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comp.lang.ruby

Damerau-Levenshtein_distance

Jeremy Wells

5/18/2007 1:36:00 AM

I wrote the method below by copying the algorithm from
http://en.wikipedia.org/wiki/Damerau-Levenshtei... (and matrix is
a really simple 2d array implementation). But the problem is that it
slows wat down as the string size gets bigger. At string length of about
150 it takes 1s, at 500 10s. Is there any way to recode this to get
better performance without rewriting it in C, and would rewrting it in C
even help or is this just a slow algorithm?

def self.distance(string1, string2)
string1 = string1.unpack('C*')
string2 = string2.unpack('C*')
s1n = string1.length
s2n = string2.length

m = DLDiff::Matrix.new(s1n+1, s2n+1)
cost = 0

(0..s1n).each {|i| m[i,0] = i}
(1..s2n).each {|j| m[0,j] = j}

(1..s1n).each do |i|
(1..s2n).each do |j|
cost = string1[i] == string2[j] ? 0 : 1
m[i, j] = [ m[i-1, j] + 1,
m[i, j-1] + 1,
m[i-1, j-1] + cost ].min
m[i, j] = [ m[i,j], m[i-2,j-2] + cost].min if(i > 1 && j > 1 &&
string1[i] == string2[j-1] && string1[i-1] == string2[j])
end
end

m[s1n, s2n]
end

class Matrix
def initialize(columns, rows)
ac = Array.new(columns, 0)
@am = Array.new(rows, 0)
@am = @am.collect{|r| ac.dup}
end

def [](c, r)
@am[r][c]
end

def []=(c,r,value)
@am[r][c] = value
end

def inspect
@am.collect{|a| a.inspect}.join("\n")
end

def to_s
@am.to_s
end
end
2 Answers

Daniel Martin

5/18/2007 5:40:00 AM

0

Jeremy <jwells@servalsystems.co.uk> writes:

> I wrote the method below by copying the algorithm from
> http://en.wikipedia.org/wiki/Damerau-Levenshtei... (and matrix
> is a really simple 2d array implementation). But the problem is that
> it slows wat down as the string size gets bigger. At string length of
> about 150 it takes 1s, at 500 10s. Is there any way to recode this to
> get better performance without rewriting it in C, and would rewrting
> it in C even help or is this just a slow algorithm?

Well, the algorithm itself is O(n*m), where n and m are the size of
the strings involved, so on large strings it's going to get slow.

I was able to shave about 40% off the time for your method, and fix a
bug.

The bug was caused because the wikipedia article indexes the strings
starting from 1, but indexes the array starting form 0. In ruby both
start at 0, of course. To fix this, I added a fake element at the
start of both strings.

Anyway, here's my slightly faster version:

def self.distance(string1, string2)
string1 = string1.unpack('C*')
string2 = string2.unpack('C*')
s1n = string1.length
s2n = string2.length
string1.unshift -1
string2.unshift -1
m = Array.new(s1n+1){Array.new(s2n+1,0)}
cost = 0
a = nil; b = nil
(0..s1n).each {|i| m[i][0] = i}
(1..s2n).each {|j| m[0][j] = j}
(1..s1n).each do |i|
(1..s2n).each do |j|
cost = string1[i] == string2[j] ? 0 : 1

a = m[i-1][j] + 1
a = b if ((b = m[i][j-1]+1) < a)
a = b if ((b = m[i-1][j-1]+cost) < a)

if(i > 1 && j > 1 &&
string1[i] == string2[j-1] &&
string1[i-1] == string2[j]) then
a = b if ((b = m[i-2][j-2] + 1) < a)
end
m[i][j] = a
end
end
m[s1n][s2n]
end


--
s=%q( Daniel Martin -- martin@snowplow.org
puts "s=%q(#{s})",s.to_a.last )
puts "s=%q(#{s})",s.to_a.last

Jeremy Wells

5/18/2007 8:41:00 AM

0

Daniel Martin wrote:
> Well, the algorithm itself is O(n*m), where n and m are the size of
> the strings involved, so on large strings it's going to get slow.
>
> I was able to shave about 40% off the time for your method, and fix a
> bug.
>
>
Thanks, thats really useful. I guess that now the data i'm comparing
has grown it would be better to use a diff type method than this.