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).
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>
Java
<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>
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>
Python
<lang python>from itertools import groupby
def soundex(word):
codes = ("bfpv","cgjkqsxz", "dt", "l", "mn", "r")
cmap = lambda kar: .join( str(ix+1) for ix,cod in enumerate(codes) if kar in cod )
cmap2 = lambda kar: cmap(kar) or '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'
Tcl
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>