(I neglected to cc the list)
---------- Forwarded message ---------
From: Bert Gunter <bgunter.4...@gmail.com>
Date: Mon, Mar 24, 2025 at 11:00 AM
Subject: Re: [R] how to create model matrix of second order terms
To: Stephen Bond <stephen.cb...@yahoo.com>
In view of the fact that your claim that:
"The current setup for formulas does not allow for I(x^2) terms"
is incorrect , e.g.
> x <- 1:5
> model.matrix(~ x + I(x^2))
(Intercept) x I(x^2)
1 1 1 1
2 1 2 4
3 1 3 9
4 1 4 16
5 1 5 25
attr(,"assign")
[1] 0 1 2
(it explicitly works for a numeric x and is nonsense for a factor, as I
explicitly noted in my response)
So my idea is that you should:
a) Study ?formula *carefully*, including the examples.
b) Do as I and Dan Nordlund suggested in our responses
I won't trouble you with any further replies.
Cheers,
Bert
On Mon, Mar 24, 2025 at 8:28 AM Stephen Bond <stephen.cb...@yahoo.com>
wrote:
> Folks,
>
> I appreciate your effort, but maybe I was not explicit enough, so let
> me try again.
>
> The current setup for formulas does not allow for I(x^2) terms as
> explained in the MASS book at the end of Section 6.2 the x:x
> interaction is treated as x.
>
> So I need to write my own code, which is clumsy unless you can refer me
> to a package that already exists or give me an idea how to improve my
> code. Also, writing out terms is not feasible when there are 1000
> variables, so the code needs to deal with taking a wide data frame or
> matrix with column names for convenience.
>
> Let me know your ideas :-)
>
> On Mon, 2025-03-24 at 02:43 -0700, Bert Gunter wrote:
> > Full disclosure: I did not attempt to decipher your code.
> >
> > But ~(A+B +C)^2 - (A + B + C)
> > gives all 2nd order interactions whether the terms are factors or
> > numeric.
> >
> > ~I(A^2) + I(B^2) gives quadratics in A and B, which must be numeric,
> > not factors, of course
> >
> > You can combine these as necessary to get a formula expression for
> > just 2nd order terms. Wrapping this in model.matrix() should then
> > give you the model matrix using "treatment" contrasts for the
> > contrasts involving factors (you can change the contrast types using
> > the 'contrasts.arg' argument of model.matrix())
> >
> > 1. Does this help?
> > 2. Do check this to make sure I'm correct
> >
> > Cheers,
> > Bert
> >
> > "An educated person is one who can entertain new ideas, entertain
> > others, and entertain herself."
> >
> >
> >
> > On Mon, Mar 24, 2025 at 12:22 AM Stephen Bond via R-help
> > <r-help@r-project.org> wrote:
> > > I am sending to this forum as stackoverflow has devolved into sth
> > > pretty bad.
> > > Below code shows how to get what I want in a clumsy way.
> > >
> > > cols <- letters[1:4]
> > > a1 <- outer(cols,cols,paste0)
> > > b1 <- a1[!lower.tri(a1)]
> > >
> > > X <- matrix(rnorm(80),ncol=4)
> > > colnames(X) <- cols
> > > X <- as.data.frame(X)
> > > XX <- matrix(0,nrow=nrow(X),ncol=length(b1))
> > > colnames(XX) <- b1
> > >
> > > for (k in 1:length(b1)){
> > > XX[,k] <- X[,substr(b1[k],1,1)]*X[,substr(b1[k],2,2)]
> > > }
> > >
> > >
> > >
> > > Is there a way to get that using a formula or some neat trick? The
> > > above will not work for factors, so I will need to create the
> > > factor
> > > crossings using formula a*b*c and then cross with the numerics,
> > > which
> > > is even more clumsy.
> > > Thanks everybody
> > >
> > > ______________________________________________
> > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > PLEASE do read the posting guide
> > > https://www.R-project.org/posting-guide.html
> > > and provide commented, minimal, self-contained, reproducible code.
>
>
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