Elixir v1.5.2 Keyword View Source

A set of functions for working with keywords.

A keyword is a list of two-element tuples where the first element of the tuple is an atom and the second element can be any value.

For example, the following is a keyword list:

[{:exit_on_close, true}, {:active, :once}, {:packet_size, 1024}]

Elixir provides a special and more concise syntax for keyword lists that looks like this:

[exit_on_close: true, active: :once, packet_size: 1024]

This is also the syntax that Elixir uses to inspect keyword lists:

iex> [{:active, :once}]
[active: :once]

The two syntaxes are completely equivalent. Note that when keyword lists are passed as the last argument to a function, if the short-hand syntax is used then the square brackets around the keyword list can be omitted as well. For example, the following:

String.split("1-0", "-", trim: true, parts: 2)

is equivalent to:

String.split("1-0", "-", [trim: true, parts: 2])

A keyword may have duplicated keys so it is not strictly a key-value store. However most of the functions in this module behave exactly as a dictionary so they work similarly to the functions you would find in the Map module.

For example, Keyword.get/3 will get the first entry matching the given key, regardless if duplicated entries exist. Similarly, Keyword.put/3 and Keyword.delete/3 ensure all duplicated entries for a given key are removed when invoked. Note that operations that require keys to be found in the keyword list (like Keyword.get/3) need to traverse the list in order to find keys, so these operations may be slower than their map counterparts.

A handful of functions exist to handle duplicated keys, in particular, Enum.into/2 allows creating new keywords without removing duplicated keys, get_values/2 returns all values for a given key and delete_first/2 deletes just one of the existing entries.

The functions in Keyword do not guarantee any property when it comes to ordering. However, since a keyword list is simply a list, all the operations defined in Enum and List can be applied too, especially when ordering is required.

Link to this section Summary

Functions

Deletes the entries in the keyword list for a specific key

Deletes the entries in the keyword list for a key with value

Deletes the first entry in the keyword list for a specific key

Drops the given keys from the keyword list

Checks if two keywords are equal

Fetches the value for a specific key and returns it in a tuple

Fetches the value for specific key

Gets the value for a specific key

Gets the value from key and updates it, all in one pass

Gets the value from key and updates it. Raises if there is no key

Gets the value for a specific key

Gets all values for a specific key

Returns whether a given key exists in the given keywords

Returns all keys from the keyword list

Returns true if term is a keyword list; otherwise returns false

Merges two keyword lists into one

Merges two keyword lists into one

Returns an empty keyword list, i.e. an empty list

Creates a keyword from an enumerable

Creates a keyword from an enumerable via the transformation function

Returns and removes all values associated with key in the keyword list

Returns and removes the first value associated with key in the keyword list

Lazily returns and removes all values associated with key in the keyword list

Puts the given value under key

Puts the given value under key unless the entry key already exists

Evaluates fun and puts the result under key in keyword list unless key is already present

Alters the value stored under key to value, but only if the entry key already exists in the keyword list

Similar to replace/3, but will raise a KeyError if the entry key does not exist

Takes all entries corresponding to the given keys and extracts them into a separate keyword list

Takes all entries corresponding to the given keys and returns them in a new keyword list

Returns the keyword list itself

Updates the key in keywords with the given function

Updates the key with the given function

Returns all values from the keyword list

Link to this section Types

Link to this section Functions

Link to this function delete(keywords, key) View Source
delete(t(), key()) :: t()

Deletes the entries in the keyword list for a specific key.

If the key does not exist, returns the keyword list unchanged. Use delete_first/2 to delete just the first entry in case of duplicated keys.

Examples

iex> Keyword.delete([a: 1, b: 2], :a)
[b: 2]
iex> Keyword.delete([a: 1, b: 2, a: 3], :a)
[b: 2]
iex> Keyword.delete([b: 2], :a)
[b: 2]
Link to this function delete(keywords, key, value) View Source
delete(t(), key(), value()) :: t()

Deletes the entries in the keyword list for a key with value.

If no key with value exists, returns the keyword list unchanged.

Examples

iex> Keyword.delete([a: 1, b: 2], :a, 1)
[b: 2]
iex> Keyword.delete([a: 1, b: 2, a: 3], :a, 3)
[a: 1, b: 2]
iex> Keyword.delete([a: 1], :a, 5)
[a: 1]
iex> Keyword.delete([a: 1], :b, 5)
[a: 1]
Link to this function delete_first(keywords, key) View Source
delete_first(t(), key()) :: t()

Deletes the first entry in the keyword list for a specific key.

If the key does not exist, returns the keyword list unchanged.

Examples

iex> Keyword.delete_first([a: 1, b: 2, a: 3], :a)
[b: 2, a: 3]
iex> Keyword.delete_first([b: 2], :a)
[b: 2]
Link to this function drop(keywords, keys) View Source
drop(t(), [key()]) :: t()

Drops the given keys from the keyword list.

Duplicated keys are preserved in the new keyword list.

Examples

iex> Keyword.drop([a: 1, b: 2, c: 3], [:b, :d])
[a: 1, c: 3]
iex> Keyword.drop([a: 1, b: 2, b: 3, c: 3, a: 5], [:b, :d])
[a: 1, c: 3, a: 5]
Link to this function equal?(left, right) View Source
equal?(t(), t()) :: boolean()

Checks if two keywords are equal.

Two keywords are considered to be equal if they contain the same keys and those keys contain the same values.

Examples

iex> Keyword.equal?([a: 1, b: 2], [b: 2, a: 1])
true
iex> Keyword.equal?([a: 1, b: 2], [b: 1, a: 2])
false
iex> Keyword.equal?([a: 1, b: 2, a: 3], [b: 2, a: 3, a: 1])
true
Link to this function fetch(keywords, key) View Source
fetch(t(), key()) :: {:ok, value()} | :error

Fetches the value for a specific key and returns it in a tuple.

If the key does not exist, returns :error.

Examples

iex> Keyword.fetch([a: 1], :a)
{:ok, 1}
iex> Keyword.fetch([a: 1], :b)
:error
Link to this function fetch!(keywords, key) View Source
fetch!(t(), key()) :: value() | no_return()

Fetches the value for specific key.

If key does not exist, a KeyError is raised.

Examples

iex> Keyword.fetch!([a: 1], :a)
1
iex> Keyword.fetch!([a: 1], :b)
** (KeyError) key :b not found in: [a: 1]
Link to this function get(keywords, key, default \\ nil) View Source
get(t(), key(), value()) :: value()

Gets the value for a specific key.

If key does not exist, return the default value (nil if no default value).

If duplicated entries exist, the first one is returned. Use get_values/2 to retrieve all entries.

Examples

iex> Keyword.get([], :a)
nil
iex> Keyword.get([a: 1], :a)
1
iex> Keyword.get([a: 1], :b)
nil
iex> Keyword.get([a: 1], :b, 3)
3

With duplicated keys:

iex> Keyword.get([a: 1, a: 2], :a, 3)
1
iex> Keyword.get([a: 1, a: 2], :b, 3)
3
Link to this function get_and_update(keywords, key, fun) View Source
get_and_update(t(), key(), (value() -> {get, value()} | :pop)) :: {get, t()} when get: term()

Gets the value from key and updates it, all in one pass.

This fun argument receives the value of key (or nil if key is not present) and must return a two-element tuple: the “get” value (the retrieved value, which can be operated on before being returned) and the new value to be stored under key. The fun may also return :pop, implying the current value shall be removed from the keyword list and returned.

The returned value is a tuple with the “get” value returned by fun and a new keyword list with the updated value under key.

Examples

iex> Keyword.get_and_update([a: 1], :a, fn current_value ->
...>   {current_value, "new value!"}
...> end)
{1, [a: "new value!"]}

iex> Keyword.get_and_update([a: 1], :b, fn current_value ->
...>   {current_value, "new value!"}
...> end)
{nil, [b: "new value!", a: 1]}

iex> Keyword.get_and_update([a: 1], :a, fn _ -> :pop end)
{1, []}

iex> Keyword.get_and_update([a: 1], :b, fn _ -> :pop end)
{nil, [a: 1]}
Link to this function get_and_update!(keywords, key, fun) View Source
get_and_update!(t(), key(), (value() -> {get, value()})) ::
  {get, t()} |
  no_return() when get: term()

Gets the value from key and updates it. Raises if there is no key.

This fun argument receives the value of key and must return a two-element tuple: the “get” value (the retrieved value, which can be operated on before being returned) and the new value to be stored under key.

The returned value is a tuple with the “get” value returned by fun and a new keyword list with the updated value under key.

Examples

iex> Keyword.get_and_update!([a: 1], :a, fn current_value ->
...>   {current_value, "new value!"}
...> end)
{1, [a: "new value!"]}

iex> Keyword.get_and_update!([a: 1], :b, fn current_value ->
...>   {current_value, "new value!"}
...> end)
** (KeyError) key :b not found in: [a: 1]

iex> Keyword.get_and_update!([a: 1], :a, fn _ ->
...>   :pop
...> end)
{1, []}
Link to this function get_lazy(keywords, key, fun) View Source
get_lazy(t(), key(), (() -> value())) :: value()

Gets the value for a specific key.

If key does not exist, lazily evaluates fun and returns its result.

This is useful if the default value is very expensive to calculate or generally difficult to setup and teardown again.

If duplicated entries exist, the first one is returned. Use get_values/2 to retrieve all entries.

Examples

iex> keyword = [a: 1]
iex> fun = fn ->
...>   # some expensive operation here
...>   13
...> end
iex> Keyword.get_lazy(keyword, :a, fun)
1
iex> Keyword.get_lazy(keyword, :b, fun)
13
Link to this function get_values(keywords, key) View Source
get_values(t(), key()) :: [value()]

Gets all values for a specific key.

Examples

iex> Keyword.get_values([], :a)
[]
iex> Keyword.get_values([a: 1], :a)
[1]
iex> Keyword.get_values([a: 1, a: 2], :a)
[1, 2]
Link to this function has_key?(keywords, key) View Source
has_key?(t(), key()) :: boolean()

Returns whether a given key exists in the given keywords.

Examples

iex> Keyword.has_key?([a: 1], :a)
true
iex> Keyword.has_key?([a: 1], :b)
false
Link to this function keys(keywords) View Source
keys(t()) :: [key()]

Returns all keys from the keyword list.

Duplicated keys appear duplicated in the final list of keys.

Examples

iex> Keyword.keys([a: 1, b: 2])
[:a, :b]
iex> Keyword.keys([a: 1, b: 2, a: 3])
[:a, :b, :a]
Link to this function keyword?(term) View Source
keyword?(term()) :: boolean()

Returns true if term is a keyword list; otherwise returns false.

Examples

iex> Keyword.keyword?([])
true
iex> Keyword.keyword?([a: 1])
true
iex> Keyword.keyword?([{Foo, 1}])
true
iex> Keyword.keyword?([{}])
false
iex> Keyword.keyword?([:key])
false
iex> Keyword.keyword?(%{})
false
Link to this function merge(keywords1, keywords2) View Source
merge(t(), t()) :: t()

Merges two keyword lists into one.

All keys, including duplicated keys, given in keywords2 will be added to keywords1, overriding any existing one.

There are no guarantees about the order of keys in the returned keyword.

Examples

iex> Keyword.merge([a: 1, b: 2], [a: 3, d: 4])
[b: 2, a: 3, d: 4]

iex> Keyword.merge([a: 1, b: 2], [a: 3, d: 4, a: 5])
[b: 2, a: 3, d: 4, a: 5]

iex> Keyword.merge([a: 1], [2, 3])
** (ArgumentError) expected a keyword list as the second argument, got: [2, 3]
Link to this function merge(keywords1, keywords2, fun) View Source
merge(t(), t(), (key(), value(), value() -> value())) :: t()

Merges two keyword lists into one.

All keys, including duplicated keys, given in keywords2 will be added to keywords1. The given function will be invoked to solve conflicts.

If keywords2 has duplicate keys, the given function will be invoked for each matching pair in keywords1.

There are no guarantees about the order of keys in the returned keyword.

Examples

iex> Keyword.merge([a: 1, b: 2], [a: 3, d: 4], fn _k, v1, v2 ->
...>   v1 + v2
...> end)
[b: 2, a: 4, d: 4]

iex> Keyword.merge([a: 1, b: 2], [a: 3, d: 4, a: 5], fn :a, v1, v2 ->
...>  v1 + v2
...> end)
[b: 2, a: 4, d: 4, a: 5]

iex> Keyword.merge([a: 1, b: 2, a: 3], [a: 3, d: 4, a: 5], fn :a, v1, v2 ->
...>  v1 + v2
...> end)
[b: 2, a: 4, d: 4, a: 8]

iex> Keyword.merge([a: 1, b: 2], [:a, :b], fn :a, v1, v2 ->
...>  v1 + v2
...> end)
** (ArgumentError) expected a keyword list as the second argument, got: [:a, :b]

Returns an empty keyword list, i.e. an empty list.

Examples

iex> Keyword.new()
[]

Creates a keyword from an enumerable.

Duplicated entries are removed, the latest one prevails. Unlike Enum.into(enumerable, []), Keyword.new(enumerable) guarantees the keys are unique.

Examples

iex> Keyword.new([{:b, 1}, {:a, 2}])
[b: 1, a: 2]

iex> Keyword.new([{:a, 1}, {:a, 2}, {:a, 3}])
[a: 3]
Link to this function new(pairs, transform) View Source
new(Enum.t(), (term() -> {key(), value()})) :: t()

Creates a keyword from an enumerable via the transformation function.

Duplicated entries are removed, the latest one prevails. Unlike Enum.into(enumerable, [], fun), Keyword.new(enumerable, fun) guarantees the keys are unique.

Examples

iex> Keyword.new([:a, :b], fn(x) -> {x, x} end)
[a: :a, b: :b]
Link to this function pop(keywords, key, default \\ nil) View Source
pop(t(), key(), value()) :: {value(), t()}

Returns and removes all values associated with key in the keyword list.

All duplicated keys are removed. See pop_first/3 for removing only the first entry.

Examples

iex> Keyword.pop([a: 1], :a)
{1, []}
iex> Keyword.pop([a: 1], :b)
{nil, [a: 1]}
iex> Keyword.pop([a: 1], :b, 3)
{3, [a: 1]}
iex> Keyword.pop([a: 1, a: 2], :a)
{1, []}
Link to this function pop_first(keywords, key, default \\ nil) View Source
pop_first(t(), key(), value()) :: {value(), t()}

Returns and removes the first value associated with key in the keyword list.

Duplicated keys are not removed.

Examples

iex> Keyword.pop_first([a: 1], :a)
{1, []}
iex> Keyword.pop_first([a: 1], :b)
{nil, [a: 1]}
iex> Keyword.pop_first([a: 1], :b, 3)
{3, [a: 1]}
iex> Keyword.pop_first([a: 1, a: 2], :a)
{1, [a: 2]}
Link to this function pop_lazy(keywords, key, fun) View Source
pop_lazy(t(), key(), (() -> value())) :: {value(), t()}

Lazily returns and removes all values associated with key in the keyword list.

This is useful if the default value is very expensive to calculate or generally difficult to setup and teardown again.

All duplicated keys are removed. See pop_first/3 for removing only the first entry.

Examples

iex> keyword = [a: 1]
iex> fun = fn ->
...>   # some expensive operation here
...>   13
...> end
iex> Keyword.pop_lazy(keyword, :a, fun)
{1, []}
iex> Keyword.pop_lazy(keyword, :b, fun)
{13, [a: 1]}
Link to this function put(keywords, key, value) View Source
put(t(), key(), value()) :: t()

Puts the given value under key.

If a previous value is already stored, all entries are removed and the value is overridden.

Examples

iex> Keyword.put([a: 1], :b, 2)
[b: 2, a: 1]
iex> Keyword.put([a: 1, b: 2], :a, 3)
[a: 3, b: 2]
iex> Keyword.put([a: 1, b: 2, a: 4], :a, 3)
[a: 3, b: 2]
Link to this function put_new(keywords, key, value) View Source
put_new(t(), key(), value()) :: t()

Puts the given value under key unless the entry key already exists.

Examples

iex> Keyword.put_new([a: 1], :b, 2)
[b: 2, a: 1]
iex> Keyword.put_new([a: 1, b: 2], :a, 3)
[a: 1, b: 2]
Link to this function put_new_lazy(keywords, key, fun) View Source
put_new_lazy(t(), key(), (() -> value())) :: t()

Evaluates fun and puts the result under key in keyword list unless key is already present.

This is useful if the value is very expensive to calculate or generally difficult to setup and teardown again.

Examples

iex> keyword = [a: 1]
iex> fun = fn ->
...>   # some expensive operation here
...>   3
...> end
iex> Keyword.put_new_lazy(keyword, :a, fun)
[a: 1]
iex> Keyword.put_new_lazy(keyword, :b, fun)
[b: 3, a: 1]
Link to this function replace(keywords, key, value) View Source
replace(t(), key(), value()) :: t()

Alters the value stored under key to value, but only if the entry key already exists in the keyword list.

In the case a value is stored multiple times in the keyword list, later occurrences are removed.

Examples

iex> Keyword.replace([a: 1], :b, 2)
[a: 1]
iex> Keyword.replace([a: 1, b: 2, a: 4], :a, 3)
[a: 3, b: 2]
Link to this function replace!(keywords, key, value) View Source
replace!(t(), key(), value()) :: t()

Similar to replace/3, but will raise a KeyError if the entry key does not exist.

Examples

iex> Keyword.replace!([a: 1, b: 2, a: 4], :a, 3)
[a: 3, b: 2]
iex> Keyword.replace!([a: 1], :b, 2)
** (KeyError) key :b not found in: [a: 1]
Link to this function split(keywords, keys) View Source
split(t(), [key()]) :: {t(), t()}

Takes all entries corresponding to the given keys and extracts them into a separate keyword list.

Returns a tuple with the new list and the old list with removed keys.

Keys for which there are no entries in the keyword list are ignored.

Entries with duplicated keys end up in the same keyword list.

Examples

iex> Keyword.split([a: 1, b: 2, c: 3], [:a, :c, :e])
{[a: 1, c: 3], [b: 2]}
iex> Keyword.split([a: 1, b: 2, c: 3, a: 4], [:a, :c, :e])
{[a: 1, c: 3, a: 4], [b: 2]}
Link to this function take(keywords, keys) View Source
take(t(), [key()]) :: t()

Takes all entries corresponding to the given keys and returns them in a new keyword list.

Duplicated keys are preserved in the new keyword list.

Examples

iex> Keyword.take([a: 1, b: 2, c: 3], [:a, :c, :e])
[a: 1, c: 3]
iex> Keyword.take([a: 1, b: 2, c: 3, a: 5], [:a, :c, :e])
[a: 1, c: 3, a: 5]
Link to this function to_list(keyword) View Source
to_list(t()) :: t()

Returns the keyword list itself.

Examples

iex> Keyword.to_list([a: 1])
[a: 1]
Link to this function update(keywords, key, initial, fun) View Source
update(t(), key(), value(), (value() -> value())) :: t()

Updates the key in keywords with the given function.

If the key does not exist, inserts the given initial value.

If there are duplicated keys, they are all removed and only the first one is updated.

Examples

iex> Keyword.update([a: 1], :a, 13, &(&1 * 2))
[a: 2]
iex> Keyword.update([a: 1, a: 2], :a, 13, &(&1 * 2))
[a: 2]
iex> Keyword.update([a: 1], :b, 11, &(&1 * 2))
[a: 1, b: 11]
Link to this function update!(keywords, key, fun) View Source
update!(t(), key(), (value() -> value())) :: t() | no_return()

Updates the key with the given function.

If the key does not exist, raises KeyError.

If there are duplicated keys, they are all removed and only the first one is updated.

Examples

iex> Keyword.update!([a: 1], :a, &(&1 * 2))
[a: 2]
iex> Keyword.update!([a: 1, a: 2], :a, &(&1 * 2))
[a: 2]

iex> Keyword.update!([a: 1], :b, &(&1 * 2))
** (KeyError) key :b not found in: [a: 1]
Link to this function values(keywords) View Source
values(t()) :: [value()]

Returns all values from the keyword list.

Values from duplicated keys will be kept in the final list of values.

Examples

iex> Keyword.values([a: 1, b: 2])
[1, 2]
iex> Keyword.values([a: 1, b: 2, a: 3])
[1, 2, 3]