Soundex

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).

Haskell

Using module Text.PhoneticCode.Soundex from package phonetic-code hackageDB <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>

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)
            .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>

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>

Python

<lang Python>def soundex(word):

  codes = ("bfpv","cgjkqsxz", "dt", "l", "mn", "r")
  sdx = .join( .join(str(1+ix) for ix,k in enumerate(codes) if kar in k) for kar in word.lower())
  sdx2 = word[0].upper() + .join( sdx[k] for k in range(1,len(sdx)) if sdx[k] != sdx[k-1])
  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>

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>