Elixir Enum成员？对于多个元素

Question

Enum.member/2只能检查一个元素的成员资格。像

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

是否有使用匿名功能等检查多个项目的会员资格，如果不包括保持干爽，避免像这样的一个元素返回false的方法吗？

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组合（如果你希望所有的项目是存在），或者Enum.any?/2（如果你想要的任何一个项目存在）+ Enum.member?/2（或in运营商，这不相同） ：

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 

Answer 2

另一种选择是带套的工作，然后用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