enums - Elixir 枚举成员？对于多个元素

Question

Enum.member/2只能检查一个元素成员资格。喜欢
Enum.member ["abc", "def", "ghi", "123", "hello"], "abc" -> true
有没有办法使用匿名函数等来检查多个项目的成员资格，如果不包含其中一个元素以保持 DRY 并避免此类情况，则返回 false ？

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

Answer 1

您可以使用 Enum.all?/2 的组合(如果您希望所有项目都存在)或 Enum.any?/2 (如果您想要任何一项存在)+ Enum.member?/2 (或 in 运算符，其作用相同):

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