Filter array of arrays with an array in Ruby

Question

Looking for a more efficient way of filtering an array of arrays using another array in Ruby. Let me demonstrate. Starting with this:

core = [[1, "apple", "James Bond"],
        [5, "orange", "Thor"],
        [10, "banana", "Wolverine"],
        [15, "orange", "Batman"],
        [20, "apple", "Mickey Mouse"],
        [25, "orange", "Lee Adama"],
        [30, "banana", "Luke Skywalker"]]

filter = ["apple", "banana"]

result = core.magical_function(filter)

# result == [[5, "orange", "Thor"],
#            [15, "orange", "Batman"],
#            [25, "orange", "Lee Adama"]]

The only thing I can think of is looping through the filter elements, but this slows down my code a lot when this toy example gets more complicated.

The most efficient way to filter an array of arrays is to not start with an array of arrays to begin with. A hash with your second column as keys would be the fastest way to process it. Can you construct such a hash instead? — Mark Thomas
– Mark Thomas, Commented Jul 25, 2017 at 16:38
@MarkThomas It would probably be more annoying to do it that way. This is a simplified example to demonstrate what I need. Converting to a hash would be really tough. — RedOster
– RedOster, Commented Jul 25, 2017 at 16:51
@sagarpandya82 I looped through each element of the filter criteria and rejected the array element if core[x][1] == i — RedOster
– RedOster, Commented Jul 25, 2017 at 16:51

Cary Swoveland · Accepted Answer · 2017-07-25 17:03:21Z

7

Use Array#reject and Array#include?:

core.reject { |_,fruit,_| filter.include?(fruit) }
  # => [[5, "orange", "Thor"], [15, "orange", "Batman"], [25, "orange", "Lee Adama"]]

If filter is large, first convert it to a set for faster lookups:

require 'set'
filter_set = filter.to_set
core.reject { |_,fruit,_| filter_set.include?(fruit) }

See Set and its instance methods. When required, set adds the instance method to_set to the module Enumerable which is inclueded by Array. Ruby implements sets with unseen hashes.

edited Jul 25, 2017 at 17:03

answered Jul 25, 2017 at 16:58

Cary Swoveland

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1 Comment

max pleaner Over a year ago

i always forget about set :(

max pleaner · Accepted Answer · 2017-07-25 16:29:11Z

3

You can change the filter array to a hash for faster look up. Then the whole thing requires just a single iteration through the core array:

filters = ["apple", "banana"].each_with_object({}) do |term, obj|
  obj[term] = true
end
# filters == {"apple" => true, "banana" => true}

filtered = core.reject do |array|
  filters[array[1]]
end

answered Jul 25, 2017 at 16:29

max pleaner

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Comments

sequielo · Accepted Answer · 2018-12-18 23:42:52Z

0

Why don't you use the intersection operator & ?

core.reject { |k| (k & filter).empty? }

answered Dec 18, 2018 at 23:42

sequielo

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