Tucker wrote:
I don't know why it doesn't work but I think people generally recommend that
It has never been implemented, and I believe the main reason is that the
labels are plotted at an angle other than a multiple of 90 degrees. Not
all devices can do that,
Hi,
I would like to write something similar to:
for(t in 1:100)
v[x[t]] - v[y[t]] + v[z[t]]
in a vectorized form. The x, y, and z vectors may contain duplicates (so v[x]
- v[y] + v[z] has different semantics). The for loop is not efficient enough
for my purposes and I
Thank you for these answers.
I ended up modifying the ps file directly. But next time I will consider
wireframe.
Nathalie
Prof Brian Ripley wrote:
On Tue, 24 Jul 2007, Stephen Tucker wrote:
I don't know why it doesn't work but I think people generally
recommend that
It has never been
On 7/24/07, Stephen Tucker [EMAIL PROTECTED] wrote:
Hope these help for alternatives to lm()? I show the use of a 2nd order
polynomial as an example to generalize a bit.
Great, thanks. If I want to demonstrate that a non-linear curve fits
better than an exponential, what's the best measure for
Andrew Clegg:
Great, thanks. If I want to demonstrate that a non-linear curve fits
better than an exponential, what's the best measure for that? Given
that neither of nls() or optim() provide R-squared.
You really need to *very* careful when trying to interprete R² (which can
be defined in
Hi all,
I am a student of MRes Bioinformatics and Comp. Biology at University of Leeds,
UK.
I am trying to compare two matrices in R. One is the matrix containing distance
and the other one contains the correlation values. I would like to create a XY
plot of these matrices and do some linear
Hi all, I am trying to fit t distribution using the function tFit in the
library(fBasics).
I am using the code tFit(datac[[2]]) and it returns the following list.
Title:
Student-t Parameter Estimation
Call:
tFit(x = datac[[2]])
Model:
Student-t Distribution
Estimated Parameter(s):
Hi,
tFit(datac[[2]])@fit$estimate
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 24/07/07, livia [EMAIL PROTECTED] wrote:
Hi all, I am trying to fit t distribution using the function tFit in the
library(fBasics).
I am using the code tFit(datac[[2]]) and it
It works. Many thanks
Henrique Dallazuanna wrote:
Hi,
tFit(datac[[2]])@fit$estimate
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 24/07/07, livia [EMAIL PROTECTED] wrote:
Hi all, I am trying to fit t distribution using the function tFit in
the
Hi,
I've created the following two matrices (mat1 and mat2) and a function
(f) to calculate the correlations between the two on a row by row basis.
mat1 - matrix(sample(1:500,50), ncol = 5,
dimnames=list(paste(row, 1:10, sep=),
paste(col, 1:5, sep=)))
I've been fixing some problems in the combine() function, but that's
only for regression data. Looks like you are doing classification, and
I don't see the problem:
R library(randomForest)
randomForest 4.5-19
Type rfNews() to see new features/changes/bug fixes.
R set.seed(1)
R rflist -
Your apply is trying to take the correlations of the rows of mat1 with the
columns of mat2 which, of course, does not work if they have different
numbers of columns. I think you mean to take the correlations of the columns
of mat1 with the columns of mat2. For example, to take the correlations
of
I'm calculating correlations between two matrices
mat1 - matrix(sample(1:500,25), ncol = 5,
dimnames=list(paste(mat1row, 1:5, sep=),
paste(mat1col, 1:5, sep=)))
mat2 - matrix(sample(501:1000,25), ncol = 5,
dimnames=list(paste(mat2row, 1:5, sep=),
paste(mat2col, 1:5, sep=)))
are you positive that your function is doing what you expect it to do?
it looks like you want something like:
sapply(1:10, function(i) cor(mat1[i,], mat2[i,]))
b
On Jul 24, 2007, at 11:05 AM, Bernzweig, Bruce ((Consultant)) wrote:
Hi,
I've created the following two matrices (mat1 and mat2)
Thanks Gabor!
You state that my apply is taking rows of mat1 with columns of mat2.
Is this because I have the y=mat2 parameter?
apply(mat1, 1, f, y=mat2)
Actually, what I would like is to run the correlations on a row against
row basis: mat1 row1 x mat2 row1, etc.
Thanks again,
- Bruce
Then try this:
cor(t(mat1), t(mat2))
Also note
1. the above implies that mat1 and mat2 must have the same
number of columns since if x and y are vectors cor(x,y) only makes
sense if they have the same length.
2. the usual convention is that variables are stored as columns
andt that rows
that's garbor's suggestion then.
sorry for the misunderstanding. :-)
b
On Jul 24, 2007, at 11:35 AM, Bernzweig, Bruce ((Consultant)) wrote:
Thanks Benilton,
I know what I want to do, just not sure how to do it using R. The
help
documentation is not very clear.
What I am trying to do is
Thanks for the explanation.
As for the rows/columns thing, the data I receive is given to me in that
way. I currently read it in using read.csv. Is there a function I
should look at that can take that and transpose it or should I just
process the data first outside of R?
Thanks,
- Bruce
Based on the examples I've seen in using statistical analysis
packages such as lmer, it seems that people usually tabulate all the
input data into one file with the first line indicating the variable
names (or labels), and then read the file inside R. However, in my
case I can't do that
Sorry. I looked up t after writing the previous email and realized that
was what I was looking for!
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Sent: Tuesday, July 24, 2007 11:48 AM
To: Bernzweig, Bruce (Consultant)
Cc: r-help@stat.math.ethz.ch
Subject: Re:
I think this is really answered already in my previous post but just in case
try this:
res1 - t(apply(mat1, 1, cor, t(mat2)))
res2 - cor(t(mat1), t(mat2))
all.equal(res1, res2, check.attributes = FALSE)
[1] TRUE
On 7/24/07, Bernzweig, Bruce (Consultant) [EMAIL PROTECTED] wrote:
I'm
Hello,
I am having problems adjusting the plot output from the quantreg
package. Anyone know what I'm doing wrong?
For example (borrowing from the help files):
plot(summary(rq(foodexp~income,tau = 1:49/50,data=engel)), nrow=1,
ncol=2,alpha = .4, ols = TRUE, xlab=test)
The alpha= parameter
Andrew Clegg andrew.clegg at gmail.com writes:
... If I want to demonstrate that a non-linear curve fits
better than an exponential, what's the best measure for that? Given
that neither of nls() or optim() provide R-squared.
To supplement Karl's comment, try Douglas Bates' (author of nls)
Dear all,
the dataset documented under ?crimtab was also used in:
@article{TreloarAE1934,
title = {The adequacy of {S}tudent's criterion of
deviations in small sample means},
author = {Treloar, A.E. and Wilder, M.A.},
journal = {The Annals of Mathematical
Hi all,
If the question below as been answered before I
apologize for the posting.
I would like to get the frequencies of occurrence of
all values in a given variable in a multivariate
dataset. In short for each variable (or field) a
summary of values contained with in a value:frequency
pair,
Hi all,
Thanks for your help in generating the matrix of distance vs correlation. I did
it using
plot(as.vector(as.matrix(cormat)), as.vector(as.matrix(distmat)))
Now I want to quantitate the same. May be on linear regression or some other
statistical functions. I have tried using linmod
Hi folks,
I am trying to generate 95% confidence intervals for a gls model
using predict.nlme
with
R version 2.5.1 (2007-06-27)
. nlme: Linear and Nonlinear Mixed Effects Models. R package version
3.1-83.
I have looked in help, and I can do it for lm and glm models, and I
can generate
Hello all,
I am sure I am missing something obvious but I cannot find the
function I am looking for. I have a data frame with three columns: X,
Y and Z, with X and Y being grid coordinates and Z the value
associated with these coordinates. I want to transform this data
frame in a matrix
Package questions to package maintainers please.
The short answer is that your alpha = .4 parameter needs to
be passed to summary not to plot. Try this:
plot(summary(rq(foodexp~income,tau = 1:49/50,data=engel),alpha =.
4), nrow=1,
ncol=2, ols = TRUE)
A longer answer would involve a
Hi all,
Situation:
- I have two matrices each w/ 4 rows and 20 columns.
mat1 - matrix(sample(1:500,80), ncol = 20,
dimnames=list(paste(mat1row, 1:4, sep=),
paste(mat1col, 1:20, sep=)))
mat2 - matrix(sample(501:1000,80), ncol = 20,
Thanks Benilton,
I know what I want to do, just not sure how to do it using R. The help
documentation is not very clear.
What I am trying to do is calculate correlations on a row against row
basis: mat1 row1 x mat2 row1, mat1 row1 x mat2 row2, ... mat1 row1 x
mat2 row-n, mat1 row-n, mat2 row-n
RV == RAVI VARADHAN [EMAIL PROTECTED]
on Sat, 07 Apr 2007 18:39:36 -0400 writes:
this is a bit a late reply... better late than never
RV Hi Martin,
Hi Ravi, thanks for your research.
RV I played a bit with rankMat. I will first state the
RV
Hello again R-experts and novices (like me),
This seems like a bug to me - or maybe it's intentional...can anyone
confirm? Up to 1000 reps, summary() of a rq object gives different
output and subtly different confidence interval estimates.
ThanksJeff
testx=runif(1200)
testy=rnorm(1200,
When in doubt: RTFM -- Quoting from ?summary.rq
se: specifies the method used to compute standard standard
errors. There are currently five available methods:
1. 'rank' which produces confidence intervals for the
estimated parameters by inverting a
To: [EMAIL PROTECTED]
Subject: New package: pomp, inference for partially-observed Markov processes
The new package 'pomp' is built around a very general realization of nonlinear
partially-observed Markov processes (AKA state-space models, nonlinear
stochastic dynamical systems). The user
Which is the proper way to rename a factor?
If I do:
test$Parc[test$Parc==OlÞrdola]-Olèrdola
R complains that
Warning message:
invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, test$Parc
== OlÞrdola, value = Olèrdola)
Thanks
Agus
--
Dr. Agustin Lobo
Institut de Ciencies de la Terra
Hi,
I need a package to run Latent Class Analysis with ordinal manifest
variables but i have not found it. LCA is for dichotomous variables and
poLCA is for nominal ones. Can anyone tell me where i can find such a
package if it exists.
Thanks.
Eduard
[[alternative HTML version
jiho jo.irisson at gmail.com writes:
Hello all,
I am sure I am missing something obvious but I cannot find the
function I am looking for. I have a data frame with three columns: X,
Y and Z, with X and Y being grid coordinates and Z the value
associated with these coordinates. I
Dear R experts,
I am plotting the population of students who live in a city, and in
successive circular bands made of the contiguous districts that surround
the city. This is a stylized figure, where I specify the area of each
successive circle based on the cumulative population of students. I
Thank you very much for the response, Prof. Ripley.
I think I am missing something here: how do you make this 'huge'
and 'gigantic'? You have not told us how many subjects you have,
but in imaging experiments it is usually no more than 50 and often
less.
Usually we have 10-30
I am sure I am missing something obvious but I cannot find the
function I am looking for. I have a data frame with three columns: X,
Y and Z, with X and Y being grid coordinates and Z the value
associated with these coordinates. I want to transform this data
frame in a matrix of Z
Try:
levels(test$Parc)[levels(test$Parc)==OlÞrdola] - Olèrdola
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 24/07/07, Agustin Lobo [EMAIL PROTECTED] wrote:
Which is the proper way to rename a factor?
If I do:
test$Parc[test$Parc==OlÞrdola]-Olèrdola
R
Hi:
Does anyone have an idea on how to color the axis and
labels using ggplot2? This is what I got:
library(ggplot2)
p - qplot(total_bill, tip, data = tips)
NewPlot- p + geom_abline(slope=c(0.1,0.15,0.2),
colour=c(red,blue,yellow),size=c(2,5,2))
NewPlot + geom_smooth(colour=green,
I doubt its any faster than using a loop but probably less code is:
library(zoo)
z1 - zoo(t(mat1)); z2 - zoo(t(mat2))
idx - 1:ncol(z1)
out - rollapply(cbind(z1, z2), 11, by.column = FALSE,
FUN = function(x) cor(x[,idx],x[,-idx]))
will give you a 10 x 16 multivariate zoo series such that:
The name of the table should give you the value. And if you have a
matrix, you just need to convert it into a vector first.
m - matrix( LETTERS[ c(1:3, 3:5, 2:4) ], nc=3 )
m
[,1] [,2] [,3]
[1,] A C B
[2,] B D C
[3,] C E D
tb - table( as.vector(m) )
tb
A B C D E
1 2 3 2 1
Hi
I have two questions:
1)
I would like to know if there is a package in R that constructs a
regression tree using the 'goodness-of-split' algorithm for survival
analysis proposed by Le Blanc and Crowley (1993) (rather than the usual
CART algorithm that uses within-node difference and impurity
Hi Felipe,
Looks like a bug! I'll try and get it fixed for the next version. In
the meantime, you can read the last chapter of the ggplot book to see
how to fix it with grid.
Hadley
On 7/24/07, Felipe Carrillo [EMAIL PROTECTED] wrote:
Hi:
Does anyone have an idea on how to color the axis
On Tue, 24 Jul 2007, Gang Chen wrote:
Thank you very much for the response, Prof. Ripley.
I think I am missing something here: how do you make this 'huge' and
'gigantic'? You have not told us how many subjects you have, but in
imaging experiments it is usually no more than 50 and often
Hi Martin,
I just realized (courtesy: ?qr) that LAPACK=TRUE always gives full rank, no
matter what the matrix and tolerance are. So, clearly my results using
LAPACK=TRUE should be ignored. So, the real comparison is only between
rankMat and qr(., LAPACK=FALSE)$rank. I cant help but fell that
I think I am missing something here: how do you make this 'huge' and
'gigantic'? You have not told us how many subjects you have, but in
imaging experiments it is usually no more than 50 and often less.
For each subject you have 3 x 30,000 responses plus an age. That is under
1Mb of data per
Hi Jean,
You haven't yet had a reply from an authoratitive source, so here is my
tuppence worth to part of your enquiry.
It's almost certain that the receiving box is a receptacle into which tags
were placed after they had been drawn and the inscribed measurement noted
down. Measurements on
While on the subject of mechanical methods of statistical research I
can't
resist quoting Doob's (1997) Statistical Science interview:
My system, complicated by my inaccurate typing, led to retyping
material over and over, and for some time I had an electric drill
on my desk, provided
Friends,
I'm trying to pass an equation as an argument to a function. The idea is as
follows. Let us say i write an independent function
Ideal Situation:
ifunc - function(x)
{
return((x*x)-2)
}
mainfunc - function(a,b)
{
evala - ifunc(a)
evalb - ifunc(b)
if (evalaevalb){return(evala)}
else
On 7/24/07, Jenny Barnes [EMAIL PROTECTED] wrote:
Dear R-Help community,
I am trying to overlay a single contour line over a correlation plot using
levelplot in the lattice package. These are the two arrays:
1) a correlation plot over Africa - so each grid square is a different colour
Here is one possible solution:
ifun -
function(a, b, FUN){
evala - FUN(a)
evalb - FUN(b)
if (evala evalb) return(evala) else return(evalb)
}
ifun(1,2,function(x) (x*x) - 2)
On 7/24/07, Anup Nandialath [EMAIL PROTECTED] wrote:
Friends,
I'm trying to pass an equation as an
On Wed, 25 Jul 2007, laimonis wrote:
Consider the following scrap of code:
...slightly modified to
x1 - ts(1:24, start = c(2000, 10), freq = 12)
x2 - ts(1:24, start = c(2000, 11), freq = 12)
and then
y1 - aggregate(x1, nfreq = 4)
gives the desired result while
y2 - aggregate(x2, nfreq
Consider the following scrap of code:
x- ts(1:50,start=c(1,11),freq=12)
y - aggregate(x,nfreq=4)
c(y)
[1] 6 15 24 33 42 51 60 69 78 87 96 105 114 123 132 141
y
Error in rep.int(, start.pad) : invalid number of copies in rep.int()
tsp(y)
[1] 1.83 5.58 4.00
So we
Dear useRs and wizaRds,
I am currently developing a set of functions using S4 classes. On the way I
encountered the problem exemplified with the code below. For some reason the
'validity' method does not seem to work, i.e. does not check for errors in the
specification of the slots of the
Hi Michal --
Add validObject to your initialize method:
setMethod(initialize, someclass,
+ function(.Object, v=numeric(0), l=character(0))
+ {
+ # strip the vector names
+ cv - v
+ cl - l
+ names(cv) - NULL
+ names(cl) - NULL
+ rval - .Object
+ [EMAIL PROTECTED] - cv
Hi,
I tried several settings by using the family=gaussian
in gl1ce, but none of them works.
For the case glm can work.
Here is the error message I got:
glm(Petal.Width~Sepal.Length+Sepal.Width+Petal.Length
,data=iris,family=gaussian())
gl1ce(Petal.Width~Sepal.Length+Sepal.Width+Petal.Length
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