Module: Build::Text::Merge

Defined in:
lib/build/text/merge.rb

Defined Under Namespace

Classes: Difference, LCSNode

Class Method Summary collapse

Class Method Details

.combine(old_text, new_text) ⇒ Object 



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# File 'lib/build/text/merge.rb', line 27

def self.combine(old_text, new_text)
	lcs = lcs(old_text, new_text)
	changes = []
	
	n = 0; o = 0; l = 0
	while o < old_text.size and n < new_text.size and l < lcs.size
		if !similar(old_text[o], lcs[l])
			changes << Difference.new(:old, old_text[o])
			o += 1
		elsif !similar(new_text[n], lcs[l])
			changes << Difference.new(:new, new_text[n])
			n += 1
		else
			changes << Difference.new(:both, lcs[l])
			o += 1; n += 1; l += 1
		end
	end
	
	while o < old_text.size
		changes << Difference.new(:old, old_text[o])
		o += 1
	end
	
	while n < new_text.size
		changes << Difference.new(:old, new_text[n])
		n += 1
	end
	
	changes.map do |change|
		change.value
	end
end

.lcs(x, y) ⇒ Object

Find the Longest Common Subsequence in the given sequences x, y.



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# File 'lib/build/text/merge.rb', line 107

def self.lcs(x, y)
	# Create the lcs matrix:
	m = Array.new(x.length + 1) do
		Array.new(y.length + 1) do
			LCSNode.new(nil, nil)
		end
	end

	# LCS(i, 0) and LCS(0, j) are always 0:
	for i in 0..x.length do m[i][0].value = 0 end
	for j in 0..y.length do m[0][j].value = 0 end

	# Main algorithm, solve row by row:
	for i in 1..x.length do
		for j in 1..y.length do
			if similar(x[i-1], y[j-1])
				# Value is based on maximizing the length of the matched strings:
				m[i][j].value = m[i-1][j-1].value + (x[i-1].chomp.length + y[j-1].chomp.length) / 2.0
				m[i][j].previous = [-1, -1]
			else
				if m[i-1][j].value >= m[i][j-1].value
					m[i][j].value = m[i-1][j].value
					m[i][j].previous = [-1, 0]
				else
					m[i][j].value = m[i][j-1].value
					m[i][j].previous = [0, -1]
				end
			end
		end
	end

	# Get the solution by following the path backwards from m[x.length][y.length]
	lcs = []
	
	i = x.length; j = y.length
	until m[i][j].previous == nil do
		if m[i][j].previous == [-1, -1]
			lcs << x[i-1]
		end
		
		i, j = i + m[i][j].previous[0], j + m[i][j].previous[1]
	end

	return lcs.reverse!
end

.levenshtein_distance(s, t) ⇒ Object

This code is based directly on the Text gem implementation Returns a value representing the “cost” of transforming str1 into str2



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# File 'lib/build/text/merge.rb', line 62

def self.levenshtein_distance(s, t)
	n = s.length
	m = t.length

	return m if n == 0
	return n if m == 0

	d = (0..m).to_a
	x = nil

	n.times do |i|
		e = i+1

		m.times do |j|
			cost = (s[i] == t[j]) ? 0 : 1
			x = [
				d[j+1] + 1, # insertion
				e + 1,      # deletion
				d[j] + cost # substitution
			].min
			d[j] = e
			e = x
		end

		d[m] = x
	end

	return x
end

.similar(s, t, factor = 0.15) ⇒ Object

Calculate the similarity of two sequences, return true if they are with factor% similarity.



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# File 'lib/build/text/merge.rb', line 93

def self.similar(s, t, factor = 0.15)
	return true if s == t
	
	distance = levenshtein_distance(s, t)
	average_length = (s.length + t.length) / 2.0

	proximity = (distance.to_f / average_length)
	
	return proximity <= factor
end