CODEWITHSUNDEEP
enumselixir
sams
sams

Reputation: 188

Elixir Enum member? for multiple elements

Enum.member/2 is only able to check for one elements membership. Like

Enum.member ["abc", "def", "ghi", "123", "hello"], "abc" -> true

Is there a way to use an anonymous function, etc. for checking membership of multiple items, and returning false if one of the elements is not included to stay DRY and avoid something like this?

Enum.member ["abc", "def", "ghi", "123", "hello"], "abc"
Enum.member ["abc", "def", "ghi", "123", "hello"], "def"
Enum.member ["abc", "def", "ghi", "123", "hello"], "ghi"

Upvotes: 5

Views: 2376

Answers (2)

Patrick Oscity
Patrick Oscity

Reputation: 54734

Another option would be to work with sets, then check with MapSet.subset?/2

iex(1)> list = ["abc", "def", "ghi", "123", "hello"]
["abc", "def", "ghi", "123", "hello"]

iex(2)> MapSet.subset?(MapSet.new(["abc", "def", "ghi"]), MapSet.new(list))
true

iex(3)> MapSet.subset?(MapSet.new(["abc", "def", "jkl"]), MapSet.new(list))
false

Upvotes: 3

Dogbert
Dogbert

Reputation: 222428

You can use a combination of Enum.all?/2 (if you want all items to be present) or Enum.any?/2 (if you want any one item to be present) + Enum.member?/2 (or the in operator, which does the same):

iex(1)> list = ["abc", "def", "ghi", "123", "hello"]
["abc", "def", "ghi", "123", "hello"]
iex(2)> Enum.all?(["abc", "def", "ghi"], fn x -> x in list end)
true
iex(3)> Enum.any?(["abc", "def", "ghi"], fn x -> x in list end)
true
iex(4)> Enum.all?(["abc", "z"], fn x -> x in list end)
false
iex(5)> Enum.any?(["abc", "z"], fn x -> x in list end)
true

Upvotes: 9

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