tuple variable in r regression model - r

Question

I'm trying to use Time of the Day as an independent variable in my model. As time is a circular variable, I'm transforming it to (sin(pi * hour / 12), cos(pi * hour / 12)).
I googled around and I still don't know how to create a column in R with the (sin, cos) formatted vector/tuple values. I don't know if the models lm, glm, glm.nb (MASS) and glmer (lme4) can support this kind of data.
Excuse me for being a novice here. If vector-type variable should not be included in a regression model, I'll go to Cross Validated (stats) for suggestion on dealing with circular variables. Please help and share your experience, thanks!

Answer 1

This is an excellent question.
Internally, R uses matrices to fit models to data. Instead of thinking about your model as a functions of tuples, you need to generate a block matrix; this is how splines are implemented, for example.
In your example, one column contains sin(x) and the other cox(x), and the blocks is tagged with a class attribute; the functions that manage this object internally are makepredictcall and predict.
Any models using the standard model.frame / model.matrix processing should be compatible with this.
sincos <- function(x, period=168/2/pi) {
structure(cbind(`_sin`=sin(x/period),
`_cos`=cos(x/period)),
class="sincos",
period=period)
}
Here we set the class and period as attributes.
makepredictcall.sincos <- function(var, call){
if (as.character(call)[1L] != "sincos")
return(call)
call["period"] <- attr(var, "period")
call
}
Set the period as needed on the call.
predict.sincos <- function(object, newx, ...)
{
if(missing(newx))
return(object)
sincos(newx, period=attr(object, "period"))
}
Invoke our function using the period from the fit model.
Here is a short example using lm:
#FAKE DATA EXAMPLE
N <- 1000
hr <- sample(168, N, replace = TRUE)
Y = 5 + sinpi(hr * 2/168) + cospi(hr * 2/168) + rnorm(N)
lm(Y~sinpi(hr*2/168)+cospi(hr*2/168))
#>
#> Call:
#> lm(formula = Y ~ sinpi(hr * 2/168) + cospi(hr * 2/168))
#>
#> Coefficients:
#> (Intercept) sinpi(hr * 2/168) cospi(hr * 2/168)
#> 5.0078 1.0243 0.9637
Our custom function matches exactly:
lm(Y~sincos(hr))
#>
#> Call:
#> lm(formula = Y ~ sincos(hr))
#>
#> Coefficients:
#> (Intercept) sincos(hr)_sin sincos(hr)_cos
#> 5.0078 1.0243 0.9637
Other functions will also be able to tell that the two columns are a single term in the model:
anova(lm(Y~sincos(hr)))
#> Analysis of Variance Table
#>
#> Response: Y
#> Df Sum Sq Mean Sq F value Pr(>F)
#> sincos(hr) 2 1051.05 525.53 553.24 < 2.2e-16 ***
#> Residuals 997 947.06 0.95
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
I'll add this to the stackoverflow package in the next few days, should anyone else find this helpful.