Predict polynomial problem

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 
<cberry at 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.

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

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Charles C. Berry                            (858) 534-2098
                                           Dept of Family/Preventive 
Medicine
E mailto:cberry at 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:cberry at tajo.ucsd.edu                UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901


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