Soundex: Difference between revisions

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Line 118: Line 118:
nLen2:=LEN(cSound2)
nLen2:=LEN(cSound2)


// make the two words the same length
// make the two words the same length. This is a safety. They both should be 4 characters long.
IF nLen1 > nLen2
IF nLen1 > nLen2
cSound2:=PADR(cSound2, nLen1-nLen2, "0")
cSound2:=PADR(cSound2, nLen1-nLen2, "0")

Revision as of 23:17, 4 November 2010

Task
Soundex
You are encouraged to solve this task according to the task description, using any language you may know.

Soundex is an algorithm for creating indices for words based on their pronunciation. The goal is for homophones to be encoded to the same representation so that they can be matched despite minor differences in spelling (from the WP article).

C

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>

char chr(int i) {

   return ((0<=i) && (i<=9)) ? '0'+i : '-';

}

char *soundex(const char *word, char *bufr) {

   static const char *lset[] = {
       "", "BFPVbfpv", "CGJKQSXZcgjkqsxz", "DTdt", "Ll","MNmn", "Rr"  };
  1. define LSIZE (sizeof(lset)/sizeof(char *))
   const char *ch;
   char *bp;
   char prev = '-';
   int j;
   bp = bufr;
   ch = word;
   while (*ch && (bp-bufr<4)) {
       for (j=1; j< LSIZE; j++)
           if (strchr( lset[j], *ch) ) break;
       if ((j<LSIZE) && (chr(j) != prev))
           *bp++ = prev = chr(j);
       else if (!strchr("WHwh", *ch)) 
           prev = '7';
       if (ch == word) {
           bp = bufr;
           *bp++ = toupper(*ch);
       }
       ch++;
   }
   while (bp-bufr < 4)
       *bp++ = '0';
   *bp = 0;
    return bufr;

}

int main() {

   const char *w, **wp;
   static const char *testwords[] = {
       "Euler","Gauss","Hilbert","Knuth","Lloyd","Lukasiewicz","Ellery",
       "Ghosh", "Heilbronn","Kant","Ladd","Lissajous","Wheaton","Ashcraft",
       "Burroughs","Burrows","O'Hara", NULL};
   char buffer[8];
   w= "Soundex";
   printf("soundex(%s) = %s\n", w, soundex(w,buffer));
   w= "Example";
   printf("soundex(%s) = %s\n", w, soundex(w,buffer));
   w= "Sownteks";
   printf("soundex(%s) = %s\n", w, soundex(w,buffer));
   w= "Ekzampul";
   printf("soundex(%s) = %s\n", w, soundex(w,buffer));
   for (wp=testwords; *wp; wp++ ) 
       printf("soundex(%s) = %s\n", *wp, soundex(*wp, buffer));
   
   return 0;

}</lang>

Clipper/XBase++

<lang Clipper/XBase++>FUNCTION Soundex(cWord)

/*

This is a Clipper/XBase++ implementation of the standard American Soundex procedure. 
*/

LOCAL cSoundex, i, nLast, cChar, nCode

cWord:=ALLTRIM(UPPER(cWord)) cSoundex:=LEFT(cWord, 1) // first letter is first char nLast:=-1 FOR i:=2 TO LEN(cWord)

  cChar:=SUBSTR(cWord, i, 1)   // get char
  nCode:=SoundexCode(cChar)    // get soundex code for char
  IF nCode=0                   // if 0, ignore
     LOOP
  ENDIF
  IF nCode#nLast               // if not same code, add to soundex
     nLast:=nCode              // and replace the last one
     cSoundex+=STR(nCode, 1)
  ENDIF

NEXT cSoundex:=PADR(cSoundex, 4, "0")

RETURN(cSoundex)

STATIC FUNCTION SoundexCode(cLetter) LOCAL aCodes:={"BFPV", "CGJKQSXZ", "DT", "L", "MN", "R"}, i, nRet:=0

FOR i:=1 TO LEN(aCodes)

  IF cLetter $ aCodes[i]
     nRet:=i
     EXIT
  ENDIF

NEXT

RETURN(nRet)

FUNCTION SoundexDifference(cSound1, cSound2) LOCAL nMatch:=0, nLen1, nLen2, i

nLen1:=LEN(cSound1) nLen2:=LEN(cSound2)

// make the two words the same length. This is a safety. They both should be 4 characters long. IF nLen1 > nLen2

  cSound2:=PADR(cSound2, nLen1-nLen2, "0")

ELSEIF nLen1 < nLen2

  cSound1:=PADR(cSound1, nLen2-nLen1, "0")

ENDIF

// compare the corresponding characters between the two words FOR i:=1 TO LEN(cSound1)

  IF SUBSTR(cSound1, i, 1) == SUBSTR(cSound2, i, 1)
     ++nMatch
  ENDIF

NEXT

RETURN(nMatch)

                                                                                                                                                              • </lang>

--Clippersolutions 23:14, 4 November 2010 (UTC)--Clippersolutions 23:14, 4 November 2010 (UTC)

Forth

This implements the US Census rules, where W and H are ignored but, unlike vowels, aren't separators. Further corner cases welcome...

<lang forth>: alpha-table create does> swap 32 or [char] a - 0 max 26 min + 1+ c@ ;

alpha-table soundex-code

 ,"  123 12. 22455 12623 1.2 2 "
  \ ABCDEFGHIJKLMNOPQRSTUVWXYZ
soundex ( name len -- pad len )
 over c@ pad c!                  \ First character verbatim
 pad 1+ 3 [char] 0 fill          \ Pad to four characters with zeros
 1 pad c@ soundex-code  ( count code )
 2swap bounds do
   i c@ soundex-code   ( count code next )
   2dup = if         drop else   \ runs are ignored
   dup [char] . = if drop else   \ W, H don't separate runs of consonants
   dup bl = if        nip else   \ vowels separate consonants but aren't coded
     nip
     2dup swap pad + c!
     swap 1+
     tuck 4 = if leave then
   then then then
 loop
 2drop pad 4 ;

\ Knuth's test cases s" Euler" soundex cr type \ E460 s" Gauss" soundex cr type \ G200 s" Hilbert" soundex cr type \ H416 s" Knuth" soundex cr type \ K530 s" Lloyd" soundex cr type \ L300 s" Lukasiewicz" soundex cr type \ L222 (W test) s" Ellery" soundex cr type \ E460 s" Ghosh" soundex cr type \ G200 s" Heilbronn" soundex cr type \ H416 s" Kant" soundex cr type \ K530 s" Ladd" soundex cr type \ L300 s" Lissajous" soundex cr type \ L222

s" Wheaton" soundex cr type \ W350 s" Ashcraft" soundex cr type \ A261 (H tests) s" Burroughs" soundex cr type \ B620 s" Burrows" soundex cr type \ B620 (W test) (any Welsh names?) s" O'Hara" soundex cr type \ O600 (punctuation test)</lang>

Haskell

<lang haskell>import Text.PhoneticCode.Soundex import Control.Arrow</lang> Example: <lang haskell>*Main> mapM_ print $ map (id &&& soundexSimple) ["Soundex", "Example", "Sownteks", "Ekzampul"] ("Soundex","S532") ("Example","E251") ("Sownteks","S532") ("Ekzampul","E251")</lang>

Icon and Unicon

Icon

<lang icon>procedure main(arglist) # computes soundex of each argument every write(x := !arglist, " => ",soundex(x)) end

procedure soundex(name)

  local  dig,i,x
  static con
  initial {                                   # construct mapping x[i] => i all else .
     x := ["bfpv","cgjkqsxz","dt","l","mn","r"]
     every ( dig := con := "") ||:= repl(i := 1 to *x,*x[i]) do con ||:= x[i]
     con := map(map(&lcase,con,dig),&lcase,repl(".",*&lcase))
     } 
 
  name := map(name)                           # lower case
  name[1] := map(name[1],&lcase,&ucase)       # upper case 1st
  name := map(name,&lcase,con)                # map cons
  every x := !"123456" do
      while name[find(x||x,name)+:2] := x     # kill duplicates
  while name[upto('.',name)] := ""            # kill .
  return left(name,4,"0")                     

end</lang>

implements soundex. The above version is an adaptation of that procedure.

Unicon

This Icon solution works in Unicon.

J

Solution <lang j>removeDups =: {.;.1~ (1 , }. ~: }: ) codes =: ;: 'BFPV CGJKQSXZ DT L MN R HW'

soundex =: 3 : 0

if. 0=# k=.toupper y do. '0' return. end.
({.k), ,": ,. 3 {. 0-.~ }. removeDups 7 0:`(I.@:=)`]} , k >:@I.@:(e. &>)"0 _ codes

)</lang> Usage <lang j>names=: 'Lloyd Woolcock Donnell Baragwanath Williams Ashcroft Euler Ellery Gauss Ghosh Hilbert Heilbronn Knuth Kant Ladd Lukasiewicz Lissajous' soundexNames=: 'L300 W422 D540 B625 W452 A226 E460 E460 G200 G200 H416 H416 K530 K530 L300 L222 L222'

  soundex &> ;: names

L300 W422 D540 B625 W452 ....</lang> Test <lang j> soundexNames-:(soundex &.>) &. ;: names 1</lang>

Java

Translation of: VBScript

<lang java>public static void main(String[] args){

   System.out.println(soundex("Soundex"));
   System.out.println(soundex("Example"));
   System.out.println(soundex("Sownteks"));
   System.out.println(soundex("Ekzampul"));
 }

private static String getCode(char c){

 switch(c){
   case 'B': case 'F': case 'P': case 'V':
     return "1";
   case 'C': case 'G': case 'J': case 'K':
   case 'Q': case 'S': case 'X': case 'Z':
     return "2";
   case 'D': case 'T':
     return "3";
   case 'L':
     return "4";
   case 'M': case 'N':
     return "5";
   case 'R':
     return "6";
   default:
     return "";
 }

}

public static String soundex(String s){

 String code, previous, soundex;
 code = s.toUpperCase().charAt(0) + "";
 previous = "7";
 for(int i = 1;i < s.length();i++){
   String current = getCode(s.toUpperCase().charAt(i));
   if(current.length() > 0 && !current.equals(previous)){
     code = code + current;
   }
   previous = current;
 }
 soundex = (code + "0000").substring(0, 4);
 return soundex;

}</lang> Output:

S532
E251
S532
E251

JavaScript

<lang javascript>var soundex = function (s) {

    var a = s
            .substring(1, s.length)
            .toLowerCase()
            .split(),
        r = ,
        codes = {
            a: , e: , i: , o: , u: ,
            b: 1, f: 1, p: 1, v: 1,
            c: 2, g: 2, j: 2, k: 2, q: 2, s: 2, x: 2, z: 2,
            d: 3, t: 3,
            l: 4,
            m: 5, n: 5,
            r: 6
        };
    r = s[0].toUpperCase() +
        a
        .filter(function (v, i, a) { return v !== a[i + 1]; })
        .map(function (v, i, a) { return codes[v] }).join();
    return (r + '000').slice(0, 4);

};</lang>

OCaml

Here is an implementation:

<lang ocaml>let c2d = function

 | 'B' | 'F' | 'P' | 'V' -> "1"
 | 'C' | 'G' | 'J' | 'K' | 'Q' | 'S' | 'X' | 'Z' -> "2"
 | 'D' | 'T' -> "3"
 | 'L' -> "4"
 | 'M' | 'N' -> "5"
 | 'R' -> "6"
 | _ -> ""

let rec dbl acc = function

 | [] -> (List.rev acc)
 | [c] -> List.rev(c::acc)
 | c1::(c2::_ as tl) ->
     if c1 = c2
     then dbl acc tl
     else dbl (c1::acc) tl

let pad s =

 match String.length s with
 | 0 -> s ^ "000"
 | 1 -> s ^ "00"
 | 2 -> s ^ "0"
 | 3 -> s
 | _ -> String.sub s 0 3

let soundex_aux rem =

 pad(String.concat "" (dbl [] (List.map c2d rem)))

let soundex s =

 let s = String.uppercase s in
 let cl = ref [] in
 String.iter (fun c -> cl := c :: !cl) s;
 match dbl [] (List.rev !cl) with
 | c::rem -> (String.make 1 c) ^ (soundex_aux rem)
 | [] -> invalid_arg "soundex"</lang>

Test our implementation:

<lang ocaml>let tests = [

 "Soundex",     "S532";
 "Example",     "E251";
 "Sownteks",    "S532";
 "Ekzampul",    "E251";
 "Euler",       "E460";
 "Gauss",       "G200";
 "Hilbert",     "H416";
 "Knuth",       "K530";
 "Lloyd",       "L300";
 "Lukasiewicz", "L222";
 "Ellery",      "E460";
 "Ghosh",       "G200";
 "Heilbronn",   "H416";
 "Kant",        "K530";
 "Ladd",        "L300";
 "Lissajous",   "L222";
 "Wheaton",     "W350";
 "Ashcraft",    "A226";
 "Burroughs",   "B622";
 "Burrows",     "B620";
 "O'Hara",      "O600";
 ]

let () =

 print_endline " Word   \t Code  Found Status";
 List.iter (fun (word, code1) ->
   let code2 = soundex word in
   let status = if code1 = code2 then "OK " else "Arg" in
   Printf.printf " \"%s\" \t %s  %s  %s\n" word code1 code2 status
 ) tests</lang>

This test outputs:

Word            Code  Found Status
"Soundex"       S532  S532  OK 
"Example"       E251  E251  OK 
"Sownteks"      S532  S532  OK 
"Ekzampul"      E251  E251  OK 
"Euler"         E460  E460  OK 
"Gauss"         G200  G200  OK 
"Hilbert"       H416  H416  OK 
"Knuth"         K530  K530  OK 
"Lloyd"         L300  L300  OK 
"Lukasiewicz"   L222  L222  OK 
"Ellery"        E460  E460  OK 
"Ghosh"         G200  G200  OK 
"Heilbronn"     H416  H416  OK 
"Kant"          K530  K530  OK 
"Ladd"          L300  L300  OK 
"Lissajous"     L222  L222  OK 
"Wheaton"       W350  W350  OK 
"Ashcraft"      A226  A226  OK 
"Burroughs"     B622  B622  OK 
"Burrows"       B620  B620  OK 
"O'Hara"        O600  O600  OK

See Soundex/OCaml for a version that can switch the language (English, French...) with a type which definition is hidden in the interface.

Perl

The Text::Soundex core module supports various soundex algorithms. <lang perl>use Text::Soundex; print soundex("Soundex"), "\n"; # S532 print soundex("Example"), "\n"; # E251 print soundex("Sownteks"), "\n"; # S532 print soundex("Ekzampul"), "\n"; # E251</lang>

PHP

PHP already has a built-in soundex() function: <lang php><?php echo soundex("Soundex"), "\n"; // S532 echo soundex("Example"), "\n"; // E251 echo soundex("Sownteks"), "\n"; // S532 echo soundex("Ekzampul"), "\n"; // E251 ?></lang>

PicoLisp

<lang PicoLisp>(de soundex (Str)

  (pack
     (pad -4
        (cons
           (uppc (char (char Str)))
           (let Last NIL
              (extract
                 '((C)
                    (and
                       (setq C
                          (case (uppc C)
                             (`(chop "BFPV") "1")
                             (`(chop "CGJKQSXZ") "2")
                             (("D" "T") "3")
                             ("L" "4")
                             (("M" "N") "5")
                             ("R" "6") ) )
                       (<> Last C)
                       (setq Last C) ) )
                 (head 3 (cdr (chop Str))) ) ) ) ) ) )</lang>

Output:

: (mapcar soundex '("Soundex" "Example" "Sownteks" "Ekzampul"))
-> ("S500" "E250" "S500" "E200")

PureBasic

<lang PureBasic>Procedure.s getCode(c.s)

   Protected  getCode.s = ""
    
   If FindString("BFPV", c ,1)     : getCode = "1" : EndIf
   If FindString("CGJKQSXZ", c ,1) : getCode = "2" : EndIf 
   If FindString("DT", c ,1)       : getCode = "3" : EndIf
   If "L" = c                      : getCode = "4" : EndIf
   If FindString("MN", c ,1)       : getCode = "5" : EndIf
   If "R" = c                      : getCode = "6" : EndIf 
   If FindString("HW", c ,1)       : getCode = "." : EndIf
   ProcedureReturn getCode

EndProcedure

Procedure.s soundex(word.s)

   Protected.s previous.s = "" , code.s , current , soundex
   Protected.i i
   
   word = UCase(word)
   code = Mid(word,1,1)
   previous = ""
   For i = 2 To (Len(word) + 1)
       current = getCode(Mid(word, i, 1))
       If current = "." : Continue : EndIf
       If Len(current) > 0 And current <> previous
           code + current
       EndIf
       previous = current
       If Len(code) = 4
         Break
       EndIf  
   Next
   If Len(code) < 4 
       code = LSet(code, 4,"0")
   EndIf
   ProcedureReturn code

EndProcedure

OpenConsole()

PrintN (soundex("Lukasiewicz")) PrintN("Press any key to exit"): Repeat: Until Inkey() <> ""</lang>

Python

<lang python>from itertools import groupby

def soundex(word):

  codes = ("bfpv","cgjkqsxz", "dt", "l", "mn", "r")
  soundDict = dict((ch, str(ix+1)) for ix,cod in enumerate(codes) for ch in cod)
  cmap2 = lambda kar: soundDict.get(kar, '9')
  sdx =  .join(cmap2(kar) for kar in word.lower())
  sdx2 = word[0].upper() + .join(k for k,g in list(groupby(sdx))[1:] if k!='9')
  sdx3 = sdx2[0:4].ljust(4,'0')
  return sdx3

</lang> Example Output <lang Python>>>>print soundex("soundex") S532 >>>print soundex("example") E251 >>>print soundex("ciondecks") C532 >>>print soundex("ekzampul") E251</lang>

Ruby

Courtesy http://snippets.dzone.com/posts/show/4530 <lang ruby>class String

 SoundexChars = 'BFPVCGJKQSXZDTLMNR'
 SoundexNums  = '111122222222334556'
 SoundexCharsEx = '^' + SoundexChars
 SoundexCharsDel = '^A-Z'
 # desc: http://en.wikipedia.org/wiki/Soundex
 def soundex(census = true)
   str = self.upcase.delete(SoundexCharsDel)
   str[0,1] + str[1..-1].delete(SoundexCharsEx).
                         tr(SoundexChars, SoundexNums).
                         squeeze[0 .. (census ? 2 : -1)].
                         ljust(3, '0') rescue 
 end
 def sounds_like(other)
   self.soundex == other.soundex
 end

end

%w(Soundex Sownteks Example Ekzampul foo bar).each_slice(2) do |word1, word2|

 [word1, word2].each {|word| puts '%-8s -> %s' % [word, word.soundex]}
 print "'#{word1}' "
 print word1.sounds_like(word2) ? "sounds" : "does not sound"
 print " like '#{word2}'\n"

end</lang>

Soundex  -> S532
Sownteks -> S532
'Soundex' sounds like 'Sownteks'
Example  -> E251
Ekzampul -> E251
'Example' sounds like 'Ekzampul'
foo      -> F000
bar      -> B600
'foo' does not sound like 'bar'

Scala

<lang scala>def soundex(s:String)={

  var code=s.head.toUpper.toString
  var previous=getCode(code.head)
  for(ch <- s.drop(1); current=getCode(ch.toUpper)){
     if (!current.isEmpty && current!=previous)
        code+=current
     previous=current
  }
  code+="0000"
  code.slice(0,4)

}

def getCode(c:Char)={

  val code=Map("1"->List('B','F','P','V'),
     "2"->List('C','G','J','K','Q','S','X','Z'),
     "3"->List('D', 'T'),
     "4"->List('L'),
     "5"->List('M', 'N'),
     "6"->List('R'))
  code.find(_._2.exists(_==c)) match {
     case Some((k,_)) => k
     case _ => ""
  }

}</lang>

<lang scala>def main(args: Array[String]): Unit = {

  val tests=Map(
     "Soundex"     -> "S532",
     "Euler"	    -> "E460",
     "Gauss"	    -> "G200",
     "Hilbert"	    -> "H416",
     "Knuth"	    -> "K530",
     "Lloyd"	    -> "L300",
     "Lukasiewicz" -> "L222",
     "Ellery"	    -> "E460",
     "Ghosh"	    -> "G200",
     "Heilbronn"   -> "H416",
     "Kant"	    -> "K530",
     "Ladd"	    -> "L300",
     "Lissajous"   -> "L222",
     "Wheaton"	    -> "W350",
     "Ashcraft"    -> "A226",
     "Burroughs"   -> "B622",
     "Burrows"	    -> "B620",
     "O'Hara"	    -> "O600")
  tests.foreach{(v)=>
     val code=soundex(v._1)
     val status=if (code==v._2) "OK" else "ERROR"
     printf("Name: %-20s  Code: %s   Found: %s  - %s\n", v._1, v._2, code, status)
  }

}</lang>

Scheme

This implements American Soundex as described at [1].

Works with: any R6RS Scheme

<lang scheme>;; The American Soundex System

The soundex code consist of the first letter of the name followed
by three digits. These three digits are determined by dropping the
letters a, e, i, o, u, h, w and y and adding three digits from the
remaining letters of the name according to the table below. There
are only two additional rules. (1) If two or more consecutive
letters have the same code, they are coded as one letter. (2) If
there are an insufficient numbers of letters to make the three
digits, the remaining digits are set to zero.
Soundex Table
1 b,f,p,v
2 c,g,j,k,q,s,x,z
3 d, t
4 l
5 m, n
6 r
Examples
Miller M460
Peterson P362
Peters P362
Auerbach A612
Uhrbach U612
Moskowitz M232
Moskovitz M213

(define (char->soundex c)

 (case (char-upcase c)
   ((#\B #\F #\P #\V) #\1)
   ((#\C #\G #\J #\K #\Q #\S #\X #\Z) #\2)
   ((#\D #\T) #\3)
   ((#\L) #\4)
   ((#\M #\N) #\5)
   ((#\R) #\6)
   (else #\nul)))

(define (collapse-dups lst)

 (if (= (length lst) 1) lst
     (if (equal? (car lst) (cadr lst))

(collapse-dups (cdr lst)) (cons (car lst) (collapse-dups (cdr lst))))))

(define (remove-nul lst)

 (filter (lambda (c)

(not (equal? c #\nul))) lst))

(define (force-len n lst)

 (cond ((= n 0) '())

((null? lst) (force-len n (list #\0))) (else (cons (car lst) (force-len (- n 1) (cdr lst))))))

(define (soundex s)

 (let ((slst (string->list s)))
   (force-len 4 (cons (char-upcase (car slst))

(remove-nul (collapse-dups (map char->soundex (cdr slst))))))))

(soundex "miller") (soundex "Peterson") (soundex "PETERS") (soundex "auerbach") (soundex "Uhrbach") (soundex "Moskowitz") (soundex "Moskovitz")</lang>

Sample Output

> "M460"
> "P362"
> "P362"
> "A612"
> "U612"
> "M232"
> "M213"

SNOBOL4

Works with: Macro Spitbol
Works with: Snobol4+
Works with: CSnobol

US National Archives (NARA) Soundex. Includes the "HW" rule omitted by Knuth and many other implementations.

<lang SNOBOL4>* # Soundex coding

  • # ABCDEFGHIJKLMNOPQRSTUVWXYZ
  • # 01230127022455012623017202
       define('soundex(str)init,ch') :(soundex_end)

soundex sdxmap = '01230127022455012623017202'

       str = replace(str,&lcase,&ucase)

sdx1 str notany(&ucase) = :s(sdx1)

       init = substr(str,1,1)
       str = replace(str,&ucase,sdxmap)

sdx2 str len(1) $ ch span(*ch) = ch :s(sdx2)

  • # Omit next line for Knuth's simple Soundex

sdx3 str len(1) $ ch ('7' *ch) = ch :s(sdx3)

       str len(1) = init

sdx4 str any('07') = :s(sdx4)

       str = substr(str,1,4)
       str = lt(size(str),4) str dupl('0',4 - size(str))
       soundex = str :(return)

soundex_end

  • # Test and display
       test = " Washington Lee Gutierrez Pfister Jackson Tymczak"

+ " Ashcroft Swhgler O'Connor Rhys-Davies" loop test span(' ') break(' ') . name = :f(end)

       output = soundex(name) ' ' name :(loop)

end</lang>

Output:

W252 Washington
L000 Lee
G362 Gutierrez
P236 Pfister
J250 Jackson
T522 Tymczak
A261 Ashcroft
S460 Swhgler
O256 O'Connor

Tcl

Library: tcllib

contains an implementation of Knuth's soundex algorithm in the soundex package.

<lang tcl>package require soundex

foreach string {"Soundex" "Example" "Sownteks" "Ekzampul"} {

   set soundexCode [soundex::knuth $string]
   puts "\"$string\" has code $soundexCode"

}</lang> Which produces this output:

"Soundex" has code S532
"Example" has code E251
"Sownteks" has code S532
"Ekzampul" has code E251

VBScript

<lang vbscript>Function getCode(c)

   Select Case c
       Case "B", "F", "P", "V"
           getCode = "1"
       Case "C", "G", "J", "K", "Q", "S", "X", "Z"
           getCode = "2"
       Case "D", "T"
           getCode = "3"
       Case "L"
           getCode = "4"
       Case "M", "N"
           getCode = "5"
       Case "R"
           getCode = "6"
   End Select

End Function

Function soundex(s)

   Dim code, previous
   code = UCase(Mid(s, 1, 1))
   previous = 7
   For i = 2 to (Len(s) + 1)
       current = getCode(UCase(Mid(s, i, 1)))
       If Len(current) > 0 And current <> previous Then
           code = code & current
       End If
       previous = current
   Next
   soundex = Mid(code, 1, 4)
   If Len(code) < 4 Then
       soundex = soundex & String(4 - Len(code), "0")
   End If

End Function</lang>