Split array by consecutive sequences [closed]

Question

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Closed 9 years ago.

Given an array of integer such as:

array = [1,2,3,5,6,7,20,21,33,87,88,89,101]

This array contains k consecutive subsequences (in this case k = 6), such as "1,2,3" and "87,88,89". How do I get an array containing arrays of these k subsequences?

For the above example these would be:

[[1,2,3], [5,6,7], [20,21], [33], [87,88,89], [101]]

Answer 1

def seq_sort(array)
  finished = [array.first], i = 0

  array.each_cons(2) do |(a,b)|
    if b - a == 1
      finished[i] << b
    else
      finished[i += 1] = [b]
    end
  end

  finished
end

> seq_sort([1,2,38,39,92,94,103,104,105,1002])
=> [[1,2], [38, 39], [92], [94], [103, 104, 105], [1002]]

Answer 2

The unshifted 0 is a dummy. It could be any number.

[1, 2, 3, 5, 6, 7]
.unshift(0)
.each_cons(2).slice_before{|m, n| m + 1 < n}.map{|a| a.map(&:last)}
# => [[1, 2, 3], [5, 6, 7]]

Answer 3

EDIT:

If you are using rails, there is an easy way, using the active support split method

def sort_it(arr)
  (arr.first..arr.last).to_a.split {|i| !arr.include?(i) }.select &:present?
end

or in Ruby

def sort_it(arr)
  tmp, main = [], []
  arr.each_with_index do |x, i|
    if arr[i-1]
      if arr[i-1] + 1 == x
        tmp << x
      else
        main << tmp unless tmp.empty?
        tmp = [x]
      end
    else
      tmp << x
    end
  end
  main << tmp
  main
end


> a = [1, 3, 5, 6, 8, 9, 11, 12, 13, 14, 16, 27]
> sort_it(a)
=> [[1], [3], [5, 6], [8, 9], [11, 12, 13, 14], [16], [27]]