CODEWITHSUNDEEP
ruby-on-railsarraysruby
Bitwise
Bitwise

Reputation: 8461

Return list index of matching elements in array

I have an array and it looks like this: [7, 2, 3, 2, 2, 1]. As you can see there are three elements in this array with the integer value of 2. I want to take this array and create an index that will look something like this: { 1 => 1, 2 => 3, 3 => 1, 4 => 0, 5 => 0, 6 => 0, 7 => 1 }

So, basically a bar chart/histogram/index of a how many duplicates there are in the array.

Just for a high level view of this. I'm trying to find all of the unique emails for each account within my application.

Upvotes: 1

Views: 109

Answers (3)

Eric Duminil
Eric Duminil

Reputation: 54223

Solution

This variant should be much faster than the other answers and will work for any array :

def histogram(array)
  array.each_with_object( Hash.new(0) ){|number, count| count[number] += 1 }
end

array = [7, 2, 3, 2, 2, 1, 'a', 'b', 'c', 'b']

p freq = histogram(array)
#=> {7=>1, 2=>3, 3=>1, 1=>1, "a"=>1, "b"=>2, "c"=>1}

p array.select{|number| freq[number] == 1}
#=> [7, 3, 1, "a", "c"]

Notes

element => 0

If I understand your question correctly, you don't need the 4 => 0, 5 => 0, 6 => 0 part in your Hash if all you want to do is to find elements that only appear once.

The information is here if you want to get it, though :

p freq[6]
#=> 0

SQL ?

If your array comes from a database via ActiveRecord, it might be a better idea to find the unique elements directly with a SQL query instead of retrieving all elements and applying logic to it.

Upvotes: 1

Cary Swoveland
Cary Swoveland

Reputation: 110665

arr = [8, 3, 4, 3, 3, 2]

min, max = arr.minmax
  #=> [2, 8] 
arr.each_with_object((min..max).to_a.product([0]).to_h) { |n,h| h[n] += 1 }
  #=> {2=>1, 3=>3, 4=>1, 5=>0, 6=>0, 7=>0, 8=>1}

Note:

(min..max).to_a.product([0]).to_h
  #=> {2=>0, 3=>0, 4=>0, 5=>0, 6=>0, 7=>0, 8=>0}

Upvotes: 0

Andrey Deineko
Andrey Deineko

Reputation: 52357

a = [7, 2, 3, 2, 2, 1]    
1.upto(a.max).each_with_object({}) {|number, hash| hash[number] = a.count(number)}
#=> {1=>1, 2=>3, 3=>1, 4=>0, 5=>0, 6=>0, 7=>1}

Upvotes: 3

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