Question

I have an simple array

array = ["apple", "orange", "lemon"] 

array2 = [["apple", "good taste", "red"], ["orange", "bad taste", "orange"], ["lemon" , "no taste", "yellow"]]

how can i convert in to this hash whenever element in array match the first element of each element in array2?

hash = {"apple" => ["apple" ,"good taste", "red"],
        "orange" => ["orange", "bad taste", "orange"], 
        "lemon" => ["lemon" , "no taste", "yellow"] }

I am quite new to ruby, and spend a lot to do this manipulation, but no luck, any help ?

Answer 1

If the order of the mapping between the key and pairs should be based on the first element in array2, then you don’t need array at all:

array2 = [
  ["apple", "good taste", "red"],
  ["lemon" , "no taste", "yellow"],
  ["orange", "bad taste", "orange"]
]

map = Hash[ array2.map{ |a| [a.first,a] } ]
p map
#=> {
#=>   "apple"=>["apple", "good taste", "red"],
#=>   "lemon"=>["lemon", "no taste", "yellow"],
#=>   "orange"=>["orange", "bad taste", "orange"]
#=> }

If you want to use array to select a subset of elements, then you can do this:

# Use the map created above to find values efficiently
array = %w[orange lemon]
hash  = Hash[ array.map{ |val| [val,map[val]] if map.key?(val) }.compact ]
p hash
#=> {
#=>   "orange"=>["orange", "bad taste", "orange"],
#=>   "lemon"=>["lemon", "no taste", "yellow"]
#=> }

The code if map.key?(val) and compact ensures that there is not a problem if array asks for keys that are not present in array2, and does so in O(n) time.