Simplest way to sum 2D array elements based on duplicates? - ruby-on-rails

Question

So say I have a 2D array. Maybe something like:
data = [['oct 1', 4], ['oct 2', 5], ['oct 3', 9], ['oct 1', 2]]
And I want to get a new array out of this, that removes the duplicate values (eg. oct 1), but sums the corresponding values.
So I would end up with:
data = (['oct 1', 6], ['oct 2', 5], ['oct 3', 9])
I can think of a few ways to do this but they seem pretty inefficent and usually there's always some crazy ruby code that can do anything in a few lines, any suggestions?

Answer 1

Here are a couple of ways of doing that, my preference being the first.
Use a counting hash
Code
def combine(data)
data.each_with_object(Hash.new(0)) { |(date, val), h| h[date] += val }.to_a
end
See Hash::new, the discussion of the default value in particular.
Example
data = [['oct 1', 4], ['oct 2', 5], ['oct 3', 9], ['oct 1', 2]]
combine data
#=> [["oct 1", 6], ["oct 2", 5], ["oct 3", 9]]
Explanation
enum = data.each_with_object(Hash.new(0))
#=> #<Enumerator: [["oct 1", 4], ["oct 2", 5], ["oct 3", 9],
# ["oct 1", 2]]:each_with_object({})>
We can see the values generated by this enumerator by converting it to an array.
enum.to_a
#=> [[["oct 1", 4], {}], [["oct 2", 5], {}], [["oct 3", 9], {}],
# [["oct 1", 2], {}]]
The first value of enum is passed to the block and the values of the block variables are computed, using disambiguation and parallel assignment.
(date, val), h = enum.next
#=> [["oct 1", 4], {}]
date
#=> "oct 1"
val
#=> 4
h[date] += val
#=> h[date] = h[date] + val
#=> h["oct 1"] = h["oct 1"] + 4
#=> h["oct 1"] = 0 + 4 (no key "oct 1" so default value of `0` used)
#=> h["oct 1"] = 4
Now,
h #=> {"oct 1"=>4}
The remaining three values of enum are passed to the block and the block calculations are performed.
(date, val), h = enum.next
#=> [["oct 2", 5], {"oct 1"=>4}]
h[date] += val
#=> 5 (the default value of `0` is again used)
h #=> {"oct 1"=>4, "oct 2"=>5}
(date, val), h = enum.next
#=> [["oct 3", 9], {"oct 1"=>4, "oct 2"=>5}]
h[date] += val
#=> 9 (the default value of `0` is again used)
h #=> {"oct 1"=>4, "oct 2"=>5, "oct 3"=>9}
(date, val), h = enum.next
#=> [["oct 1", 2], {"oct 1"=>4, "oct 2"=>5, "oct 3"=>9}]
h[date] += val
#=> 6
h #=> {"oct 1"=>6, "oct 2"=>5, "oct 3"=>9}
In the last calculation, the default value was not used because the hash h already had a key "oct 1":
h[date] += val
#=> h[date] = h[date] + val
#=> h["oct 1"] = h["oct 1"] + 2
#=> h["oct 1"] = 4 + 2
Lastly,
h.to_a
#=> [["oct 1", 6], ["oct 2", 5], ["oct 3", 9]]
Use Enumerable#group_by
Code
def combine(data)
data.group_by(&:first).map { |date, vals| [date, vals.map(&:last).reduce(:+)] }
end
Example
combine data
#=> [["oct 1", 6], ["oct 2", 5], ["oct 3", 9]]
Explanation
The steps:
h = data.group_by(&:first)
#=> {"oct 1"=>[["oct 1", 4], ["oct 1", 2]],
# "oct 2"=>[["oct 2", 5]], "oct 3"=>[["oct 3", 9]]}
The first key-value pair of h is passed to the block:
date, vals = h.first
#=> ["oct 1", [["oct 1", 4], ["oct 1", 2]]]
date
#=> "oct 1"
vals
#=> [["oct 1", 4], ["oct 1", 2]]
and the block calculation is performed.
a = vals.map(&:last)
#=> [4, 2]
t = a.reduce(:+)
#=> 6
So the first key-value pair of h is mapped to
[date, t]
#=> ["oct 1", 6]
The remaining calculations are similar.

Answer 2

Try following with inject method:
data = [["oct 1", 4], ["oct 2", 5], ["oct 3", 9], ["oct 1", 2]]
data.inject({}) { |sum, (n, t)| sum[n] ||= 0; sum[n] += t; sum }.to_a
=> [["oct 1", 6], ["oct 2", 5], ["oct 3", 9]]