Grouping consecutive numbers in an array - ruby

Question

I need to add consecutive numbers to a new array and, if it is not a consecutive number, add only that value to a new array:
old_array = [1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
I want to get this result:
new_array = [
[1,2,3],
[5],
[7,8,9]
[20,21]
[23],
[29]
]
Is there an easier way to do this？

Answer 1

A little late to this party but:
old_array.slice_when { |prev, curr| curr != prev.next }.to_a
# => [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]

Answer 2

This is the official answer given in RDoc (slightly modified):
actual = old_array.first
old_array.slice_before do
|e|
expected, actual = actual.next, e
expected != actual
end.to_a

Answer 3

A couple other ways:
old_array = [1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
#1
a, b = [], []
enum = old_array.each
loop do
b << enum.next
unless enum.peek.eql?(b.last.succ)
a << b
b = []
end
end
a << b if b.any?
a #=> [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
#2
def pull_range(arr)
b = arr.take_while.with_index { |e,i| e-i == arr.first }
[b, arr[b.size..-1]]
end
b, l = [], a
while l.any?
f, l = pull_range(l)
b << f
end
b #=> [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]

Answer 4

Using chunk you could do:
old_array.chunk([old_array[0],old_array[0]]) do |item, block_data|
if item > block_data[1]+1
block_data[0] = item
end
block_data[1] = item
block_data[0]
end.map { |_, i| i }
# => [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]

Answer 5

Some answers seem unnecessarily long, it is possible to do this in a very compact way:
arr = [1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
arr.inject([]) { |a,e| (a[-1] && e == a[-1][-1] + 1) ? a[-1] << e : a << [e]; a }
# [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
Alternatively, starting with the first element to get rid of the a[-1] condition (needed for the case when a[-1] would be nil because a is empty):
arr[1..-1].inject([[arr[0]]]) { |a,e| e == a[-1][-1] + 1 ? a[-1] << e : a << [e]; a }
# [[1, 2, 3], [5], [7, 8, 9], [20, 21], [23], [29]]
Enumerable#inject iterates all elements of the enumerable, building up a result value which starts with the given object. I give it an empty Array or an Array with the first value wrapped in an Array respectively in my solutions. Then I simply check if the next element of the input Array we are iterating is equal to the last value of the last Array in the resulting Array plus 1 (i.e, if it is the next consecutive element). If it is, I append it to the last list. Otherwise, I start a new list with that element in it and append it to the resulting Array.

Answer 6

You could also do it like this:
old_array=[1, 2, 3, 5, 7, 8, 9, 20, 21, 23, 29]
new_array=[]
tmp=[]
prev=nil
for i in old_array.each
if i != old_array[0]
if i - prev == 1
tmp << i
else
new_array << tmp
tmp=[i]
end
if i == old_array[-1]
new_array << tmp
break
end
prev=i
else
prev=i
tmp << i
end
end

Answer 7

Using a Hash you can do:
counter = 0
groups = {}
old_array.each_with_index do |e, i|
groups[counter] ||= []
groups[counter].push old_array[i]
counter += 1 unless old_array.include? e.next
end
new_array = groups.keys.map { |i| groups[i] }