How to find number of a specific integer in an array of Fixnums ruby

Question

I'm doing an exercise now where I'm looking for all of the zeros in an array.

The input is:

numbers = [1, 3, 500, 200, 4000, 3000, 10000, 90, 20, 500000]

I want to sort them into a hash by the number of zeros. The expected output is:

expected = {0=>[1, 3], 2=>[500, 200], 3=>[4000, 3000], 4=>[10000], 1=>[90, 20], 5=>[500000]}

I have the structure built but I'm not sure how to count the number of zeros:

grouped = Hash.new {|hash, key| hash[key] = []}
numbers.each do |num|
  grouped[num] << num
end

EDITED for clarity: Any advice would be appreciated. Also, a lot of the advice I read on this recommended converting the array of integers to a string in order to solve the problem. Is there a way to count the number of digits (not just zeros) without converting the array to a string? The expected output in this case would look like:

expected = {1=>[1, 3], 2=>[90, 20], 3=>[500, 200], 4=>[4000, 3000], 5=>[10000], 6=>[500000]

Thanks in advance.

Answer 1

If you wish to group non-negative integers by the number of zero digits they contain, you can do this:

def nbr_zeroes(n)
  return 1 if n == 0
  m = n
  i = 0
  while m > 0
    i += 1 if m % 10 == 0
    m /= 10
  end
  i
end

numbers = [1, 3, 500, 200, 4000, 3000, 10000, 90, 20, 500000]
numbers.group_by { |i| nbr_zeroes(i) }
  #=> { 0=>[1, 3], 2=>[500, 200], 3=>[4000, 3000], 4=>[10000] }

numbers = [100000, 100001, 304070, 3500040, 314073, 2000, 314873, 0]
numbers.group_by { |i| nbr_zeroes(i) }
  #=> { 5=>[100000],    4=>[100001, 3500040], 3=>[304070, 2000],
  #     1=>[314073, 0], 0=>[314873] }

Answer 2

Like many transformations you'll want to do, this one's found in Enumerable.

Grouping by number of digits:

grouped = numbers.group_by { |n| Math.log10(n).to_i + 1 }
# => {1=>[1, 3], 3=>[500, 200], 4=>[4000, 3000], 5=>[10000], 2=>[90, 20], 6=>[500000]}

Grouping by number of zeroes:

grouped = numbers.group_by { |n| n.to_s.match(/0+$/) ? $&.length : 0 }
# => {0=>[1, 3], 2=>[500, 200], 3=>[4000, 3000], 4=>[10000], 1=>[90, 20], 5=>[500000]}

The group_by method is a handy way to convert an Array to a Hash with things organized into pigeon-holes.

Answer 3

I wound up using

grouped = Hash.new {|hash, key| hash[key] = []}
numbers.each do |num|
  grouped[num.to_s.count('0')] << num
end

but I really liked the variation in responses. I didn't realize there were so many ways to go about this. Thank you everyone.

Answer 4

Group by floor of log base 10?

1.9.3-p484 :014 > numbers.each {|n| grouped[Math.log10(n).floor] << n}
 => [1, 3, 500, 200, 4000, 3000, 10000, 90, 20, 500000] 
1.9.3-p484 :016 > grouped   
 => {0=>[1, 3], 2=>[500, 200], 3=>[4000, 3000], 4=>[10000], 1=>[90, 20], 5=>[500000]} 

Or try 1 + Math.log10(n).floor if you need the keys to be the actual number of digits.