Dimitris.Kapetanakis gmail.com> writes:
> I am using glm function in order to estimate a logit model i.e. glm(Y ~
> data[,2] + data[,3], family = binomial(link = "logit")).
>
> I also created a function that estimates logit model and I would like it to
> compare it with the glm function.
>
> S
Dear all,
I am using glm function in order to estimate a logit model i.e. glm(Y ~
data[,2] + data[,3], family = binomial(link = "logit")).
I also created a function that estimates logit model and I would like it to
compare it with the glm function.
So, does anyone know what optimizer or optimi
I am looking for a way to correct for autocorrelation using for my GLM
models. I can identify the degree of autocorrelation, using PACF, but cannot
figure out a way to specify which variables are the ones that are
autocorrelated, and how to correct for this. Thank you for any help!
--
View this
him about the survey package.
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: Friday, 25 May 2012 9:37p
To: ilai
Cc: Steve Taylor; r-help@r-project.org
Subject: Re: [R] glm(weights) and standard errors
Weighting can be confusing: There are three standard forms of
gestion is the same as one of mine, and doesn't do what
>> I'm looking for.
>>
>>
>> -Original Message-----
>> From: David Winsemius [mailto:dwinsem...@comcast.net]
>> Sent: Tuesday, 22 May 2012 3:37p
>> To: Steve Taylor
>> Cc: r-help
rrected) suggestion is the same as one of mine, and doesn't do what
> I'm looking for.
>
>
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Tuesday, 22 May 2012 3:37p
> To: Steve Taylor
> Cc: r-help@r-project.org
> S
2 3:37p
To: Steve Taylor
Cc: r-help@r-project.org
Subject: Re: [R] glm(weights) and standard errors
On May 21, 2012, at 10:58 PM, Steve Taylor wrote:
> Is there a way to tell glm() that rows in the data represent a certain
> number of observations other than one? Perhaps even fractional
>
On May 21, 2012, at 10:58 PM, Steve Taylor wrote:
Is there a way to tell glm() that rows in the data represent a
certain number of observations other than one? Perhaps even
fractional values?
Using the weights argument has no effect on the standard errors.
Compare the following; is the
Is there a way to tell glm() that rows in the data represent a certain number
of observations other than one? Perhaps even fractional values?
Using the weights argument has no effect on the standard errors. Compare the
following; is there a way to get the first and last models to produce the s
On 17/05/2012 20:35, Sophie Baillargeon wrote:
Hi,
When I run the following code :
Y<- c(rep(0,35),1,2,0,6,8,16,43)
cst<- log(choose(42, 42:1))
beta<- 42:1
tau<- (beta^2)/2
fit<- glm(formula = Y ~ offset(cst) + beta + tau, family = poisson)
fit
fit$converged
glm prints a warning saying that th
Hi Sophie
It helps if you do some detective work
Try
fit1 <- glm(formula = Y ~ offset(cst) + beta + tau, family =
poisson,trace = T, maxit = 200)
and compare
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...
Hi,
When I run the following code :
Y <- c(rep(0,35),1,2,0,6,8,16,43)
cst <- log(choose(42, 42:1))
beta <- 42:1
tau <- (beta^2)/2
fit <- glm(formula = Y ~ offset(cst) + beta + tau, family = poisson)
fit
fit$converged
glm prints a warning saying that the algorithm did not converge.
However, fit$
Hi All,
I have 1 GB dataset in ffdf format. Is there any package / machine learning
algorithms available that I can apply on these ffdf format datasets?
Regards,
Indrajit
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
risks <- inv.logit(predict(model,newd))
risk.diff <- risks[2] - risks[1]
Many thanks,
Dominic C.
2012/3/20 Dominic Comtois
> Case solved. Thanks a lot Peter!
>
> Dominic C.
>
>
> -Message d'origine-
> De : peter dalgaard [mailto:pda...@gmail.com]
> Envo
Case solved. Thanks a lot Peter!
Dominic C.
-Message d'origine-
De : peter dalgaard [mailto:pda...@gmail.com]
Envoyé : 20 mars 2012 07:57
À : Dominic Comtois
Cc : r-help@r-project.org help
Objet : Re: [R] glm: getting the confidence interval for an Odds Ratio, when
using pr
[Oops, forgot cc. to list]
On Mar 20, 2012, at 04:40 , Dominic Comtois wrote:
> I apologize for the errors in the previous code. Here is a reworked example.
> It works, but I suspect problems in the se calculation. I changed, from the
> 1st prediction to the 2nd only one covariate, so that the
I apologize for the errors in the previous code. Here is a reworked
example. It works, but I suspect problems in the se calculation. I changed,
from the 1st prediction to the 2nd only one covariate, so that the OR's CI
should be equal to the exponentiated variable's coefficient and ci. And we
get s
On Mar 19, 2012, at 22:32 , Dominic Comtois wrote:
> Thanks for your answer, much appreciated.
>
> This ain't trivial indeed. I worked my way through it, until I got a "non
> conformable arguments" error when trying to calculate the new standard error.
> Since I'm not following 100% what's hap
Thanks for your answer, much appreciated.
This ain't trivial indeed. I worked my way through it, until I got a "non
conformable arguments" error when trying to calculate the new standard
error. Since I'm not following 100% what's happening, it's hard for me to
figure out what I should do next.
He
On Mar 19, 2012, at 03:32 , Dominic Comtois wrote:
> Say I fit a logistic model and want to calculate an odds ratio between 2
> sets of predictors. It is easy to obtain the difference in the predicted
> logodds using the predict() function, and thus get a point-estimate OR. But
> I can't see how
Say I fit a logistic model and want to calculate an odds ratio between 2
sets of predictors. It is easy to obtain the difference in the predicted
logodds using the predict() function, and thus get a point-estimate OR. But
I can't see how to obtain the confidence interval for such an OR.
For exa
Craig Lyon rogers.com> writes:
>
> Hi,
>
> I am trying to run a generalized linear regression using a negative binomial
> error distribution. However, I want to use an overdispersion parameter that
> varies (dependent on the length of a stretch of road) so glm.nb will not do.
>
> >From what I'
Hi,
I am trying to run a generalized linear regression using a negative binomial
error distribution. However, I want to use an overdispersion parameter that
varies (dependent on the length of a stretch of road) so glm.nb will not do.
>From what I've read I should be able to do this using GLM by s
Thank you.
On Thu, Mar 1, 2012 at 9:58 AM, Bert Gunter wrote:
> Google is your friend! -- as usual.
>
> If you had searched on "glm with regularization" you would have bumped
> into the glmnet R package, which I think is what you're looking for.
>
> -- Bert
>
> On Wed, Feb 29, 2012 at 6:22 PM, Dm
Google is your friend! -- as usual.
If you had searched on "glm with regularization" you would have bumped
into the glmnet R package, which I think is what you're looking for.
-- Bert
On Wed, Feb 29, 2012 at 6:22 PM, Dmitriy Lyubimov wrote:
> Hello,
>
> Thank you for probably not so new questio
Hello,
Thank you for probably not so new question, but i am new to R.
Does any of packages have something like glm+regularization? So far i
see probably something close to that as a ridge regression in MASS but
I think i need something like GLM, in particular binomial regularized
versions of poly
Below. -- Bert
On Thu, Feb 9, 2012 at 9:06 AM, David Winsemius wrote:
>
> On Feb 9, 2012, at 7:32 AM, wo...@posteo.de wrote:
>
>> Dear all,
>>
>> I have question regarding GLMs:
>> I have a discrete response variable and a continuous explaining variable.
>> Like this:
>> http://www.myimg.de/?img=
On Feb 9, 2012, at 7:32 AM, wo...@posteo.de wrote:
Dear all,
I have question regarding GLMs:
I have a discrete response variable and a continuous explaining
variable. Like this:
http://www.myimg.de/?img=example1db0f.jpg
I want to use a GLM to investigate. I have to specify the "familiy
o
Dear all,
I have question regarding GLMs:
I have a discrete response variable and a continuous explaining
variable. Like this:
http://www.myimg.de/?img=example1db0f.jpg
I want to use a GLM to investigate. I have to specify the "familiy of
the distribution of the response variable" - or, maybe
I've originally made 48 GLM binomial models and compare the AIC values. But
dispersion was very large:
Example: Residual deviance: 8811.6 on 118 degrees of freedom
I was suggested to do a quasibinomial afterwards but found that it did not
help the dispersion factor of models and received a warn
I want to run the glm () function for my data but instead of using the
family distributions in R, I need the 4P Burr distribution.
Can some please explain how can I go about doing that. Or please provide
me with an example.
Im new to R.
Eg.
Model1 <- glm(Postwt ~ Prewt + Treat + offset(Pr
Hi everyone,
I just did a GLM binomial regression. I am wondering how do I know when my
responses are fixed?
I have found that an interaction between two factors is the most significant
predictor amongst other factors by comparing the AIC and add1 with each
factor. Does this mean that the inter
On 08/01/12 05:54, emily wrote:
Hi Dr. Snow,
This is the r-help mailing list, not Greg Snow's private email. If
you just want to email Dr. Snow, then email *him* (his address was
given in the post to which you replied).
I am not using R at the moment (working in SPSS, have to love the GUI)
Hi Emily,
This is the R-help forum---it is for R questions, not basic
statistics. You should check out http://stats.stackexchange.com/ for
those type of questions. glm(log(y) ~ x, poisson(link = "identity"))
is not the same as glm(y ~ x, poisson(link = "log")), so I am not
surprised you are gett
Hi Dr. Snow,
I am a graduate student working on analyzing data for my thesis and came
across your post on an R forum:
The default link function for the glm poisson family is a log link, which
means that it is fitting the model:
log(mu) ~ b0 + b1 * x
But the data that you generate is b
Hi Ben,
Yes thanks you are right, I was able to fix it but first I had to fix the data
frame over which I built my model to use numeric for those and then making the
grid values also numeric it finally worked thanks!
Thank you for your help!
Best regards,
Giovanni
On Dec 26, 2011, at 4:57 PM,
Giovanni Azua gmail.com> writes:
>
> Hello,
>
> I have tried reading the documentation and googling for the answer but
reviewing the online matches I end up
> more confused than before.
>
> My problem is apparently simple. I fit a glm model (2^k experiment), and then
I would like to predict th
Hi,
This might be due to the fact that factor levels are arbitary unless
they are ordinal, even that quantitative relationships between levels
are unclear. Therefore, the model has no way to predict unseen factor
levels.
Does it make sense to treat 'No_databases' as numeric instead of a
factor va
Hello,
I have tried reading the documentation and googling for the answer but
reviewing the online matches I end up more confused than before.
My problem is apparently simple. I fit a glm model (2^k experiment), and then I
would like to predict the response variable (Throughput) for unseen fact
Put them in a list:
ModelList <- vector("list", 100)
ModelList[[i]] <- mod.step <- step(mod, direction="both",trace=T)
Then come back and use sapply() to do whatever you want to the set of
models to compare/count/etc them
Michael
On Wed, Nov 30, 2011 at 6:12 AM, Schrabauke wrote:
> Hi
Hi volks,
i have a question about the step() fkt. Is there a possibility to save the
last model generated from this method. I have a loop and so i generate 100
different models with the step fkt and i want to know which model is the
most common.
CODE:
...
missStep -> numeric(100)
for (j in 1:10
Are you sure your variables are categorical or numeric? Of course, glm
differentiates these two kinds of variables. For example, I ran the
same variable with different modes, the results are very different.
> dat<-data.frame(y=rpois(100,5),xf=as.factor(sample(1:4,100,replace=T)))
> glm(y~xf,data=d
Hey all,
I am attempting to replicate my results achieved in another program within
R (so I can expand my options for methods). I am trying to run a GLM (Family
= Poisson) for count data in R. Some of my variables are factors and I am
under the impression that the function glm() cannot run a mode
Many thanks for your replies. I appreciate that.
I tried what you suggested and it did work for the Poisson model (glm,
"poisson" familly). Unfortunately, the negative binomial (glm.nb) did not
work as I work the following message:
Warning messages:
1: In ifelse(y > mu, d.res, -d.res) :
Reache
On 21.10.2011 23:14, Ken wrote:
Your memory shouldn't be capped there,
Where? You cannot know from the output below.
try ?memory.size and ?memory.limit. Background less things.
Good luck,
Ken Hutchison
On Oct 21, 2554 BE, at 11:57 AM, D_Tomas wrote:
My apologies for my vague
Your memory shouldn't be capped there, try ?memory.size and ?memory.limit.
Background less things.
Good luck,
Ken Hutchison
On Oct 21, 2554 BE, at 11:57 AM, D_Tomas wrote:
> My apologies for my vague comment.
>
> My data comprises 400.000 x 21 (17 explanatory variables, plus response
My apologies for my vague comment.
My data comprises 400.000 x 21 (17 explanatory variables, plus response
variable, plus two offsets).
If I build the full model (only linear) I get:
Error: cannot allocate vector of size 112.3 Mb
I have a 4GB RAM laptop... Would i get any improvemnt on a 8G
D_Tomas hotmail.com> writes:
>
> Hi,
>
> I am trying to fi a glm-poisson model to 400.000 records. I have tried biglm
> and glmulti but i have problems... can it really be the case that 400.000
> are too many records???
>
> I am thinking of using random samples of my dataset.
>
"I hav
Hi,
I am trying to fi a glm-poisson model to 400.000 records. I have tried biglm
and glmulti but i have problems... can it really be the case that 400.000
are too many records???
I am thinking of using random samples of my dataset.
Many thanks,
--
View this message in context:
http://r.78
Dear Ben,
First of all, many thanks for your reply. I am highly appreciative of that.
I am still unsure about some issues
The dispersion parameter is that which is estimated by
sum(residuals(fit,type="pearson")^2)/fit$df.res. This is what a quasipoisson
model estimates. This corresponds
D_Tomas hotmail.com> writes:
>
> Hi userRs!
>
> I am trying to fit some GLM-poisson and neg.binomial. The neg. Binomial
> model is to account for over-dispersion.
>
> When I fit the poisson model i get:
> (Dispersion parameter for poisson family taken to be 1)
>
> However, if I estimate the d
Hi userRs!
I am trying to fit some GLM-poisson and neg.binomial. The neg. Binomial
model is to account for over-dispersion.
When I fit the poisson model i get:
(Dispersion parameter for poisson family taken to be 1)
However, if I estimate the dispersion coefficient by means of:
sum(residuals(fi
The minimum achievable level of significance is defined asthe minimum of
Prob(Y=y) over all y's. If I have GLM with a treatment and replicate and I
would like to find out how to compute the minimum achievable level of
significance for that GLM in R
For example, how do I do this for the following d
y is the dependent variable, not a predictor or independent variable. since
this is a binomial model, y should be 0/1 or, atypically, a proportion.
HTH,
Daniel
Samuel Okoye wrote:
>
> Dear all,
>
> I am using glm with quasibinomial. What does the following error message
> mean:
>
> Error in e
Dear all,
I am using glm with quasibinomial. What does the following error message mean:
Error in eval(expr, envir, enclos) : y values must be 0 <= y <= 1
Does it mean that the predictor variable should only have zero and one or it is
also possible to have continuous values between zero and one
to do the prediction for the hold-out data. Is there any
better way for cross-validation to learn a model on training data and test it
on test data in R?
Thanks,
Andra
--- On Mon, 8/22/11, Joshua Wiley wrote:
> From: Joshua Wiley
> Subject: Re: [R] GLM question
> To: "And
Hi Andra,
There are several problems with what you are doing (by the way, I
point them out so you can learn and improve, not to be harsh or rude).
The good news is there is a solution (#3) that is easier than what
you are doing right now!
1) glm.fit() is a function so it is a good idea not to us
Hi All,
I am trying to fit my data with glm model, my data is a matrix of size n*100.
So, I have n rows and 100 columns and my vector y is of size n which contains
the labels (0 or 1)
My question is:
instead of manually typing the model as
glm.fit = glm(y~ x[,1]+x[,2]+...+x[,100], family=bino
Thanks very much again,
I´m reading some papers and articles about this issue and I think i´m
starting to understand the problem.
And thanks for the link to Professor Fox about the non-sequential Anova.
I'll be back with more doubts. I'm sure of that.
-
Mario Garrido Escudero
PhD student
Dpto
On Jul 27, 2011 gaiarrido wrote:
> I've been reading these days about what you tell me, but i don't
> understand properly.
> How could I know, with this tests, which variables are significant?...
Mario,
You need to get in touch with a statistician at your university. You are
fitting quite a comp
Ok, thanks,
I've been reading these days about what you tell me, but i don't understand
properly.
How could I know, with this tests, which variables are significant? I know
my dependent variable depends on the lcc and on the edadysexo. but only one
per test seems to be significant.
Thanks again
-
On Jul 24, 2011 Gaiarrido wrote:
> Why the order of the factors give different results? I suppose it's
> because the order of the
> factors, i've just changed "lcc" from the first position to the last in
> the model, and the significance
> change completely
> ...snip <
> Ijow can i know what's c
I've read something about this problem, but I don't know how can i avoid this
problem.
Why the order of the factors give different results? I suppose it's because
the order of the factors, i've just changed "lcc" from the first position to
the last in the model, and the significance change complete
Peter Maclean yahoo.com> writes:
>
> In glm() you can use the summary() function to recover
> the shape parameter (the reciprocal of the
> dispersion parameter). How do you recover the scale parameter?
> Also, in the given example, how I estimate
> and save the geometric mean of the predicted v
In glm() you can use the summary() function to recover the shape parameter (the
reciprocal of the dispersion parameter). How do you recover the scale
parameter? Also, in the given example, how I estimate and save the geometric
mean of the predicted values? For a simple model you can use fitted()
On Jun 14, 2011, at 09:53 , Anna Mill wrote:
>
> Also note that success+failure is exactly 102 in fragment 1 and 105 in
> fragment 2, as is the sum of the successes for each fragment (of course it
> has to to make exactly 1/4). It is rather easy to suspect that it is actually
> a 0/1 coding o
> Also note that success+failure is exactly 102 in fragment 1 and 105 in
> fragment 2, as is the sum of the successes for each fragment (of course it
> has to to make exactly 1/4). It is rather easy to suspect that it is
> actually a 0/1 coding of the type (as in "tick exactly one box"), and not
>
On Jun 14, 2011, at 08:13 , Prof Brian Ripley wrote:
> I presume you intended 'type' and 'fragment' to be factors (see below). Such
> a model would fit exactly. The additive model
>
>> model <- glm(y ~ fragment+type, binomial)
>
> is only modestly over-dispersed, and shows that 'fragment' ha
thanks for the answer!
yes, indeed, type and fragment should be factors but it was no artificial
data!
2011/6/14 Prof Brian Ripley
> I presume you intended 'type' and 'fragment' to be factors (see below).
> Such a model would fit exactly. The additive model
>
>
> model <- glm(y ~ fragment+typ
I presume you intended 'type' and 'fragment' to be factors (see
below). Such a model would fit exactly. The additive model
model <- glm(y ~ fragment+type, binomial)
is only modestly over-dispersed, and shows that 'fragment' has zero
effect. Not 'a negligible effect', but no effect. So so
Dear all,
I am new to R and my question may be trivial to you...
I am doing a GLM with binomial errors to compare proportions of species in
different categories of seed sizes (4 categories) between 2 sites.
In the model summary the residual deviance is much higher than the degree
of freedom (Resi
On Apr 21, 2011, at 11:30 , Jeffrey Pollock wrote:
> So am I right in saying that Binary data isnt the only case where this is
> true? It would make sense to me that for a multinomial model you could have a
> unique factor for each data point and thus be able to create a likelihood of
> 1.
Ye
algorithm until the coefficients where either 'Inf' or '-Inf'.
Please let me know your thoughts on this.
Thanks again,
Jeff
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: 21 April 2011 09:32
To: Juliet Hannah
Cc: Jeffrey Pollock; r-help@r-p
On Apr 21, 2011, at 05:14 , Juliet Hannah wrote:
> As you mentioned, the deviance does not always reduce to:
>
> D = -2(loglikelihood(model))
>
> It does for ungrouped data, such as for binary logistic regression.
To be precise, it only happens when the log likelihood of the saturated model
i
As you mentioned, the deviance does not always reduce to:
D = -2(loglikelihood(model))
It does for ungrouped data, such as for binary logistic regression. So
let's stick with the original definition.
In this case, we need the log-likelihood for the saturated model.
x = rnorm(10)
y = rpois(10,l
Sacha Viquerat web.de> writes:
>
> Am 15.04.2011 20:14, schrieb Christian Hennig:
> > Normality of the predictors doesn't belong to the assumptions of the
> > GLM, so you don't have to check this.
> >
> > On Fri, 15 Apr 2011, Simone Santoro wrote:
> >
> >> I want to estimate the possible effects
Am 15.04.2011 20:14, schrieb Christian Hennig:
Normality of the predictors doesn't belong to the assumptions of the
GLM, so you don't have to check this.
Note, however, that there are all kinds of potential problems which to
detect is fairly hopeless with n=11 and three predictors, so you
should
Normality of the predictors doesn't belong to the assumptions of the GLM,
so you don't have to check this.
Note, however, that there are all kinds of potential problems which to
detect is fairly hopeless with n=11 and three predictors, so you shouldn't
be too confident about your results anywa
Hi,
I have found quite a few posts on normality checking of response variables, but
I am still in doubt about that. As it is easy to understand I'm not a
statistician so be patient please.
I want to estimate the possible effects of some predictors on my response
variable that is nº of males an
It has always been my understanding that deviance for GLMs is defined
by;
D = -2(loglikelihood(model) - loglikelihood(saturated model))
and this can be calculated by (or at least usually is);
D = -2(loglikelihood(model))
As is done so in the code for 'polr' by Brian Ripley (in the
Sascha
Thanks that works.
Dirk
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https://stat.eth
On 11-04-09 10:34, dirknbr wrote:
I am looping through various models with different combinations of
independent variables which are stored as columns in x
glm(y ~ ??, data=x)
How can I pass the colnames of the selected columns of x into ?? seperating
them with a +
ie I want to generate
glm(
I am looping through various models with different combinations of
independent variables which are stored as columns in x
glm(y ~ ??, data=x)
How can I pass the colnames of the selected columns of x into ?? seperating
them with a +
ie I want to generate
glm(y ~ x1 + x2, data=x)
glm(y ~ x2 + x3,
On 4/6/2011 2:17 PM, dirknbr wrote:
I am aware this has been asked before but I could not find a resolution.
I am doing a logit
lg<- glm(y[1:200] ~ x[1:200,1],family=binomial)
glm (and most modeling functions) are designed to work with data frames,
not raw vectors.
Then I want to predict
Dear Dirk,
You should avoid indexing in the glm call so that the name of the terms
will not contain the indexing part. (Check str(lg) in your example.)
A more preferred solution uses predefined data frames in the original calls:
n <- 250
x <- rnorm(n)
noise <- rnorm(n,0,0.3)
y <- round(exp(x+nois
I am aware this has been asked before but I could not find a resolution.
I am doing a logit
lg <- glm(y[1:200] ~ x[1:200,1],family=binomial)
Then I want to predict a new set
pred <- predict(lg,x[201:250,1],type="response")
But I get varying error messages or warnings about the different number
Update:
turns out there was a sister posting to mine two years ago:
http://r.789695.n4.nabble.com/Zinb-for-Non-interger-data-td898206.html
It was then suggested to use a zero-inflated distribution from the
gamlss package. It turns out that they do have a zero-adjusted (albeit
not strictly speakin
On Mar 30, 2011; 11:41am Mikhail wrote:
>> I'm wondering if there's any way to do the same in R (lme can't deal
>> with this, as far as I'm aware).
You can do this using the pscl package.
Regards, Mark.
--
View this message in context:
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Hi Dennis,
Thanks - these were the first things I tried, but the problem is that
they refuse to work with non-count data...
Mikhail
On Wed, Mar 30, 2011 at 12:56 PM, Dennis Murphy wrote:
> Hi:
>
> You might want to consider hurdle models in the pscl package.
>
> HTH,
> Dennis
>
> On Wed, Mar 30,
Hi:
You might want to consider hurdle models in the pscl package.
HTH,
Dennis
On Wed, Mar 30, 2011 at 2:41 AM, a11msp wrote:
> Hello,
>
> I'd like to implement a regression model for extremely zero-inflated
> continuous data using a conditional approach, whereby zeroes are
> modelled as coming
Hello,
I'd like to implement a regression model for extremely zero-inflated
continuous data using a conditional approach, whereby zeroes are
modelled as coming from a binary distribution, while non-zero values
are modelled as log-normal.
So far, I've come across two solutions for this: one, in R,
Hello,
I am analyzing a dataset where the response is count data. I have one
two-level factor that is repeated within-subjects and additional
between-subject variables that are either categorical or continuous. I have
previously modeled a comparable dataset (without the within-subjects factor)
u
Dear list,
My question to follow is not a pure R question but contains also a
more general statistical/econometrical part, but I was hoping that
perhaps someone knowledgable on this list could offer some help.
I have estimated a binary logistic regression model and would like to
calculate average
Please read the Help for predict.glm carefully to make sure you are
not confusing predicted response on the linear scale (log odds) with
that on the probability scale.
The warning is just that: a warning. It means that you have fitted
PROBABILITIES on the boundary, which might compromise the itera
Hi there,
I am encountering a problem with the GLM tool performing logistic regression.
After computing a warning appears, saying “glm.fit: fitted probabilities
numerically 0 or 1 occurred”. A prediction of new values confirms the problem
as the model does not produce regular probability estima
Hi Celine,
GLM outputs usually give the null deviance and residual deviance in the
summary() term - so you can work out % deviance explained for a variable/model
from this. Hope this helps.
Best wishes,
Clare
Dr Clare B Embling
Visiting Research Fellow
Marine Inst
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Hello,
I wanted to ask if there is an R package to fit GLM (logistic for
example) via empirical likelihood.
--
-Tony
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PLEASE
Your first model is a binomial glm witb 4 observations of 6,6,4,4
trials.
Your second model is a Bernoulli glm with 20 observations of one trial
each.
The saturated models are different, as are the likelihoods
(unsurprising given the data is different): the binomial model has
comnbinarial f
Hi,
when I apply a glm() model in two ways,
first with the response in a two column matrix specification with
successes and failures
y <- matrix(c(
5, 1,
3, 3,
2, 2,
0, 4), ncol=2, byrow=TRUE)
X <- data.frame(x1 = factor(c(1,1,0,0)),
x2 = factor(c(0,1,0,1)))
g
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