获取所有可能的子字符串及其计数

Question

I'm trying to get all possible substrings and their count in a hash. Eg

 "abc" => { a: 1, b: 1, ab: 1, bc: 1}

For that I wrote the following code:

 def get_all(b)
     (0..(b.size-1)).to_a.combination(2).inject({}) { |h, g|
        s = b[g[0],g[1]]
        h[s] ? ( h[s] += 1) : ( h[s] = 1 )
        h 
      } 
 end

But somehow It does not work correctly, because for "abchh" It returns:

{"a"=>1, "ab"=>1, "abc"=>1, "abch"=>1, "bc"=>1, "bch"=>1, "bchh"=>1, "chh"=>2, "hh"=>1}

chh is in there twice , but I can't understand why. What do I wrong?

Thank you!

Answer 1

String#[] can be called in various ways, including:

 str[start, length] → new_str or nil str[range] → new_str or nil 

The former expects start and length , whereas the latter expects a range denoting start and end .

So instead of two arguments g[0] and g[1] :

b[g[0], g[1]]

you have to pass a single argument g[0]..g[1] :

b[g[0]..g[1]]

Besides, you have to use repeated_combination in order to get the single characters as well:

(0..2).to_a.combination(2).to_a
#=> [[0, 1], [0, 2], [1, 2]]

(0..2).to_a.repeated_combination(2).to_a
#=> [[0, 0], [0, 1], [0, 2], [1, 1], [1, 2], [2, 2]]

Furthermore, your code can be simplified:

• use a...b instead of a..(b-1)
• prefer each_with_object over inject so you don't have to return the hash from the block
• set a default hash value via Hash.new(0)
• decompose the tuple array via (i, j) to have i..j instead of g[0]..g[1]

Example: (the indices variable can be inlined)

def get_all(str)
  indices = (0...str.size).to_a.repeated_combination(2)
  indices.each_with_object(Hash.new(0)) do |(i, j), h|
    h[str[i..j]] += 1
  end
end

Or, using two nested loops:

def get_all(str)
  (0...str.size).each_with_object(Hash.new(0)) do |i, h|
    (i...str.size).each do |j|
      h[str[i..j]] += 1
    end
  end
end

Maybe the method is already doing too much. I'd probably split it into two methods: one for enumerating the substrings and another one for counting them.