Hello,
Another way is to cbind the vectors 'a' and 'b', but this needs argument
'raw' set to TRUE.
poly(cbind(a, b), 6, raw=TRUE)
To the OP: is this time series related? With 6 being a lag or test
(e.g., Tsay, 1986) order? I'm asking this because package nlts has a
function for this test up to order 5 and it uses poly().
Hope this helps,
Rui Barradas
Em 02-07-2012 16:04, David Winsemius escreveu:
On Jul 2, 2012, at 10:51 AM, David Winsemius wrote:
On Jul 2, 2012, at 9:29 AM, YTP wrote:
I would like to specify a model with all polynomial interaction terms
between
two variables, say, up to degree 6. For example, terms like a^6 + (a^5 *
b^1) + (a^4 * b^2) + ... and so on. The documentation states
The ^ operator indicates crossing to the specified degree.
so I would expect a model specified as y ~ (a+b)^6 to produce these
terms.
However doing this only returns four slope coefficients, for
Intercept, a,
b, and a:b. Does anyone know how to produce the desired result?
Thanks in
advance.
You might try:
poly(a,6)*poly(b,6)
(untested ... and it looks somewhat dangerous to me.)
Well, now it's tested and succeeds at least numerically. Also tested
( poly(a,6) +poly(b,6) )^2 with identical results.
Whether this is wise practice remains in doubt:
dfrm <- data.frame(out=rnorm(100), a=rnorm(100), b=rnorm(100) )
anova(lm( out ~ (poly(a,6) +poly(b,6) )^2, data=dfrm) )
#-----------------------
Analysis of Variance Table
Response: out
Df Sum Sq Mean Sq F value Pr(>F)
poly(a, 6) 6 12.409 2.06810 3.0754 0.01202 *
poly(b, 6) 6 5.321 0.88675 1.3187 0.26596
poly(a, 6):poly(b, 6) 36 41.091 1.14142 1.6974 0.04069 *
Residuals 51 34.295 0.67246
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
