Hi,
Sorry, I should have said the bigglm() is in the biglm package.
I have tried putting the offset in the model formula as you suggested
but it doesn't make any difference. I was comparing the results of
bigglm() with glm() on a small data set before I applied it to the much
larger data set
Dennis Murphy gmail.com> writes:
>
> Hi:
>
> Did you try putting the offset in the model formula, as in
>
> bigglm( y ~ offset(z) + x, ...) ?
>
> I haven't tried bigglm() personally (BTW, it's in the biglm package, which
> wasn't mentioned), but this syntax works in the standard glm() fu
Hi:
Did you try putting the offset in the model formula, as in
bigglm( y ~ offset(z) + x, ...) ?
I haven't tried bigglm() personally (BTW, it's in the biglm package, which
wasn't mentioned), but this syntax works in the standard glm() function, so
perhaps it maps to bigglm() as well... (?)
Dear all,
I have a large data set and would like to fit a logistic regression
model using the bigglm function. I need to include an offset in the
model but when I do this the bigglm function seems to ignore it.
For example, running the two models below produces the same model and
the offset
Sorry, I had no seen your previous e-mail. Just another question. Is
there any way to obtain an R2 to have a numeric idea of how good is the
fitting?
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma de Barcelona
Edifici V, Campus UA
Actually "drm" as posted before fits a sigmoid curve (a generalized
logistic function with 4 parameters, see ?LL.4), so I didn't get the
point of your new question.
Dani Valverde schrieb:
Thank you all for your answers. If you look at the plot resulting from
my data, it seems that it is some
Thank you all for your answers. If you look at the plot resulting from
my data, it seems that it is some kind of sigmoid function, not only
polynomial. How could I fit it?
Best,
Dani
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma
you might use the drc-package (equivalently you could use nls with an
appropriate "selfstart" model like SSlogis)
library(drc)
mm<-drm(delta~ph,fct=LL.4())
plot(mm)
From your plot I was assuming that "ph" is the independent variable (as
modelled above) - so if you want to predict a ph from del
If you are looking for a parameteric form then a polynomial seems
to work:
plot(delta ~ ph)
for(i in 1:4) lines(ph, fitted(lm(delta ~ poly(ph, i))), col = i, lty = i)
legend("topright", legend = 1:4, col = 1:4, lty = 1:4)
On Thu, Nov 20, 2008 at 6:53 AM, Dani Valverde <[EMAIL PROTECTED]> wrote:
>
Hello,
This is a very basic question, but I don'y know the answer. I have these
data
delta <-
c(28.6-8.825,28.6-8.828,28.6-8.836,28.6-8.845,28.6-8.897,28.6-8.944,28.6-9.027,28.6-9.091,28.6-9.263,28.6-9.4,28.6-9.7,28.6-9.981,
28.6-10.287,28.6-10.48,28.6-10.684,28.6-10.875)
ph <- c(4.4,4.6,4.8,
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