is_formula

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Is object a formula?

is_formula() tests if x is a call to ~. is_bare_formula() tests in addition that x does not inherit from anything else than "formula". is_formulaish() returns TRUE for both formulas and definitions of the type a := b.

Usage
is_formula(x, scoped = NULL, lhs = NULL)

is_bare_formula(x, scoped = NULL, lhs = NULL)

is_formulaish(x, scoped = NULL, lhs = NULL)

Arguments
x

An object to test.

scoped

A boolean indicating whether the quosure or formula is scoped, that is, has a valid environment attribute. If NULL, the scope is not inspected.

lhs

A boolean indicating whether the formula or definition has a left-hand side. If NULL, the LHS is not inspected.

Details

The scoped argument patterns-match on whether the scoped bundled with the quosure is valid or not. Invalid scopes may happen in nested quotations like ~~expr, where the outer quosure is validly scoped but not the inner one. This is because ~ saves the environment when it is evaluated, and quoted formulas are by definition not evaluated.

See Also

is_quosure() and is_quosureish()

Aliases
  • is_formula
  • is_bare_formula
  • is_formulaish
Examples
# NOT RUN {
x <- disp ~ am
is_formula(x)

is_formula(~10)
is_formula(10)

is_formula(quo(foo))
is_bare_formula(quo(foo))

# Note that unevaluated formulas are treated as bare formulas even
# though they don't inherit from "formula":
f <- quote(~foo)
is_bare_formula(f)

# However you can specify `scoped` if you need the predicate to
# return FALSE for these unevaluated formulas:
is_bare_formula(f, scoped = TRUE)
is_bare_formula(eval(f), scoped = TRUE)


# There is also a variant that returns TRUE for definitions in
# addition to formulas:
is_formulaish(a ~ b)
is_formulaish(a := b)
# }
Documentation reproduced from package rlang, version 0.1.6, License: GPL-3

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