#ruby Array of Hashes Quiz

Found this interesting ruby quiz from AlphaSights. Given an array of hashes, collapse into an array of hashes containing one entry per day. And you can only reference the :time key and not the rest.

log = [
  {time: 201201, x: 2},
  {time: 201201, y: 7},
  {time: 201201, z: 2},
  {time: 201202, a: 3},
  {time: 201202, b: 4},
  {time: 201202, c: 0}

# result should be
  {time: 201201, x: 2, y: 7, z: 2},
  {time: 201202, a: 3, b: 4, c: 0},

The first thing came to mind is to use Enumerable#group_by

grouped = log.group_by { |i| i[:time] }
collapsed = grouped.collect do |t, a|
  no_time_h = a.inject({}) do |others, h|
    others.merge h.reject { |k, v| k.to_sym == :time }

  {time: t}.merge(no_time_h)

puts collapsed.inspect

However, after reading this a couple of times, I still find the solution hard to follow. For starter, group_by returns a hash where the values are an array of hashes which brings me back to the original problem even though it is already grouped by time. That I feel made the rest of the code more complicated.

# result of group_by
{201201=>[{:time=>201201, :x=>2}, {:time=>201201, :y=>7}, {:time=>201201, :z=>2}], 201202=>[{:time=>201202, :a=>3}, {:time=>201202, :b=>4}, {:time=>201202, :c=>0}]}

For my second version, I simply loop into the array and compose the hash using :time as the key. Afterwards, use the key-value pair to compose the resulting array. The code may be longer but it is more readable. Remember, Correct, Beautiful, Fast (in That Order).

hash_by_time = {}
log.each do |h|
  time = h[:time]
  others = h.reject { |k,v| k.to_sym == :time }

  if hash_by_time[time]
    hash_by_time[time].merge! others
    hash_by_time[time] = others

collapsed = hash_by_time.collect do |k, v|
  {time: k}.merge(v)