sort an array of arrays based on number of occurences in ruby

Question

I want this integer array to be sorted in the right order based on its number of occurrences.

question = [[1, 7, 8, 9, 10, 11, 12, 19, 20, 21, 31, 32, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 129, 132, 133, 134, 135, 136, 139], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 24, 25, 26, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 129, 130, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141], [30], [77]] 

question.flatten.uniq.size = 90

answer = sort_it(question)

answer = [77, 68, 8, 9, 10, 11, 12, 19, 20, 21, 31, 139, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 135, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 136, 66, 67, 7, 70, 71, 72, 73, 74, 75, 76, 1, 78, 79, 81, 129, 132, 133, 134, 45, 65, 32, 2, 3, 4, 5, 6, 13, 14, 15, 16, 17, 22, 23, 24, 25, 26, 27, 29, 33, 41, 69, 130, 137, 138, 140, 141, 30]

answer.uniq.size = 90

Here is my Ruby code:

def sort_it(actual)
        join=[]
        buffer = actual.dup 
        final = [ ]

                (actual.size-2).downto(0) {|j|
                join.unshift(actual.map{|i| i }.inject(:"&"))
                actual.pop
                }
        ordered_join =  join.reverse.flatten
        final << ordered_join
        final << buffer.flatten - ordered_join

        final.flatten
end

Is this approach OK? Is there a more efficient approach?

EDIT:

As a tribute to tokland and niklas, edited the answer which was in the wrong order before. Thanks!

What do you mean by "the right order based on its number of occurrences"? Can you say more explicitly what it is that you want the code to do? — Gareth McCaughan
– Gareth McCaughan, Commented May 1, 2012 at 12:46
@GarethMcCaughan : Now, I want to search for the set of books with the features, "fiction","hardcover","2011". I will have book_ids mapped to those features in a table and select them to form an array like [[1,2,3,4],[3,4,5,6],[4,9,8]]. So, book_id 4 is what the user is looking for. I want this array to be ordered as [4,3,1,2,5,6,9,8]. When I display my results in that order, user is happy. Do i make sense? — beck03076
– beck03076, Commented May 1, 2012 at 13:12
@beck: question is an array of integer arrays. Also, please include the desired output (because as tokland suggested, your answer doesn't make sense on first glance). In any case, from your description I think that Marc is spot-on with his answer. However, you should really solve this using a custom SQL query, your database is a lot faster than Ruby. — Niklas B.
– Niklas B., Commented May 1, 2012 at 13:17

Marc-André Lafortune · Accepted Answer · 2012-05-01 12:49:03Z

4

Use group_by:

question.flatten.group_by{|x| x}.sort_by{|k, v| -v.size}.map(&:first)

answered May 1, 2012 at 12:49

Marc-André Lafortune

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2 Comments

Marc-André Lafortune Over a year ago

@NiklasB. I was going to delete my solution! You were quicker to post... and quicker to delete!

Niklas B. Over a year ago

But yours is nicer, as it doesn't involve a separate hash.

UncleGene · Accepted Answer · 2012-05-01 18:36:53Z

0

Answer with sort_by{|k, v| -v.size} calls v.size every time elements are compared. More effective solution:

question.flatten.group_by(&:to_i).map{|k,v| [k, -v.size]}.sort_by(&:last).map(&:first)

Though size of array is easy to get, it is unnecessary expense (O(sorting algorithm) instead of O(n)), and this idiom is good to remember anyway for more expensive operations

answered May 1, 2012 at 18:36

UncleGene

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