How to delete a hash from an array of hashes based on a condition in Ruby

Question

I have an array of hashes:

a = [{'id'=> '1', 'subject'=> 'this is subject 1', 'orig_id'=> 123, 'parent_id'=> 123}, 
     {'id'=> '2', 'subject'=> 'this is subject 2', 'orig_id'=> 456, 'parent_id'=> 123},
     {'id'=> '3', 'subject'=> 'this is subject 3', 'orig_id'=> 789, 'parent_id'=> 980}]

I want to filter it based on the condition that if the parent_id of one object is equal to the orig_id or the parent_id of the other object then keep the first hash and remove the other.

From the above scenario, the second hash will be removed and the output would be:

a = [{'id'=> '1', 'subject'=> 'this is subject 1', 'orig_id'=> 123, 'parent_id'=> 123}, 
     {'id'=> '3', 'subject'=> 'this is subject 3', 'orig_id'=> 789, 'parent_id'=> 980}]

Suppose a = [{‘parent_id’=>1, ‘user_id’=>2}, {‘parent_id’=>2, ‘user_id’=>3}, {‘parent_id’=>3, ‘user_id’=>1}]. What is the desired return value, and why? — Cary Swoveland
– Cary Swoveland, Commented Feb 9, 2020 at 15:58
This looks like you're retrieving records from a database. If so, you should be using a query to do this; It'd be much faster. — the Tin Man
– the Tin Man, Commented Feb 9, 2020 at 18:56

Aleksei Matiushkin · Accepted Answer · 2020-02-09 09:21:32Z

1

Group by parent_id and detect the proper one, otherwise return all of grouped hashes.

a.
  group_by { |h| h["parent_id"] }.
  flat_map do |id, hs|
    hs.find { |h| h["orig_id"] == id } || hs
  end

answered Feb 9, 2020 at 9:21

Aleksei Matiushkin

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3 Comments

Cary Swoveland Over a year ago

It is my understanding that if a = [{'parent_id'=>1, 'user_id'=>3}, {'parent_id'=>2, 'user_id'=>1}], [{'parent_id'=>1, 'user_id'=>3}] is to be returned, but you return a. Are we interpreting the question differently?

Aleksei Matiushkin Over a year ago

@CarySwoveland I do not know, but this solution was accepted, so...

the Tin Man Over a year ago

The question isn't entirely clear so y'all are guessing what the goal is. Instead, get the question clarified first, then answer.

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