but why don't they give the same fitted values and adjusted R2?
Fitted values won't necessarily be the same, if you don't use all columns from poly.
set.seed(0)
y <- runif(100)
x <- runif(100)
X <- poly(x, 3)
all.equal(lm(y ~ X)$fitted, lm(y ~ x + I(x ^ 2) + I(x ^ 3))$fitted)
#[1] TRUE
all.equal(lm(y ~ X[, 1:2])$fitted, lm(y ~ x + I(x ^ 2))$fitted)
#[1] TRUE
all.equal(lm(y ~ X - 1)$fitted, lm(y ~ x + I(x ^ 2) + I(x ^ 3) - 1)$fitted) ## no intercept
#[1] "Mean relative difference: 33.023"
all.equal(lm(y ~ X[, c(1, 3)])$fitted, lm(y ~ x + I(x ^ 3))$fitted)
#[1] "Mean relative difference: 0.03008166"
all.equal(lm(y ~ X[, c(2, 3)])$fitted, lm(y ~ I(x ^ 2) + I(x ^ 3))$fitted)
#[1] "Mean relative difference: 0.03297488"
We only have ~ 1 + poly(x, degree)[, 1:k] equivalent to ~ 1 + x + I(x ^ 2) + ... + I(x ^ k), for any k <= degree. (I explicitly write out the intercept, to emphasize that we have to start from polynomial of degree 0.)
(The reason is related to how an orthogonal polynomial is generated. See How `poly()` generates orthogonal polynomials? How to understand the "coefs" returned? for great great details. Note that when doing a QR factorization X = QR, as R is an upper triangular matrix (not a diagonal matrix), Q[, ind] will not have the same column space with X[, ind] for an arbitrary subset ind, unless ind = 1:k.)
So, I(x ^ 2) is not equivalent to ploy(x, 2)[, 2], and you will get different fitted values hence (adjusted) R2.
is it valid to use only second term orthogonal polynomial and specify the model as follows?
It is really a bad idea for leaps (or generally any modeler) to drop columns from an orthogonal polynomial. An orthogonal polynomial is a factor-alike term, whose significance is determined by F-statistics (i.e., treating all columns as a whole), rather than t-statistics for individual columns.
In fact, even for raw polynomials, it is not a good idea to omit any low order term. For example, y ~ 1 + I(x ^ 2) omitting linear term is not a good idea. A basic problem here is that it is not invariant to linear shift. For example, if we shift x for x1:
shift <- runif(1) ## an arbitrary value; can be `mean(x)`
x1 <- x - shift
then y ~ 1 + I(x ^ 2) is not equivalent to y ~ 1 + I(x1 ^ 2), but still, y ~ 1 + x + I(x ^ 2) is equivalent to y ~ 1 + x1 + I(x1 ^ 2).
all.equal(lm(y ~ 1 + I(x ^ 2))$fitted, lm(y ~ 1 + I(x1 ^ 2))$fitted)
#[1] "Mean relative difference: 0.02020984"
all.equal(lm(y ~ 1 + x + I(x ^ 2))$fitted, lm(y ~ 1 + x1 + I(x1 ^ 2))$fitted)
#[1] TRUE
I briefly mentioned the issue of dropping columns at R: How to or should I drop an insignificant orthogonal polynomial basis in a linear model?, but my examples here give you more insight.
Is there more direct way to retrieve single model from regsubsets output than specifying the model by hand?
I don't know; at least I did not figure it out almost 2 years ago when answering this thread: Get all models from leaps regsubsets.
One remaining question though. Assuming that leaps returns poly(X, 2)1 I should definitely retain poly(X, 2)1 in my model. But what if only poly(X, 2)1 is returned by leaps? Can higher order term can be dropped then?
There is no problem dropping higher order terms (in this case where you originally fitted a quadratic polynomial). As I said, we have equivalence for ind = 1:j, where j <= degree. But make sure you understand this. Take the following two examples.
- If
leaps drops poly(x, 5)3 and poly(x, 5)5. you can safely remove poly(x, 5)5, but are still advised to retain poly(x, 5)3. This is, instead of fitting an 5-th order polynomial, you fit a 4-th order one.
- If
leaps drops poly(x, 6)3 and poly(x, 6)5. Since poly(x, 6)6 is not dropped, you are advised to drop no terms at all.
