MichaelChirico MichaelChirico - 2 months ago 13
R Question

Defining an infix operator for use within a formula

I'm trying to create a more parsimonious version of this solution, which entails specifying the RHS of a formula in the form

d1 + d1:d2
.

Given that
*
in the context of a formula is a pithy stand-in for full interaction (i.e.
d1 * d2
gives
d1 + d2 + d1:d2
), my approach has been to try and define an alternative operator, say
%+:%
using the infix approach I've grown accustomed to in other applications, a la:

"%+:%" <- function(d1,d2) d1 + d2 + d1:d2


However, this predictably fails because I haven't been careful about evaluation; let's introduce an example to illustrate my progress:

set.seed(1029)
v1 <- runif(1000)
v2 <- runif(1000)
y <- .8*(v1 < .3) + .2 * (v2 > .25 & v2 < .8) -
.4 * (v2 > .8) + .1 * (v1 > .3 & v2 > .8)


With this example, hopefully it's clear why simply writing out the two terms might be undesirable:

y ~ cut(v2, breaks = c(0, .25, .8, 1)) +
cut(v2, breaks = c(0, .25, .8, 1)):I(v1 < .3)


One workaround which is close to my desired output is to define the whole formula as a function:

plus.times <- function(outvar, d1, d2){
as.formula(paste0(quote(outvar), "~", quote(d1),
"+", quote(d1), ":", quote(d2)))
}


This gives the expected coefficients when passed to
lm
, but with names that are harder to interpret directly (especially in the real data where we take care to give
d1
and
d2
descriptive names, in contrast to this generic example):

out1 <- lm(y ~ cut(v2, breaks = c(0, .25, .8, 1)) +
cut(v2, breaks = c(0, .25, .8, 1)):I(v1 < .3))
out2 <- lm(plus.times(y, cut(v2, breaks = c(0, .25, .8, 1)), I(v1 < .3)))
any(out1$coefficients != out2$coefficients)
# [1] FALSE
names(out2$coefficients)
# [1] "(Intercept)" "d1(0.25,0.8]" "d1(0.8,1]" "d1(0,0.25]:d2TRUE"
# [5] "d1(0.25,0.8]:d2TRUE" "d1(0.8,1]:d2TRUE"


So this is less than optimal.

Is there any way to define the adjust the code so that the infix operator I mentioned above works as expected? How about altering the form of
plus.times
so that the variables are not renamed?

I've been poking around (
?formula
,
?"~"
,
?":"
,
getAnywhere(formula.default)
, this answer, etc.) but haven't seen how exactly R interprets
*
when it's encountered in a formula so that I can make my desired minor adjustments.

Answer

You do not need to define a new operator in this case: in a formula d1/d2 expands to d1 + d1:d2. In other words d1/d2 specifies that d2 is nested within d1. Continuing your example:

out3 <- lm(y ~ cut(v2,breaks=c(0,.25,.8,1))/I(v1 < .3))
all.equal(coef(out1), coef(out3))
# [1] TRUE

Further comments

Factors may be crossed or nested. Two factors are crossed if it possible to observe every combination of levels of the two factors, e.g. sex and treatment, temperature and pH, etc. A factor is nested within another if each level of that factor can only be observed within one of the levels of the other factor, e.g. town and country, staff member and store etc.

These relationships are reflected in the parametrization of the model. For crossed factors we use d1*d2 or d1 + d2 + d1:d2, to give the main effect of each factor, plus the interaction. For nested factors we use d1/d2 or d1 + d1:d2 to give a separate submodel of the form 1 + d2 for each level of d1.

The idea of nesting is not restricted to factors, for example we may use sex/x to fit a separate linear regression on x for males and females.

In a formula, %in% is equivalent to :, but it may be used to emphasize the nested, or hierarchical structure of the data/model. For example, a + b %in% a is the same as a + a:b, but reading it as "a plus b within a" gives a better description of the model being fitted. Even so, using / has the advantage of simplifying the model formula at the same time as emphasizing the structure.

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