Is object a call?

This function tests if x is a call. This is a pattern-matching predicate that returns FALSE if name and n are supplied and the call does not match these properties.

is_call(x, name = NULL, n = NULL, ns = NULL)

Arguments

x	An object to test. If a formula, the right-hand side is extracted.
name	An optional name that the call should match. It is passed to sym() before matching. This argument is vectorised and you can supply a vector of names to match. In this case, is_call() returns TRUE if at least one name matches.
n	An optional number of arguments that the call should match.
ns	The namespace of the call. If NULL, the namespace doesn't participate in the pattern-matching. If an empty string "" and x is a namespaced call, is_call() returns FALSE. If any other string, is_call() checks that x is namespaced within ns. Can be a character vector of namespaces, in which case the call has to match at least one of them, otherwise is_call() returns FALSE.

Life cycle

is_lang() has been soft-deprecated and renamed to is_call() in rlang 0.2.0 and similarly for is_unary_lang() and is_binary_lang(). This renaming follows the general switch from "language" to "call" in the rlang type nomenclature. See lifecycle section in call2().

See also

is_expression()

Examples

is_call(quote(foo(bar)))
#> [1] TRUE

# You can pattern-match the call with additional arguments:
is_call(quote(foo(bar)), "foo")
#> [1] TRUE
is_call(quote(foo(bar)), "bar")
#> [1] FALSE
is_call(quote(foo(bar)), quote(foo))
#> [1] TRUE

# Match the number of arguments with is_call():
is_call(quote(foo(bar)), "foo", 1)
#> [1] TRUE
is_call(quote(foo(bar)), "foo", 2)
#> [1] FALSE


# By default, namespaced calls are tested unqualified:
ns_expr <- quote(base::list())
is_call(ns_expr, "list")
#> [1] TRUE

# You can also specify whether the call shouldn't be namespaced by
# supplying an empty string:
is_call(ns_expr, "list", ns = "")
#> [1] FALSE

# Or if it should have a namespace:
is_call(ns_expr, "list", ns = "utils")
#> [1] FALSE
is_call(ns_expr, "list", ns = "base")
#> [1] TRUE

# You can supply multiple namespaces:
is_call(ns_expr, "list", ns = c("utils", "base"))
#> [1] TRUE
is_call(ns_expr, "list", ns = c("utils", "stats"))
#> [1] FALSE

# If one of them is "", unnamespaced calls will match as well:
is_call(quote(list()), "list", ns = "base")
#> [1] FALSE
is_call(quote(list()), "list", ns = c("base", ""))
#> [1] TRUE
is_call(quote(base::list()), "list", ns = c("base", ""))
#> [1] TRUE


# The name argument is vectorised so you can supply a list of names
# to match with:
is_call(quote(foo(bar)), c("bar", "baz"))
#> [1] FALSE
is_call(quote(foo(bar)), c("bar", "foo"))
#> [1] TRUE
is_call(quote(base::list), c("::", ":::", "$", "@"))
#> [1] TRUE