CODEWITHSUNDEEP
arraysrubyhashgrouping
GhostRider
GhostRider

Reputation: 2170

Group an array of integers in a hash of ranges and counts

I have an array of integers like this:

  [1, 1, 1, 2, 2, 4, 4, 5, 6, 11, 11, 12, 15, 22, 23, 23, 23, 31, 32, 32]

I am trying to convert this to a hash, grouping according to ranges at intervals of 10....

So, in this case it would be

{ [1..10] => 9, [11..20] => 4, [21..30] => 4, [31..40] => 3 }

I have tried a few things which haven't come close so it's a bit pointless putting them down here. I can convert the array to ranges 

  [1, 1, 1, 2, 2, 4, 4, 5, 6, 11, 11, 12, 15, 22, 23, 23, 23, 31, 32, 32].sort.uniq.inject([]) do |spans, n|
    if spans.empty? || spans.last.last != n - 1
      spans + [n..n]
    else
      spans[0..-2] + [spans.last.first..n]
    end
   end

But this is not what I am looking for. Any suggestions?

Upvotes: 1

Views: 302

Answers (3)

Cary Swoveland
Cary Swoveland

Reputation: 110675

I've modified the example to have no numbers between 21 and 30, so that the hash should include the key-value pair 21..30=>0.

arr = [1, 1, 1, 2, 2, 4, 4, 5, 6, 11, 11, 12, 20, 32, 33, 33, 33, 41, 42, 42]

intervals = (1..arr.last-1).step(10).each_with_object({}) { |n,h| h[n..n+9] = 0 }
  #=> {1..10=>0, 11..20=>0, 21..30=>0, 31..40=>0, 41..50=>0}

arr.each_with_object(intervals) do |n,intervals|
  interval_end = 10*((n+9)/10)     
  intervals[interval_end-9..interval_end] += 1
end
  #=> {1..10=>9, 11..20=>4, 21..30=>0, 31..40=>4, 41..50=>3}

Upvotes: 2

Aleksei Matiushkin
Aleksei Matiushkin

Reputation: 121000

arr.each_with_object(Hash.new(0)) do |e, hash|
  i = e / 10
  hash[i*10+1..i*10+10] += 1
end
#⇒ {
#    1..10 => 9,
#   11..20 => 4,
#   21..30 => 4,
#   31..40 => 3
# }

Upvotes: 3

Mladen Jablanović
Mladen Jablanović

Reputation: 44080

Hash[
  your_array.group_by{|i| i / 10}.map{|k,v|
    [(k*10+1..(k+1)*10), v.count]
  }
]
#=> {1..10=>9, 11..20=>4, 21..30=>4, 31..40=>3}

Upvotes: 3

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