Charles C. Berry wrote:
On Tue, 19 Jan 2010, Charles C. Berry wrote:
and the values in those places are different:
On Tue, 19 Jan 2010, Barry Rowlingson wrote:
On Tue, Jan 19, 2010 at 1:36 AM, Charles C. Berry
<cbe...@tajo.ucsd.edu>
wrote:
> Its the environment thing.
> > I think you want something like this:
> > � � � � � � � �models[[i]]=lm( bquote( y ~ poly(x,.(i)) ), data=d)
> > Use
> � � � � � � � �terms( mmn[[3]] )
> > both with and without this change and
> > > � � � � � � � �ls( env = environment( formula( mmn[[3]] ) ) )
> � � � � � � � �get("i",env=environment(formula(mmn[[3]])))
> � � � � � � � �sapply(mmn,function(x) environment( formula( x ) ) )
> > > to see what gives.
Think I see it now. predict involves evaluating poly, and poly here
needs 'i' for the order. If the right 'i' isn't gotten when predict is
called then I get the error. Your fix sticks the right 'i' into the
environment when predict is called.
I haven't quite got my head round _how_ it does it, and I have no
idea how I could have figured this out for myself. Oh well...
Per ?bquote, "bquote quotes its argument except that terms wrapped in
'.()' are evaluated in the specified 'where' environment.
(which by default is the parent.frame)
Note:
i <- 20
bquote(y ~ poly(x,.(i)))
y ~ poly(x, 20)
So, now 'i' is irrelevant as the expression returned by bquote has
'20' as the 'degree' arg.
The following lines are also illustrative:
d = data.frame(x=1:10,y=runif(10))
i=3
#1 naive model:
m1 = lm(y~poly(x,i),data=d)
#2,3 bquote, without or with i-wrapping:
m2 = lm(bquote(y~poly(x,i)),data=d)
m3 = lm(bquote(y~poly(x,.(i))),data=d)
#1 works, gets 'i' from global i=3 above:
predict(m1,newdata=data.frame(x=9:11))
#2 fails - why?
predict(m2,newdata=data.frame(x=9:11))
Well, the terms() objects are the same:
all.equal(terms(m1),terms(m2))
[1] TRUE
but they will look in different places for 'i':
environment(terms(m2))
<environment: 0x01b7c178>
environment(terms(m1))
<environment: R_GlobalEnv>
and the values in those places are different:
environment(terms(m2))$i
[1] 2
environment(terms(m1))$i
[1] 3
And I should have mentioned that environment(terms(m2)) happens to have
an object 'i' in it regardless of whether poly() is used.
Right. It might be worth pointing out that the 'i' in
environment(terms(m2)) or in environment(terms(m3)),
which also has an 'i', is there even if you use poly(x,j).
It is (if I'm not mistaken) the number of 'variables' in
the formula: response plus predictor terms. Thus
j <- 5
m4 <- lm(bquote(y ~ sqrt(x) + poly(x, .(j))), data=d)
environment(terms(m4))$i
[1] 3
-Peter Ehlers
Chuck
#3 works, gets 'i' from within:
predict(m3,newdata=data.frame(x=9:11))
It doesn't need 'i', because the i was evaluated and substituted by
bquote. That is, it doesn't get("i") as the expression returned by
bquote has no 'i' in it.
HTH,
Chuck
rm(i)
#1 now fails because we removed 'i' from top level:
predict(m1,newdata=data.frame(x=9:11))
#2 still fails:
predict(m2,newdata=data.frame(x=9:11))
#3 still works:
predict(m3,newdata=data.frame(x=9:11))
Thanks
--
blog: http://geospaced.blogspot.com/
web: http: //www.maths.lancs.ac.uk/~rowlings
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Charles C. Berry (858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
92093-0901
Charles C. Berry (858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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