is_formula()
tests whether x
is a call to ~
. is_bare_formula()
tests in addition that x
does not inherit from anything else than
"formula"
.
Note: When we first implemented is_formula()
, we thought it
best to treat unevaluated formulas as formulas by default (see
section below). Now we think this default introduces too many edge
cases in normal code. We recommend always supplying scoped = TRUE
. Unevaluated formulas can be handled via a is_call(x, "~")
branch.
is_formula(x, scoped = NULL, lhs = NULL) is_bare_formula(x, scoped = TRUE, lhs = NULL)
x  An object to test. 

scoped  A boolean indicating whether the quosure is scoped,
that is, has a valid environment attribute and inherits from

lhs  A boolean indicating whether the formula has a lefthand
side. If 
At parse time, a formula is a simple call to ~
and it does not
have a class or an environment. Once evaluated, the ~
call
becomes a properly structured formula. Unevaluated formulas arise
by quotation, e.g. ~~foo
, quote(~foo)
, or substitute(arg)
with arg
being supplied a formula. Use the scoped
argument to
check whether the formula carries an environment.
is_formula(~10)#> [1] TRUEis_formula(10)#> [1] FALSE# If you don't supply `lhs`, both onesided and twosided formulas # will return `TRUE` is_formula(disp ~ am)#> [1] TRUEis_formula(~am)#> [1] TRUE# You can also specify whether you expect a LHS: is_formula(disp ~ am, lhs = TRUE)#> [1] TRUEis_formula(disp ~ am, lhs = FALSE)#> [1] FALSEis_formula(~am, lhs = TRUE)#> [1] FALSEis_formula(~am, lhs = FALSE)#> [1] TRUE# Handling of unevaluated formulas is a bit tricky. These formulas # are special because they don't inherit from `"formula"` and they # don't carry an environment (they are not scoped): f < quote(~foo) f_env(f)#> NULL# By default unevaluated formulas are treated as formulas is_formula(f)#> [1] TRUE# Supply `scoped = TRUE` to ensure you have an evaluated formula is_formula(f, scoped = TRUE)#> [1] FALSE# By default unevaluated formulas not treated as bare formulas is_bare_formula(f)#> [1] FALSE# If you supply `scoped = TRUE`, they will be considered bare # formulas even though they don't inherit from `"formula"` is_bare_formula(f, scoped = TRUE)#> [1] FALSE