similar_text

(PHP 4, PHP 5)

similar_text — 二つの文字列の間の類似性を計算する

説明

int similar_text ( string $first , string $second [, float &$percent ] )

この関数は、Oliver [1993] に記述されたように二つの文字列の間の類似性を計算します。この実装は Oliver の擬似コードの様にスタックを使用せず、プロセス全体の速度が改善されるかどうかにかかわらず再帰呼び出しを行うことに注意してください。このアルゴリズムの複雑さは、O(N**3) であることにも注意してください。ただし、N は最も長い文字列の長さです。

パラメータ

first: 最初の文字列。
second: 次の文字列。
percent: 3 番目の引数としてリファレンスを渡すことにより、 similar_text() は類似性をパーセントで計算します。

返り値

両方の文字列でマッチした文字の数を返します。

参考

soundex

sha1

Last updated: Fri, 18 Jul 2008

add a note User Contributed Notes
similar_text

romain dot boyer at gmail dot com
08-Aug-2007 05:59


Like levenchtein(), You can do :



(strlen($string2) - similar_text($string,$string2))



to see how much characters have been changed.

Paul
18-Jan-2007 08:50


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.

dmitry dot polushkin at gmail dot com
07-Mar-2006 08:08


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;

}

?>

brad dot fish at gmail dot com
24-Feb-2006 02:30


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
06-Jun-2004 05:01


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
09-May-2004 09:21


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 06:46


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 &quot;".$row['reserved']."&quot;";

        break;

       }

    }

georgesk at hotmail dot com
08-Mar-2002 08:14


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
09-Oct-2001 06:30


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
01-Feb-2000 05:39


$i = similar_text($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.

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soundex

sha1

Last updated: Fri, 18 Jul 2008