Robert,
I really do understand your message, and thank you.
The details of the example did not make sense to me, among other
things I'm not using Sinatra yet. I readily confess this. But he
did say that he was looking to:
find the difference between two strings?
not just about, but EXACTLY. I would not be defensive it I had not
been called on this in such a rude fashion. Really. Being ignored is
a strong enough signal.
Logically, it still seems that there is a "work-around" solution in
that code somewhere because he could have his own overriding equal
test on string. If they Look the same - I'd sure bet that they would
pass this test whatever the deviation in the inner workings.
....and BTW this code DOES work EXACTLY as submitted in Ruby 1.9 just
as it did for me 3 years ago in whatever version we were using back
then.
It seemed like it COULD be a useful piece of code to look at for a few
folks in any event.
You guys have been VERY helpful to me, I am also looking seriously to
use Sinatra, this guy did really seem in a bit of distress.... and bla
bla bla
I did not expect to have somebody call me out and tell me the approach
was totally off the mark, and make the completely UNFOUNDED (and
untested) statement that the code does not work. It does, at least in
my version 1.9
Mr Ryan, it sure seems to me, went well over the top, and also was
just plain WRONG about the code if you should care to run it:
Way too much sideways motion, I won't let myself be provoked into any
more re-re-rebuttals they do no good and are just so much noise.
class String
def levenshtein( other, ins=2, del=2, sub=1)
#ins, del, sub are weighted costs
return nil if self.nil?
return nil if other.nil?
dm = #distance matrix
#Initialize first row values
dm[0] = (0..self.length).collect { | i | i * ins }
fill = [0] * (self.length - 1)
#initialize first column values
for i in 1..other.length
dm[i] = [i * del, fill.flatten]
end
#populate matrix
for i in 1..other.length
for j in 1..self.length
#critical comparison
dm[i][j] = [dm[i-1][j-1] + (self[j-1] == other[i-1] ?
0 : sub), dm[i] [j-1] + ins, dm[i-1][j] + del ].min
end
end
#The last value in the matrix is the Levenshtein distance betw
the strings
dm[other.length][self.length]
end
end
def ls( ar, threshold=3 )#Array must have at least 2 elements
word1, word2, nRslt, lRslt = ar.first.to_s, ar[1].to_s, 999, false
if ar.size == 2
nRslt = word1.levenshtein( word2 )
lRslt = nRslt <= threshold
elsif ar.size > 2
range = 1..ar.size - 1
range.each do | n |
word2 = ar[n]
nRslt = word1.levenshtein( word2 )
puts "word2 = " + word2.to_s + ", ls value = " +
nRslt.to_s
if nRslt <= threshold
lRslt = true
break
end
end
end
puts "word1 = " + word1.to_s + ", and word2 = " + word2.to_s +
"Result = " + nRslt.to_s + " Passed? " + lRslt.to_s
lRslt
end
puts ls( ["Davis", "Bully"] )
puts ls( ["Bully", "Bully"] )
ruby1.9 t.rb
word1 = Davis, and word2 = Bully Result = 5 Passed? false
false
word1 = Bully, and word2 = Bully Result = 0 Passed? true
true
Exit code: 0
Sincerely,
George
