similar_text

(PHP 4, PHP 5)

similar_textCalculate the similarity between two strings

Description

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

This calculates the similarity between two strings as described in Programming Classics: Implementing the World's Best Algorithms by Oliver (ISBN 0-131-00413-1). 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.

Parameters

first

The first string.

second

The second string.

percent

By passing a reference as third argument, similar_text() will calculate the similarity in percent for you.

Return Values

Returns the number of matching chars in both strings.

See Also

  • levenshtein() - Calculate Levenshtein distance between two strings
  • soundex() - Calculate the soundex key of a string

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User Contributed Notes 12 notes

up
33
SPAM HATER
2 years ago
Hey there,

Be aware when using this function, that the order of passing the strings is very important if you want to calculate the percentage of similarity, in fact, altering the variables will give a very different result, example :

<?php
$var_1
= 'PHP IS GREAT';
$var_2 = 'WITH MYSQL';

similar_text($var_1, $var_2, $percent);

echo
$percent;
// 27.272727272727

similar_text($var_2, $var_1, $percent);

echo
$percent;
// 18.181818181818
?>
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30
daniel dot karbach at localhorst dot tv
3 years ago
Please note that this function calculates a similarity of 0 (zero) for two empty strings.

<?php
similar_text
("", "", $sim);
echo
$sim; // "0"
?>
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6
ryan at derokorian dot com
8 months ago
Note that this function is case sensitive:

<?php

$var1
= 'Hello';
$var2 = 'Hello';
$var3 = 'hello';

echo
similar_text($var1, $var2);  // 5
echo similar_text($var1, $var3);  // 4
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9
julius at infoguiden dot no
11 years ago
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.

<?php
$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;
       }
    }
?>
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5
Paul
7 years ago
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.
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3
romain dot boyer at gmail dot com
7 years ago
Like levenchtein(), You can do :

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

to see how much characters have been changed.
up
3
georgesk at hotmail dot com
12 years ago
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..

<?php
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
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2
hate at spam dot com dot BR
10 years ago
In PHP4+, you don't need to pass the percent variable as reference..
Instead, use this way:

<?php
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! :)
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1
daniel at reflexionsdesign dot com
13 years ago
If performance is an issue, you may wish to use the levenshtein() function instead, which has a considerably better complexity of O(str1 * str2).
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1
louis #at# mulliemedia.com
10 years ago
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.
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0
I_HATE_SPAMMER- PAZ!
28 days ago
Actually similar_text() is not bad...
it works good. But before processing i think is a good way to make a little mod like this

$var_1 = strtoupper("doggy");
$var_2 = strtoupper("Dog");

similar_text($var_1, $var_2, $percent);

echo $percent; // output is 75 but without strtoupper output is 50
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2
mogmios at hushmail dot com
14 years ago
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.

<?php
$i
= similar_text($first_word, $second_word, &$p);
echo(
"Matched: $i  Percentage: $p%");
?>
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