Grouping data from nested arrays in ruby

Question

Assuming the following data tuple containing a person's name, age and the books he has read:

list =     [
  ["Peter", 21, ["Book 1", "Book 2", "Book 3", "Book 4"],
  ["Amy", 19, ["Book 3", "Book 4"],
  ["Sanders", 32, ["Book 1", "Book 2",],
  ["Charlie", 21, ["Book 4", "Book 5", "Book 6"],
  ["Amanda", 21, ["Book 2", "Book 5"]
]

What is the optimal way to extract names grouped by the books read, into the following format (basically a an array of arrays containing the book name and an array of names of people who read it)

results = [
   ["Book 1", ["Sanders", "Peter"]],
   ["Book 2", ["Sanders" "Amanda", "Peter"]],
   ["Book 3", ["Peter", "Amy"]],
   ["Book 4", ["Charlie", "Peter", "Amy"]],
   ["Book 5", ["Amanda","Charlie"]],
   ["Book 6", ["Charlie"]]
]

I've tried the following iterating method which extracts the lists of names and puts them into a hash, with the book title as the keys.

book_hash = Hash.new([])
list.each { |name,age,books| 
books { |x|  book_hash[x] = book_hash[x] + [name] }
}
results = book_hash.to_a.sort

However, the above method seems rather inefficient when handling large datasets containing millions of names. I've attempted to use the Array.group_by, but so far I'm unable to make it work with nested arrays.

Does anyone have any idea about the above?

Why are you expecting an array where your solution gives a hash? A hash makes more sense in this case. — sawa
– sawa, Commented Mar 3, 2015 at 13:43
Don't be lazy. You should provide at least enough sample data for your list that would correspond to your sample result. — sawa
– sawa, Commented Mar 3, 2015 at 13:44
@sawa My apologies, I've just added some sample data and results. — Rod
– Rod, Commented Mar 4, 2015 at 5:39

humza · Accepted Answer · 2015-03-04 08:13:06Z

2

Hash output. More suitable.

list.each_with_object({}) do |(name, age, books), hash|
  books.each do |book|
    (hash[book] ||= []) << name    
  end
end

If you must make it an array, then append a .to_a to the output of the above.

edited Mar 4, 2015 at 8:13

answered Mar 3, 2015 at 13:42

humza

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1 Comment

Rod Over a year ago

This did the trick, but since I'm running 1.8.7 I opted for for .each instead of .each_with_object. Thanks!

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