In Ruby how do I find the index of one of an array of elements?

You can use find_index and pass the needed value from the array:

a = ["a", "b", "c"]
p a.find_index('a')
p a.find_index('b')
p a.find_index('c')
# => 0
# => 1
# => 2

You can use map to get every element inside your a array and then to get the index corresponding to each element:

p a.map{|e| a.find_index(e)}
# => [0, 1, 2]

Another possible way to handle it could be to use the Enumerable#each_with_index:

a.each_with_index{|e,i| puts "Element: #{e}, Index: #{i}"}
# => Element: a, Index: 0
# => Element: b, Index: 1
# => Element: c, Index: 2

If you want to check the indexes for each element in ["b", "c"] using the ["a", "b", "c"] array, you can map the first one, get the array values, and then use the a,b,c to check those indexes:

p ["b", "c"].map{|e| ["a", "b", "c"].find_index(e) }
# => [1, 2]

You can also see Array#index and Enumerable#find_index.

find_index takes a single element. You could find the minimum by doing something like

a = ["a", "b", "c"]
to_find = ["b", "c"]
to_find.map {|i| a.find_index(i) } .compact.min # => 1

You can use Array#index with a block.

a = ['a', 'b', 'c']
a.index { |x| ['b', 'c'].include?(x) }
#=> 1

Quote from the docs:

If a block is given instead of an argument, returns the index of the first object for which the block returns true. Returns nil if no match is found.

---

As Cary pointed out in his comment it is not the most performant algorithm to compare all elements in a against all elements in ['b', 'c'] (this would lead to O(n*m)). Depending on the size of both arrays if might sense to build a more efficient data structure first. Using a Set instead of an Array has some cost in creating the set upfront, but makes the comparison in the block much faster (overall O(n+m)):

require 'set'

a = ['a', 'b', 'c']
set = Set.new(['b', 'c'])
a.index { |x| set.include?(x) }
#=> 1