How to split an array by a condition on adjacent elements into a limited number of partitions

Question

Let's say I have an array of numbers, e.g.

ary = [1, 3, 6, 7, 10, 9, 11, 13, 7, 24]

I would like to split the array between the first point where a smaller number follows a larger one. My output should be:

[[1, 3, 6, 7, 10], [9, 11, 13, 7, 24]]

I've tried slice_when and it comes quite close:

ary.slice_when { |i, j| i > j }.to_a
#=> [[1, 3, 6, 7, 10], [9, 11, 13], [7, 24]]

But it also splits between 13 and 7, so I have to join the remaining arrays:

first, *rest = ary.slice_when { |i, j| i > j }.to_a
[first, rest.flatten(1)]
#=> [[1, 3, 6, 7, 10], [9, 11, 13, 7, 24]]

which looks a bit cumbersome. It also seems inefficient to keep comparing items when the match was already found.

I am looking for a general solution based on an arbitrary condition. Having numeric elements and i > j is just an example.

Is there a better way to approach this?

Answer 1

There's probably a better way to do this but my first thought is...

break_done = false
ary.slice_when { |i, j| (break_done = i > j) unless break_done }.to_a
#=> [[1, 3, 6, 7, 10], [9, 11, 13, 7, 24]]

Answer 2

I'm not sure you'll find this more elegant, but it prevents the split-and-rejoin maneuver:

def split(ary)
  split_done = false
  ary.slice_when do |i, j|
    !split_done && (split_done = yield(i, j))
  end.to_a
end

Usage:

ary = [1, 3, 6, 7, 10, 9, 11, 13, 7, 24]    
split(ary){ |i, j| i > j }
#=> [[1, 3, 6, 7, 10], [9, 11, 13, 7, 24]]

---

Update:

Some may find this variant more readable. #chunk_while is the inverse of #split_when and then I just applied De Morgan's Law to the block.

def split(ary)
  split_done = false
  ary.chunk_while do |i, j|
    split_done || !(split_done = yield(i, j))
  end.to_a
end