红宝石each_slice具有不同的大小

Question

I have:

stuff = [1, 2, "a", "b", "c", "d", 4, 5, "z", "l", "m", "l", 5, 4, 4, 77]

Numbers come in groups of multiples of two, and letters come in groups of multiples of four.

I want to group numbers in twos and letters in fours like this:

stuff_processed = [
  [1, 2],
  ["a", "b", "c", "d"],
  [4, 5], 
  ["z", "l", "m", "l"],
  [5, 4],
  [4, 77]
]

The order inside of an array that holds numbers or letters is important, the order between the arrays I do not care about.

I know stuff.each_slice(2).to_a will take me part of the way. I can't figure out how to get all the way to what I need though.

Answer 1

stuff
.chunk(&:class)
.flat_map{|klass, a| a.each_slice(klass == Fixnum ? 2 : 4).to_a}
# => [[1, 2], ["a", "b", "c", "d"], [4, 5], ["z", "l", "m", "l"], [5, 4], [4, 77]]

Answer 2

arr = [1, 2, "a", "b", "c", "d", 4, 5, "z", "l", "m", "l", "s", "t",
       "u", "v", 5, 4, 4, 77, 91, 65]

H = { Fixnum=>1, String=>3 }
count = 0
arr.slice_when do |a,b|
  if a.class == b.class && count < H[a.class]
    count += 1
    false
  else
    count = 0
    true
  end
end.to_a   
  # => [[1, 2], ["a", "b", "c", "d"], [4, 5], ["z", "l", "m", "l"],
  #     ["s", "t", "u", "v"], [5, 4], [4, 77], [91, 65]] 

See Enumerable#slice_when , which first appeared in Ruby v2.2.

Answer 3

This Array#conditional_slice method accepts a Block and returns an Enumerator :

stuff = [1, 2, "a", "b", "c", "d", 4, 5, "z", "l", "m", "l", 5, 4, 4, 77]

class Array
  def conditional_slice(&block)
    clone = self.dup
    Enumerator.new do |yielder|
      until clone.empty? do
        yielder << clone.shift(block_given? ? block.call(clone.first) : 1)
      end
    end
  end
end

sliced_stuff = stuff.conditional_slice{|x| x.is_a?(Numeric) ? 2 : 4}

puts sliced_stuff.to_a.inspect
# => [[1, 2], ["a", "b", "c", "d"], [4, 5], ["z", "l", "m", "l"], [5, 4], [4, 77]]