is_call: Is object a call?

In hadley/rlang: Functions for Base Types and Core R and 'Tidyverse' Features

Is object a call?

Description

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.

Usage

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

Arguments

x	An object to test. Formulas and quosures are treated literally.
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.

See Also

is_expression()

Examples

is_call(quote(foo(bar)))

# You can pattern-match the call with additional arguments:
is_call(quote(foo(bar)), "foo")
is_call(quote(foo(bar)), "bar")
is_call(quote(foo(bar)), quote(foo))

# Match the number of arguments with is_call():
is_call(quote(foo(bar)), "foo", 1)
is_call(quote(foo(bar)), "foo", 2)


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

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

# Or if it should have a namespace:
is_call(ns_expr, "list", ns = "utils")
is_call(ns_expr, "list", ns = "base")

# You can supply multiple namespaces:
is_call(ns_expr, "list", ns = c("utils", "base"))
is_call(ns_expr, "list", ns = c("utils", "stats"))

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


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

hadley/rlang documentation built on Dec. 17, 2024, 3:26 a.m.