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;
}
?>
similar_text
Description
int similar_text ( string first, string second [, float &percent] )This calculates the similarity between two strings as described in Oliver [1993]. Note that this implementation does not use a stack as in Oliver's pseudo code, but recursive calls which may or may not speed up the whole process. Note also that the complexity of this algorithm is O(N**3) where N is the length of the longest string.
By passing a reference as third argument, similar_text() will calculate the similarity in percent for you. It returns the number of matching chars in both strings.
See also levenshtein(), and soundex().
add a note
User Contributed Notes
dmitry dot polushkin at gmail dot com
08-Mar-2006 12:08
08-Mar-2006 12:08
brad dot fish at gmail dot com
25-Feb-2006 06:30
25-Feb-2006 06:30
The link below to the Oliver document appears to be broken. Here is one that works: http://citeseer.ist.psu.edu/oliver93decision.html
louis #at# mulliemedia.com
07-Jun-2004 08:01
07-Jun-2004 08:01
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.
hate at spam dot com dot BR
10-May-2004 12:21
10-May-2004 12:21
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! :)
julius at infoguiden dot no
06-Feb-2003 10:46
06-Feb-2003 10:46
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;
}
}
georgesk at hotmail dot com
09-Mar-2002 12:14
09-Mar-2002 12:14
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("[]","e", $s);
$s = ereg_replace("[]","a", $s);
$s = ereg_replace("[]","i", $s);
$s = ereg_replace("[]","o", $s);
$s = ereg_replace("[]","u", $s);
$s = ereg_replace("[]","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
daniel at reflexionsdesign dot com
10-Oct-2001 09:30
10-Oct-2001 09:30
If performance is an issue, you may wish to use the levenshtein() function instead, which has a considerably better complexity of O(str1 * str2).
mogmios at hushmail dot com
02-Feb-2000 09:39
02-Feb-2000 09:39
$i = similar_text($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.