Levenshtein distance/Alignment
The Levenshtein distance algorithm returns the number of atomic operations (insertion, deletion or edition) that must be performed on a string in order to obtain an other one, but it does not say anything about the actual operations used or their order.
You are encouraged to solve this task according to the task description, using any language you may know.
An alignment is a notation used to describe the operations used to turn a string into an other. At some point in the strings, the minus character ('-') is placed in order to signify that a character must be added at this very place. For instance, an alignment between the words 'place' and 'palace' is:
P-LACE PALACE
- Task
Write a function that shows the alignment of two strings for the corresponding levenshtein distance.
As an example, use the words "rosettacode" and "raisethysword".
You can either implement an algorithm, or use a dedicated library (thus showing us how it is named in your language).
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
typedef struct edit_s edit_t, *edit; struct edit_s { char c1, c2; int n; edit next; };
void leven(char *a, char *b) { int i, j, la = strlen(a), lb = strlen(b); edit *tbl = malloc(sizeof(edit) * (1 + la)); tbl[0] = calloc((1 + la) * (1 + lb), sizeof(edit_t)); for (i = 1; i <= la; i++) tbl[i] = tbl[i-1] + (1+lb);
for (i = la; i >= 0; i--) { char *aa = a + i; for (j = lb; j >= 0; j--) { char *bb = b + j; if (!*aa && !*bb) continue;
edit e = &tbl[i][j]; edit repl = &tbl[i+1][j+1]; edit dela = &tbl[i+1][j]; edit delb = &tbl[i][j+1];
e->c1 = *aa; e->c2 = *bb; if (!*aa) { e->next = delb; e->n = e->next->n + 1; continue; } if (!*bb) { e->next = dela; e->n = e->next->n + 1; continue; }
e->next = repl; if (*aa == *bb) { e->n = e->next->n; continue; }
if (e->next->n > delb->n) { e->next = delb; e->c1 = 0; } if (e->next->n > dela->n) { e->next = dela; e->c1 = *aa; e->c2 = 0; } e->n = e->next->n + 1; } }
edit p = tbl[0]; printf("%s -> %s: %d edits\n", a, b, p->n);
while (p->next) { if (p->c1 == p->c2) printf("%c", p->c1); else { putchar('('); if (p->c1) putchar(p->c1); putchar(','); if (p->c2) putchar(p->c2); putchar(')'); }
p = p->next; } putchar('\n');
free(tbl[0]); free(tbl); }
int main(void) { leven("raisethysword", "rosettacode"); return 0; }</lang>
- Output:
raisethysword -> rosettacode: 8 edits r(a,o)(i,)set(h,t)(y,a)(s,c)(w,)o(r,d)(d,e)
D
Using the standard library. <lang d>void main() {
import std.stdio, std.algorithm;
immutable s1 = "rosettacode"; immutable s2 = "raisethysword";
string s1b, s2b; size_t pos1, pos2;
foreach (immutable c; levenshteinDistanceAndPath(s1, s2)[1]) {
final switch (c) with (EditOp) {
case none, substitute:
s1b ~= s1[pos1++];
s2b ~= s2[pos2++];
break;
case insert:
s1b ~= "_";
s2b ~= s2[pos2++];
break;
case remove:
s1b ~= s1[pos1++];
s2b ~= "_";
break;
}
}
writeln(s1b, "\n", s2b);
}</lang>
- Output:
r_oset_tacode raisethysword
Eiffel
<lang eiffel> distance (source, target: STRING): INTEGER -- Minimum number of operations to turn `source' into `target'. local l_distance: ARRAY2 [INTEGER] del, ins, subst: INTEGER do create l_distance.make (source.count, target.count) ⟳ ic:(1 |..| source.count) ¦ l_distance [ic, 1] := ic - 1 ⟲ ⟳ ij:(1 |..| target.count) ¦ l_distance [1, ij] := ij - 1 ⟲
⟳ ic:(2 |..| source.count) ¦ ⟳ jc:(2 |..| target.count) ¦ if source [ic] = target [jc] then -- same char l_distance [ic, jc] := l_distance [ic - 1, jc - 1] else -- diff char del := l_distance [ic - 1, jc] -- delete? ins := l_distance [ic, jc - 1] -- insert? subst := l_distance [ic - 1, jc -1] -- substitute/swap? l_distance [ic, jc] := del.min (ins.min (subst)) + 1 end ⟲ ⟲ Result:= l_distance [source.count, target.count] end
</lang>
- Output:
Result = 8
The ⟳ ¦ ⟲ represents the "symbolic form" of an Eiffel across loop. In the case above, the first two ⟳ (loops) initialize the first column and the first row, and then two nested ⟳ (loops) to walk the distance computation.
Also—the `ic' is this programmers shorthand for ITERATION_CURSOR. Therefore, in the nested loops, the `ic' is the iteration cursor following `i' and the `ij' is following `j' as indexes on the source, target, and l_distance.
Go
Alignment computed by the Needleman-Wunch algorithm, which is used in bioinformatics. <lang go>package main
import (
"fmt"
"github.com/biogo/biogo/align" ab "github.com/biogo/biogo/alphabet" "github.com/biogo/biogo/feat" "github.com/biogo/biogo/seq/linear"
)
func main() {
// Alphabets for things like DNA are predefined in biogo, but we
// define our own here.
lc := ab.Must(ab.NewAlphabet("-abcdefghijklmnopqrstuvwxyz",
feat.Undefined, '-', 0, true))
// Construct scoring matrix for Needleman-Wunch algorithm.
// We leave zeros on the diagonal for the Levenshtein distance of an
// exact match and put -1s everywhere else for the Levenshtein distance
// of an edit.
nw := make(align.NW, lc.Len())
for i := range nw {
r := make([]int, lc.Len())
nw[i] = r
for j := range r {
if j != i {
r[j] = -1
}
}
}
// define input sequences
a := &linear.Seq{Seq: ab.BytesToLetters([]byte("rosettacode"))}
a.Alpha = lc
b := &linear.Seq{Seq: ab.BytesToLetters([]byte("raisethysword"))}
b.Alpha = lc
// perform alignment
aln, err := nw.Align(a, b)
// format and display result
if err != nil {
fmt.Println(err)
return
}
fa := align.Format(a, b, aln, '-')
fmt.Printf("%s\n%s\n", fa[0], fa[1])
aa := fmt.Sprint(fa[0])
ba := fmt.Sprint(fa[1])
ma := make([]byte, len(aa))
for i := range ma {
if aa[i] == ba[i] {
ma[i] = ' '
} else {
ma[i] = '|'
}
}
fmt.Println(string(ma))
}</lang>
- Output:
The lines after the alignment point out the 8 edits.
r-oset-tacode raisethysword || |||| ||
Java
<lang java>public class LevenshteinAlignment {
public static String[] alignment(String a, String b) {
a = a.toLowerCase();
b = b.toLowerCase();
// i == 0
int[][] costs = new int[a.length()+1][b.length()+1];
for (int j = 0; j <= b.length(); j++)
costs[0][j] = j;
for (int i = 1; i <= a.length(); i++) {
costs[i][0] = i;
for (int j = 1; j <= b.length(); j++) {
costs[i][j] = Math.min(1 + Math.min(costs[i-1][j], costs[i][j-1]), a.charAt(i - 1) == b.charAt(j - 1) ? costs[i-1][j-1] : costs[i-1][j-1] + 1);
}
}
// walk back through matrix to figure out path StringBuilder aPathRev = new StringBuilder(); StringBuilder bPathRev = new StringBuilder(); for (int i = a.length(), j = b.length(); i != 0 && j != 0; ) { if (costs[i][j] == (a.charAt(i - 1) == b.charAt(j - 1) ? costs[i-1][j-1] : costs[i-1][j-1] + 1)) { aPathRev.append(a.charAt(--i)); bPathRev.append(b.charAt(--j)); } else if (costs[i][j] == 1 + costs[i-1][j]) { aPathRev.append(a.charAt(--i)); bPathRev.append('-'); } else if (costs[i][j] == 1 + costs[i][j-1]) { aPathRev.append('-'); bPathRev.append(b.charAt(--j)); } }
return new String[]{aPathRev.reverse().toString(), bPathRev.reverse().toString()};
}
public static void main(String[] args) {
String[] result = alignment("rosettacode", "raisethysword"); System.out.println(result[0]); System.out.println(result[1]);
}
}</lang>
- Output:
r-oset-tacode raisethysword
Julia
<lang julia>function levenshteinalign(a::AbstractString, b::AbstractString)
a = lowercase(a) b = lowercase(b) len_a = length(a) len_b = length(b)
costs = Matrix{Int}(len_a + 1, len_b + 1)
costs[1, :] .= 0:len_b
@inbounds for i in 2:(len_a + 1)
costs[i, 1] = i
for j in 2:(len_b + 1)
tmp = ifelse(a[i-1] == b[j-1], costs[i-1, j-1], costs[i-1, j-1] + 1)
costs[i, j] = min(1 + min(costs[i-1, j], costs[i, j-1]), tmp)
end
end
apathrev = IOBuffer()
bpathrev = IOBuffer()
local i = len_a + 1
local j = len_b + 1
@inbounds while i != 1 && j != 1
tmp = ifelse(a[i-1] == b[j-1], costs[i-1, j-1], costs[i-1, j-1] + 1)
if costs[i, j] == tmp
i -= 1
j -= 1
print(apathrev, a[i])
print(bpathrev, b[j])
elseif costs[i, j] == 1 + costs[i-1, j]
i -= 1
print(apathrev, a[i])
print(bpathrev, '-')
elseif costs[i, j] == 1 + costs[i, j-1]
j -= 1
print(apathrev, '-')
print(bpathrev, b[j])
end
end
return reverse(String(take!(apathrev))), reverse(String(take!(bpathrev)))
end
foreach(println, levenshteinalign("rosettacode", "raisethysword")) foreach(println, levenshteinalign("place", "palace"))</lang>
- Output:
r-oset-tacode raisethysword p-lace palace
Kotlin
<lang scala>// version 1.1.3
fun levenshteinAlign(a: String, b: String): Array<String> {
val aa = a.toLowerCase()
val bb = b.toLowerCase()
val costs = Array(a.length + 1) { IntArray(b.length + 1) }
for (j in 0..b.length) costs[0][j] = j
for (i in 1..a.length) {
costs[i][0] = i
for (j in 1..b.length) {
val temp = costs[i - 1][j - 1] + (if (aa[i - 1] == bb[j - 1]) 0 else 1)
costs[i][j] = minOf(1 + minOf(costs[i - 1][j], costs[i][j - 1]), temp)
}
}
// walk back through matrix to figure out path
val aPathRev = StringBuilder()
val bPathRev = StringBuilder()
var i = a.length
var j = b.length
while (i != 0 && j != 0) {
val temp = costs[i - 1][j - 1] + (if (aa[i - 1] == bb[j - 1]) 0 else 1)
when (costs[i][j]) {
temp -> {
aPathRev.append(aa[--i])
bPathRev.append(bb[--j])
}
1 + costs[i-1][j] -> {
aPathRev.append(aa[--i])
bPathRev.append('-')
}
1 + costs[i][j-1] -> {
aPathRev.append('-')
bPathRev.append(bb[--j])
}
}
}
return arrayOf(aPathRev.reverse().toString(), bPathRev.reverse().toString())
}
fun main(args: Array<String>) {
var result = levenshteinAlign("place", "palace")
println(result[0])
println(result[1])
println()
result = levenshteinAlign("rosettacode","raisethysword")
println(result[0])
println(result[1])
}</lang>
- Output:
p-lace palace r-oset-tacode raisethysword
Mathematica / Wolfram Language
<lang Mathematica>DamerauLevenshteinDistance["rosettacode", "raisethysword"]</lang>
- Output:
8
Nim
<lang Nim>import algorithm, sequtils, strutils
proc levenshteinAlign(a, b: string): tuple[a, b: string] =
let a = a.toLower()
let b = b.toLower()
var costs = newSeqWith(a.len + 1, newSeq[int](b.len + 1))
for j in 0..b.len: costs[0][j] = j
for i in 1..a.len:
costs[i][0] = i
for j in 1..b.len:
let tmp = costs[i - 1][j - 1] + ord(a[i - 1] != b[j - 1])
costs[i][j] = min(1 + min(costs[i - 1][j], costs[i][j - 1]), tmp)
# Walk back through matrix to figure out path.
var aPathRev, bPathRev: string
var i = a.len
var j = b.len
while i != 0 and j != 0:
let tmp = costs[i - 1][j - 1] + ord(a[i - 1] != b[j - 1])
if costs[i][j] == tmp:
dec i
dec j
aPathRev.add a[i]
bPathRev.add b[j]
elif costs[i][j] == 1 + costs[i-1][j]:
dec i
aPathRev.add a[i]
bPathRev.add '-'
elif costs[i][j] == 1 + costs[i][j-1]:
dec j
aPathRev.add '-'
bPathRev.add b[j]
else:
quit "Internal error"
result = (reversed(aPathRev).join(), reversed(bPathRev).join())
when isMainModule:
var result = levenshteinAlign("place", "palace")
echo result.a
echo result.b
echo ""
result = levenshteinAlign("rosettacode","raisethysword")
echo result.a
echo result.b</lang>
- Output:
p-lace palace r-oset-tacode raisethysword
Perl
<lang perl>use strict; use warnings;
use List::Util qw(min);
sub levenshtein_distance_alignment {
my @s = ('^', split //, shift);
my @t = ('^', split //, shift);
my @A;
@{$A[$_][0]}{qw(d s t)} = ($_, join(, @s[1 .. $_]), ('~' x $_)) for 0 .. $#s;
@{$A[0][$_]}{qw(d s t)} = ($_, ('-' x $_), join , @t[1 .. $_]) for 0 .. $#t;
for my $i (1 .. $#s) {
for my $j (1 .. $#t) {
if ($s[$i] ne $t[$j]) { $A[$i][$j]{d} = 1 + ( my $min = min $A[$i-1][$j]{d}, $A[$i][$j-1]{d}, $A[$i-1][$j-1]{d} ); @{$A[$i][$j]}{qw(s t)} = $A[$i-1][$j]{d} == $min ? ($A[$i-1][$j]{s}.$s[$i], $A[$i-1][$j]{t}.'-') : $A[$i][$j-1]{d} == $min ? ($A[$i][$j-1]{s}.'-', $A[$i][$j-1]{t}.$t[$j]) : ($A[$i-1][$j-1]{s}.$s[$i], $A[$i-1][$j-1]{t}.$t[$j]); }
else {
@{$A[$i][$j]}{qw(d s t)} = ( $A[$i-1][$j-1]{d}, $A[$i-1][$j-1]{s}.$s[$i], $A[$i-1][$j-1]{t}.$t[$j] );
}
}
}
return @{$A[-1][-1]}{'s', 't'};
}
print join "\n", levenshtein_distance_alignment "rosettacode", "raisethysword";</lang>
- Output:
ro-settac-o-de raisethysword-
Phix
plus the indicator from Go
<lang Phix>function LevenshteinAlignment(string a, b)
integer la = length(a)+1,
lb = length(b)+1
sequence costs = repeat(repeat(0,lb),la)
for j=1 to lb do
costs[1][j] = j
end for
for i=2 to la do
costs[i][1] = i
for j=2 to lb do
integer tmp1 = 1+min(costs[i-1][j], costs[i][j-1]),
tmp2 = costs[i-1][j-1] + (a[i-1]!=b[j-1])
costs[i][j] = min(tmp1, tmp2)
end for
end for
-- walk back through matrix to figure out the path
string arev = "",
brev = ""
integer i = la, j = lb
while i>1 and j>1 do
integer tmp = costs[i-1][j-1] + (a[i-1]!=b[j-1])
switch costs[i][j] do
case tmp: i -= 1; arev &= a[i]
j -= 1; brev &= b[j]
case 1 + costs[i-1][j]: i -= 1; arev &= a[i]
brev &= '-'
case 1 + costs[i][j-1]: arev &= '-'
j -= 1; brev &= b[j]
end switch
end while
return {reverse(arev),reverse(brev)}
end function
procedure test(string a,b)
{a,b} = LevenshteinAlignment(a,b)
string c = sq_add(repeat(' ',length(a)),sq_mul(sq_ne(a,b),'|'-' '))
printf(1,"%s\n%s\n%s\n\n",{a,b,c})
end procedure test("rosettacode", "raisethysword") test("place", "palace")</lang>
- Output:
r-oset-tacode raisethysword || |||| || p-lace palace |
Python
<lang Python>from difflib import ndiff
def levenshtein(str1, str2):
result = ""
pos, removed = 0, 0
for x in ndiff(str1, str2):
if pos<len(str1) and str1[pos] == x[2]:
pos += 1
result += x[2]
if x[0] == "-":
removed += 1
continue
else:
if removed > 0:
removed -=1
else:
result += "-"
print(result)
levenshtein("place","palace") levenshtein("rosettacode","raisethysword") </lang>
- Output:
p-lace ro-settac-o-de
Racket
Simple version (no aligment)
First we will analyze this solution that only computes the distance. See http://blog.racket-lang.org/2012/08/dynamic-programming-versus-memoization.html for a discussion of the code.
<lang racket>#lang racket
(define (memoize f)
(local ([define table (make-hash)])
(lambda args
(dict-ref! table args (λ () (apply f args))))))
(define levenshtein
(memoize
(lambda (s t)
(cond
[(and (empty? s) (empty? t)) 0]
[(empty? s) (length t)]
[(empty? t) (length s)]
[else
(if (equal? (first s) (first t))
(levenshtein (rest s) (rest t))
(min (add1 (levenshtein (rest s) t))
(add1 (levenshtein s (rest t)))
(add1 (levenshtein (rest s) (rest t)))))]))))</lang>
Demonstration: <lang racket>(levenshtein (string->list "rosettacode")
(string->list "raisethysword"))</lang>
- Output:
8
Complete version
Now we extend the code from http://blog.racket-lang.org/2012/08/dynamic-programming-versus-memoization.html to show also the alignment. The code is very similar, but it stores the partial results (number of edits and alignment of each substring) in a lev structure. <lang Racket>#lang racket
(struct lev (n s t))
(define (lev-add old n sx tx)
(lev (+ n (lev-n old))
(cons sx (lev-s old))
(cons tx (lev-t old))))
(define (list-repeat n v)
(build-list n (lambda (_) v)))
(define (memoize f)
(local ([define table (make-hash)])
(lambda args
(dict-ref! table args (λ () (apply f args))))))
(define levenshtein/list
(memoize
(lambda (s t)
(cond
[(and (empty? s) (empty? t))
(lev 0 '() '())]
[(empty? s)
(lev (length t) (list-repeat (length t) #\-) t)]
[(empty? t)
(lev (length s) s (list-repeat (length s) #\-))]
[else
(if (equal? (first s) (first t))
(lev-add (levenshtein/list (rest s) (rest t))
0 (first s) (first t))
(argmin lev-n (list (lev-add (levenshtein/list (rest s) t)
1 (first s) #\-)
(lev-add (levenshtein/list s (rest t))
1 #\- (first t))
(lev-add (levenshtein/list (rest s) (rest t))
1 (first s) (first t)))))]))))
(define (levenshtein s t)
(let ([result (levenshtein/list (string->list s)
(string->list t))])
(values (lev-n result)
(list->string (lev-s result))
(list->string (lev-t result)))))</lang>
Demonstration: <lang racket>(let-values ([(dist exp-s exp-t)
(levenshtein "rosettacode" "raisethysword")]) (printf "levenshtein: ~a edits\n" dist) (displayln exp-s) (displayln exp-t))</lang>
- Output:
levenshtein: 8 edits r-oset-taco-de raisethysword-
Raku
(formerly Perl 6)
<lang perl6>sub align ( Str $σ, Str $t ) {
my @s = flat *, $σ.comb;
my @t = flat *, $t.comb;
my @A;
@A[$_][ 0]<d s t> = $_, @s[1..$_].join, '-' x $_ for ^@s;
@A[ 0][$_]<d s t> = $_, '-' x $_, @t[1..$_].join for ^@t;
for 1 ..^ @s X 1..^ @t -> (\i, \j) {
if @s[i] ne @t[j] {
@A[i][j]<d> = 1 + my $min =
min @A[i-1][j]<d>, @A[i][j-1]<d>, @A[i-1][j-1]<d>;
@A[i][j] =
@A[i-1][j]<d> == $min ?? (@A[i-1][j] Z~ @s[i], '-') !!
@A[i][j-1]<d> == $min ?? (@A[i][j-1] Z~ '-', @t[j]) !!
(@A[i-1][j-1] Z~ @s[i], @t[j]);
} else {
@A[i][j]<d s t> = @A[i-1][j-1]<d s t> Z~ , @s[i], @t[j];
}
}
return @A[*-1][*-1];
}
.say for align 'rosettacode', 'raisethysword';</lang>
- Output:
ro-settac-o-de raisethysword-
Ruby
uses "lcs" from here <lang ruby>require 'lcs'
def levenshtein_align(a, b)
apos, bpos = LCS.new(a, b).backtrack2
c = ""
d = ""
x0 = y0 = -1
dx = dy = 0
apos.zip(bpos) do |x,y|
diff = x + dx - y - dy
if diff < 0
dx -= diff
c += "-" * (-diff)
elsif diff > 0
dy += diff
d += "-" * diff
end
c += a[x0+1..x]
x0 = x
d += b[y0+1..y]
y0 = y
end
c += a[x0+1..-1]
d += b[y0+1..-1]
diff = a.length + y0 - b.length - x0
if diff < 0
c += "-" * (-diff)
elsif diff > 0
d += "-" * diff
end
[c, d]
end
puts levenshtein_align("rosettacode", "raisethysword")</lang>
- Output:
r-oset-taco-de raisethysword-
Rust
Cargo.toml
[dependencies] edit-distance = "^1.0.0"
src/main.rs <lang Rust>extern crate edit_distance;
edit_distance("rosettacode", "raisethysword");</lang>
Scala
- Output:
Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or [1].
<lang Scala>import scala.collection.mutable import scala.collection.parallel.ParSeq
object LevenshteinAlignment extends App {
val vlad = new Levenshtein("rosettacode", "raisethysword")
val alignment = vlad.revLevenstein()
class Levenshtein(s1: String, s2: String) {
val memoizedCosts = mutable.Map[(Int, Int), Int]()
def revLevenstein(): (String, String) = {
def revLev: (Int, Int, String, String) => (String, String) = {
case (_, 0, revS1, revS2) => (revS1, revS2)
case (0, _, revS1, revS2) => (revS1, revS2)
case (i, j, revS1, revS2) =>
if (memoizedCosts(i, j) == (memoizedCosts(i - 1, j - 1)
+ (if (s1(i - 1) != s2(j - 1)) 1 else 0)))
revLev(i - 1, j - 1, s1(i - 1) + revS1, s2(j - 1) + revS2)
else if (memoizedCosts(i, j) == 1 + memoizedCosts(i - 1, j))
revLev(i - 1, j, s1(i - 1) + revS1, "-" + revS2)
else
revLev(i, j - 1, "-" + revS1, s2(j - 1) + revS2)
}
revLev(s1.length, s2.length, "", "") }
private def levenshtein: Int = {
def lev: ((Int, Int)) => Int = {
case (k1, k2) =>
memoizedCosts.getOrElseUpdate((k1, k2), (k1, k2) match {
case (i, 0) => i
case (0, j) => j
case (i, j) =>
ParSeq(1 + lev((i - 1, j)),
1 + lev((i, j - 1)),
lev((i - 1, j - 1))
+ (if (s1(i - 1) != s2(j - 1)) 1 else 0)).min
})
}
lev((s1.length, s2.length)) }
levenshtein }
println(alignment._1) println(alignment._2)
}</lang>
Sidef
<lang ruby>func align(s, t) {
s.chars!.prepend!('^')
t.chars!.prepend!('^')
var A = []
} = (A[i-1][j]{:d} == min
? [A[i-1][j]{:s}+s[i], A[i-1][j]{:t}+'-']
: (A[i][j-1]{:d} == min
? [A[i][j-1]{:s}+'-', A[i][j-1]{:t}+t[j]]
: [A[i-1][j-1]{:s}+s[i], A[i-1][j-1]{:t}+t[j]]))...
}
else {
A[i][j]{@|<d s t>} = (
A[i-1][j-1]{:d},
A[i-1][j-1]{:s}+s[i],
A[i-1][j-1]{:t}+t[j],
)
}
}
}
return [A[-1][-1]{@|}]
}
align("rosettacode", "raisethysword").each { .say }</lang>
- Output:
ro-settac-o-de raisethysword-
Tcl
<lang tcl>package require struct::list proc levenshtein/align {a b} {
lassign [struct::list longestCommonSubsequence [split $a ""] [split $b ""]]\
apos bpos
set c ""
set d ""
set x0 [set y0 -1]
set dx [set dy 0]
foreach x $apos y $bpos {
if {$x+$dx < $y+$dy} { set n [expr {($y+$dy)-($x+$dx)}] incr dx $n append c [string repeat "-" $n] } elseif {$x+$dx > $y+$dy} { set n [expr {($x+$dx)-($y+$dy)}] incr dy $n append d [string repeat "-" $n] } append c [string range $a $x0+1 $x] set x0 $x append d [string range $b $y0+1 $y] set y0 $y
}
append c [string range $a $x0+1 end]
append d [string range $b $y0+1 end]
set al [string length $a]
set bl [string length $b]
if {$al+$y0 < $bl+$x0} {
append c [string repeat "-" [expr {$bl+$x0-$y0-$al}]]
} elseif {$bl+$x0 < $al+$y0} {
append d [string repeat "-" [expr {$al+$y0-$x0-$bl}]]
} return $c\n$d
}
puts [levenshtein/align "rosettacode" "raisethysword"]</lang>
- Output:
r-oset-taco-de raisethysword-
Wren
<lang ecmascript>import "/str" for Str import "/math" for Math
var levenshteinAlign = Fn.new { |a, b|
a = Str.lower(a)
b = Str.lower(b)
var costs = List.filled(a.count+1, null)
for (i in 0..a.count) costs[i] = List.filled(b.count+1, 0)
for (j in 0..b.count) costs[0][j] = j
for (i in 1..a.count) {
costs[i][0] = i
for (j in 1..b.count) {
var temp = costs[i - 1][j - 1] + ((a[i - 1] == b[j - 1]) ? 0 : 1)
costs[i][j] = Math.min(1 + Math.min(costs[i - 1][j], costs[i][j - 1]), temp)
}
}
// walk back through matrix to figure out path
var aPathRev = ""
var bPathRev = ""
var i = a.count
var j = b.count
while (i != 0 && j != 0) {
var temp = costs[i - 1][j - 1] + ((a[i - 1] == b[j - 1]) ? 0 : 1)
var cij = costs[i][j]
if (cij == temp) {
i = i - 1
aPathRev = aPathRev + a[i]
j = j - 1
bPathRev = bPathRev + b[j]
} else if (cij == 1 + costs[i-1][j]) {
i = i - 1
aPathRev = aPathRev + a[i]
bPathRev = bPathRev + "-"
} else if (cij == 1 + costs[i][j-1]) {
aPathRev = aPathRev + "-"
j = j - 1
bPathRev = bPathRev+ b[j]
}
}
return [aPathRev[-1..0], bPathRev[-1..0]]
}
var result = levenshteinAlign.call("place", "palace") System.print(result[0]) System.print(result[1]) System.print() result = levenshteinAlign.call("rosettacode","raisethysword") System.print(result[0]) System.print(result[1])</lang>
- Output:
p-lace palace r-oset-tacode raisethysword
zkl
<lang zkl>fcn alignment(a,b){
a,b = a.toLower(), b.toLower();
costs := (a.len()+1).pump(List(),'wrap(a){ [1..b.len()].pump(List(a)) });
foreach i,j in (a.len()+1, [1..b.len()]){
costs[i][j] = ( 1 + costs[i-1][j].min(costs[i][j-1]) )
.min(if(a[i-1] == b[j-1]) costs[i-1][j-1] else costs[i-1][j-1] + 1);
}
// walk back through matrix to figure out path
aPathRev,bPathRev := Data(),Data(); // byte buckets
i,j := a.len(), b.len();
while(i!=0 and j!= 0){
if (costs[i][j] ==
( if(a[i-1]==b[j-1]) costs[i-1][j-1] else costs[i-1][j-1]+1 )){
aPathRev.append(a[i-=1]);
bPathRev.append(b[j-=1]);
} else if(costs[i][j] == 1+costs[i-1][j]){
aPathRev.append(a[i-=1]); bPathRev.append("-");
} else if (costs[i][j] == 1+costs[i][j-1]){
aPathRev.append("-"); bPathRev.append(b[j-=1]);
} } return(aPathRev.text.reverse(), bPathRev.text.reverse())
}</lang> <lang zkl>result := alignment("rosettacode", "raisethysword"); println(result[0]); println(result[1]);</lang>
- Output:
r-oset-tacode raisethysword