符合条件的数组的第一个元素的Ruby抓取索引

Question

I ran across this issue and am wondering if there is a better way to do it.

Given the array:

options = [
  SelectItem.new(-1, "String 1"),
  SelectItem.new(0, "String 2"),
  SelectItem.new(7, "String 3"),
  SelectItem.new(14, "String 4")
]

What is the best way to grab the index of the first element that matches a certain criteria?

My solution was:

my_index = 0
options.each_with_index do |o, index|
  if o.value == some_value
    my_index = index
    break
  end
end

Is there another way to do this? Enumerable#find returns the first object that satisfies the condition, but I want something that returns the index of the first object that satisfies the condition.

Answer 1

a=[100,200,300]
a.index{ |x| x%3==0 }   # returns 2

for your case:

options.index{ |o| o.value == some_value }

Answer 2

Use Array#index

> a = ['asdf', 'qwer', '1234']
> a.index { |e| e =~ /\d/ }
=> 2