Like levenchtein(), You can do :
(strlen($string2) - similar_text($string,$string2))
to see how much characters have been changed.
similar_text
(PHP 4, PHP 5)
similar_text — Calcule la similarité de deux chaînes
Description
Calcule la similarité entre les deux chaînes first et second , selon la méthode d'Oliver [1993]. Notez que cette implémentation n'utilise pas la méthode de pile comme dans le pseudo code d'Oliver, mais des appels récursifs, ce qui accélère ou pas le processus. Notez que la complexité de l'algorithme est en O(N**3) où N est la taille de la plus grande chaîne.
Liste de paramètres
- first
-
La première chaîne.
- second
-
La seconde chaîne.
- percent
-
En passant une référence en tant que troisième argument, similar_text() va calculer la similarité en pourcentage automatiquement.
Valeurs de retour
Retourne le nombre de caractères identiques dans les deux chaînes.
similar_text
08-Aug-2007 02:59
19-Jan-2007 05:50
The speed issues for similar_text seem to be only an issue for long sections of text (>20000 chars).
I found a huge performance improvement in my application by just testing if the string to be tested was less than 20000 chars before calling similar_text.
20000+ took 3-5 secs to process, anything else (10000 and below) took a fraction of a second.
Fortunately for me, there was only a handful of instances with >20000 chars which I couldn't get a comparison % for.
08-Mar-2006 05:08
Well... hard to explain, why I have written this function, but maybe it will be usefull.
<?php
// returns the percentage of the string "similarity"
function str_compare($str1, $str2) {
$count = 0;
$str1 = ereg_replace("[^a-z]", ' ', strtolower($str1));
while(strstr($str1, ' ')) {
$str1 = str_replace(' ', ' ', $str1);
}
$str1 = explode(' ', $str1);
$str2 = ereg_replace("[^a-z]", ' ', strtolower($str2));
while(strstr($str2, ' ')) {
$str2 = str_replace(' ', ' ', $str2);
}
$str2 = explode(' ', $str2);
if(count($str1)<count($str2)) {
$tmp = $str1;
$str1 = $str2;
$str2 = $tmp;
unset($tmp);
}
for($i=0; $i<count($str1); $i++) {
if(in_array($str1[$i], $str2)) {
$count++;
}
}
return $count/count($str2)*100;
}
?>
24-Feb-2006 11:30
The link below to the Oliver document appears to be broken. Here is one that works: http://citeseer.ist.psu.edu/oliver93decision.html
07-Jun-2004 02:01
Note that this function will calculate the percentage blindly, without regard to the LENGHT of the string.
This may become important if you try to print similar names to SMALL strings :
e.g.
I want to print out the value if it is 90 percent similar to the other one : the value is HE, the correct value is HEC
The similar_text() function will return approximately 66.7 %, and it will not print it because it is smaller than 90 %, although almost all of the string was matched.
10-May-2004 06:21
In PHP4+, you don't need to pass the percent variable as reference..
Instead, use this way:
<?
similar_text($string1, $string2, $p);
echo "Percent: $p%";
?>
In PHP5, you'll get a ugly warning message when passing this variable as reference.. But it's configurable in php.ini (allow_call_time_pass_reference = Off)
That's it... Another great function! :)
06-Feb-2003 03:46
If you have reserved names in a database that you don't want others to use, i find this to work pretty good.
I added strtoupper to the variables to validate typing only. Taking case into consideration will decrease similarity.
$query = mysql_query("select * from $table") or die("Query failed");
while ($row = mysql_fetch_array($query)) {
similar_text(strtoupper($_POST['name']), strtoupper($row['reserved']), $similarity_pst);
if (number_format($similarity_pst, 0) > 90){
$too_similar = $row['reserved'];
print "The name you entered is too similar the reserved name "".$row['reserved'].""";
break;
}
}
09-Mar-2002 05:14
Well, as mentioned above the speed is O(N^3), i've done a longest common subsequence way that is O(m.n) where m and n are the length of str1 and str2, the result is a percentage and it seems to be exactly the same as similar_text percentage but with better performance... here's the 3 functions i'm using..
function LCS_Length($s1, $s2)
{
$m = strlen($s1);
$n = strlen($s2);
//this table will be used to compute the LCS-Length, only 128 chars per string are considered
$LCS_Length_Table = array(array(128),array(128));
//reset the 2 cols in the table
for($i=1; $i < $m; $i++) $LCS_Length_Table[$i][0]=0;
for($j=0; $j < $n; $j++) $LCS_Length_Table[0][$j]=0;
for ($i=1; $i <= $m; $i++) {
for ($j=1; $j <= $n; $j++) {
if ($s1[$i-1]==$s2[$j-1])
$LCS_Length_Table[$i][$j] = $LCS_Length_Table[$i-1][$j-1] + 1;
else if ($LCS_Length_Table[$i-1][$j] >= $LCS_Length_Table[$i][$j-1])
$LCS_Length_Table[$i][$j] = $LCS_Length_Table[$i-1][$j];
else
$LCS_Length_Table[$i][$j] = $LCS_Length_Table[$i][$j-1];
}
}
return $LCS_Length_Table[$m][$n];
}
function str_lcsfix($s)
{
$s = str_replace(" ","",$s);
$s = ereg_replace("[éèêëËÊÉÈ]","e", $s);
$s = ereg_replace("[àáâãäåÄÅÃÂÁÀ]","a", $s);
$s = ereg_replace("[ìíîïÏÎÍÌ]","i", $s);
$s = ereg_replace("[òóôõöÖÕÔÓ]","o", $s);
$s = ereg_replace("[ÜÛÚÙùúûü]","u", $s);
$s = ereg_replace("[Ç]","c", $s);
return $s;
}
function get_lcs($s1, $s2)
{
//ok, now replace all spaces with nothing
$s1 = strtolower(str_lcsfix($s1));
$s2 = strtolower(str_lcsfix($s2));
$lcs = LCS_Length($s1,$s2); //longest common sub sequence
$ms = (strlen($s1) + strlen($s2)) / 2;
return (($lcs*100)/$ms);
}
you can skip calling str_lcsfix if you don't worry about accentuated characters and things like that or you can add up to it or modify it for faster performance, i think ereg is not the fastest way?
hope this helps.
Georges
10-Oct-2001 03:30
If performance is an issue, you may wish to use the levenshtein() function instead, which has a considerably better complexity of O(str1 * str2).
02-Feb-2000 02:39
$i = similar_text($first_word, $second_word, &$p);
echo("Matched: $i Percentage: $p%");
Don't forget your passing the double as a reference. If you use this and soundex() together you can get a pretty good guess as to how well two words match. Is useful for simple bot-like programs.
