Distribute items into containers in twos - Rails - ruby-on-rails

Question

I have a list of 10 items -- it is an array of hashes.
[{ id: 1, name: 'one'}, { id: 2, name: 'two' } .. { id: 10, name: 'ten' }]
I also have a random number of containers -- let's say 3, in this case. These containers are hashes with array values.
{ one: [], two: [], three: [] }
What I want to do, is iterate over the containers and drop 2 items at a time resulting in:
{
one: [{id:1}, {id:2}, {id:7}, {id:8}],
two: [{id:3}, {id:4}, {id:9}, {id:10}],
three: [{id:5}, {id:6}]
}
Also, if the item list is an odd number (11), the last item is still dropped into the next container.
{
one: [{id:1}, {id:2}, {id:7}, {id:8}],
two: [{id:3}, {id:4}, {id:9}, {id:10}],
three: [{id:5}, {id:6}, {id:11}]
}
note: the hashes are snipped here so it's easier to read.
My solution is something like this: (simplified)
x = 10
containers = { one: [], two: [], three: [] }
until x < 1 do
containers.each do |c|
c << 'x'
c << 'x'
end
x -= 2
end
puts containers
I'm trying to wrap my head around how I can achieve this but I can't seem to get it to work.

Answer 1

Round-robin pair distribution into three bins:
bins = 3
array = 10.times.map { |i| i + 1 }
# => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
array.
each_slice(2). # divide into pairs
group_by. # group into bins
with_index { |p, i| i % bins }. # round-robin style
values. # get rid of bin indices
each(&:flatten!) # join pairs in each bin
Completely different approach, stuffing bins in order:
base_size, bins_with_extra = (array.size / 2).divmod(bins)
pos = 0
bins.times.map { |i|
length = 2 * (base_size + (i < bins_with_extra ? 1 : 0)) # how much in this bin?
array[pos, length].tap { pos += length } # extract and advance
}
# => [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
If you absolutely need to have this in a hash,
Hash[%i(one two three).zip(binned_array)]
# => {:one=>[1, 2, 7, 8], :two=>[3, 4, 9, 10], :three=>[5, 6]}
The lovely (but likely not as performant) solution hinted at by Stefan Pochmann:
bins.times.with_object(array.to_enum).map { |i, e|
Array.new(2 * (base_size + (i < bins_with_extra ? 1 : 0))) { e.next }
}

Answer 2

This is just to show a different approach (and I would probably not use this one myself).
Given an array of items and the containers hash:
items = (1..10).to_a
containers = { one: [], two: [], three: [] }
You could dup the array (in order not to modify the original one) and build an enumerator that cycles each_value in the hash:
array = items.dup
enum = containers.each_value.cycle
Using the above, you can shift 2 items off the array and push them to the next container until the array is emtpy?:
enum.next.push(*array.shift(2)) until array.empty?
Result:
containers
#=> {:one=>[1, 2, 7, 8], :two=>[3, 4, 9, 10], :three=>[5, 6]}

Answer 3

You can use Enumerable#each_slice to iterate over a range from 0 to 10 in 3s and then append to an array of arrays:
containers = [
[],
[],
[]
]
(1...10).each_slice(3) do |slice|
containers[0] << slice[0]
containers[1] << slice[1]
containers[2] << slice[2]
end
p containers
# [[1, 4, 7], [2, 5, 8], [3, 6, 9]]