Extract Keywords From A Text String With PHP

A common issue I have come across in the past is that I have a CMS system, or an old copy of Wordpress, and I need to create a set of keywords to be used in the meta keywords field. To solve this I put together a simple function that runs through a string and picks out the most commonly used words in that list as an array. This is currently set to be 10, but you can change that quite easily.

The first thing the function defines is a list of "stop" words. This is a list of words that occur quite a bit in English text and would therefore interfere with the outcome of the function. The function also uses a variant of the slug function to remove any odd characters that might be in the text.

function extractCommonWords($string){
      $stopWords = array('i','a','about','an','and','are','as','at','be','by','com','de','en','for','from','how','in','is','it','la','of','on','or','that','the','this','to','was','what','when','where','who','will','with','und','the','www');
   
      $string = preg_replace('/\s\s+/i', '', $string); // replace whitespace
      $string = trim($string); // trim the string
      $string = preg_replace('/[^a-zA-Z0-9 -]/', '', $string); // only take alphanumerical characters, but keep the spaces and dashes too…
      $string = strtolower($string); // make it lowercase
   
      preg_match_all('/\b.*?\b/i', $string, $matchWords);
      $matchWords = $matchWords[0];
      
      foreach ( $matchWords as $key=>$item ) {
          if ( $item == '' || in_array(strtolower($item), $stopWords) || strlen($item) <= 3 ) {
              unset($matchWords[$key]);
          }
      }   
      $wordCountArr = array();
      if ( is_array($matchWords) ) {
          foreach ( $matchWords as $key => $val ) {
              $val = strtolower($val);
              if ( isset($wordCountArr[$val]) ) {
                  $wordCountArr[$val]++;
              } else {
                  $wordCountArr[$val] = 1;
              }
          }
      }
      arsort($wordCountArr);
      $wordCountArr = array_slice($wordCountArr, 0, 10);
      return $wordCountArr;
}

The function returns the 10 most commonly occurring words as an array, with the key as the word and the amount of times it occurs as the value. To extract the words just use the implode() function in conjunction with the array_keys() function. To change the number of words returned just alter the value in the third parameter of the array_slice() function near the return statement, currently set to 10. Here is an example of the function in action.

$text = "This is some text. This is some text. Vending Machines are great.";
$words = extractCommonWords($text);
echo implode(',', array_keys($words));

This produces the following output.

some,text,machines,vending

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Comments

split a command stored in a variable

hello can any one help me with a problem? I have a variable containing a html commando like this:
$str="<a id=home href=http://d-grund.dk/ title=D-Grund.DK target=_blank><img id=home src=http://d_grund.dk/images/homelogo.gif alt=D-Grund.DK border=1 width=166 height=90 /></a>";
and i will split it up so i can change the values and sample it again to work with echo $str; sorry for my bad english. i have searched the entire net and nothing is what i need so i hope anyone can help me. my solution is not good because it bet big and time consuming. and i have find some on the internet that look a like that i need but i cant sample it again without destroying anything. sincerely LAT D-Grund.dk Administrator my email is lat@ofir.dk
philipnorton42's picture

I would love to help but I'm

I would love to help but I'm not sure what you are trying to do. Passing this string through the extractCommonWords() function probably won't produce any meaningful output as the string is just HTML containing a link and an image.

the script doesnt want to work:(:( and it is exactly what i need

<?php function extractCommonWords($string){ $stopWords = array('i','a','about','an','and','are','as','at','be','by','com','de','en','for','from','how','in','is','it','la','of','on','or','that','the','this','to','was','what','when','where','who','will','with','und','the','www'); $string = preg_replace('/ss+/i', '', $string); $string = trim($string); // trim the string $string = preg_replace('/[^a-zA-Z0-9 -]/', '', $string); // only take alphanumerical characters, but keep the spaces and dashes too… $string = strtolower($string); // make it lowercase preg_match_all('/\b.*?\b/i', $string, $matchWords); $matchWords = $matchWords[0]; foreach ( $matchWords as $key=>$item ) { if ( $item == '' || in_array(strtolower($item), $stopWords) || strlen($item) <= 3 ) { unset($matchWords[$key]); } } $wordCountArr = array(); if ( is_array($matchWords) ) { foreach ( $matchWords as $key => $val ) { $val = strtolower($val); if ( isset($wordCountArr[$val]) ) { $wordCountArr[$val]++; } else { $wordCountArr[$val] = 1; } } } arsort($wordCountArr); $wordCountArr = array_slice($wordCountArr, 0, 10); return $wordCountArr; } } $text = "This is some text. This is some text. Vending Machines are great."; $words = extractCommonWords($text); echo implode(',', array_keys($words)); ?>
philipnorton42's picture

Thanks for the input

Thanks for the input, after running the script it was clear that it wouldn't work as there was one too many curly braces in the function. I have removed this in the example so it will run now.

Do you maybe know, what would

Do you maybe know, what would be the problem, .. when I get my keywords with your script.. I always get letter "p" added right before first keyword and right after last keyword.. os the keyword list looks like this: pkeyword1, keyword2...... keyword10p .. and I dont know what causes this.. thanks
philipnorton42's picture

Are you passing it a HTML

Are you passing it a HTML string with p tags at the start and end? Try using strip_tags() first.

Tit for tat

Your approach of stripping punctuation was a heck of a lot better than my method. I went the opposite approach and used a stock snippet of code with several preg_replace patterns, except for WordPress curly smart quotes. This is much more elegant and simple. Thanks for that, and here is how I would count word frequency:
    $wordCountArray = array_count_values( $matchWords );
No need of a foreach loop. The strtolower() call inside both foreach loops is really not needed either since you have already converted $string to lowercase beforehand.

Help calling a certain key of the array

Hey I’ve found your code very useful. Thanks a lot! However I’m finding an issue and that is that I want to divide the number of times a word repeats by the total number of words. The problem I have is that I cannot access each item of the array since the keys change every time I change the text. I was wondering if there is a say around this or if a new dimension can be added to the array so that things like this will work: echo $wordCountArr[2]/$totalwords; Any help will be much appreciated! Thanks in advance! :)

follow up

By the way, I've also tryed this with now luck... $numrepeats = print_r (array_values($words), true); for ($h=0; $h<=5; $h+=1) { $density[$h] = $numrepeats[$h]*2; echo $density[$h]; echo ""; }
philipnorton42's picture

@Guillermo I think to get the

@Guillermo I think to get the effect you are looking for you'll need to modify the function in the following way:
function extractCommonWords($string){
      $stopWords = array('i','a','about','an','and','are','as','at','be','by','com','de','en','for','from','how','in','is','it','la','of','on','or','that','the','this','to','was','what','when','where','who','will','with','und','the','www');
   
      $string = preg_replace('/ss+/i', '', $string);
      $string = trim($string); // trim the string
      $string = preg_replace('/[^a-zA-Z0-9 -]/', '', $string); // only take alphanumerical characters, but keep the spaces and dashes too…
      $string = strtolower($string); // make it lowercase
   
      preg_match_all('/\b.*?\b/i', $string, $matchWords);
      $matchWords = $matchWords[0];

      $totalWords = count($matchWords[0]);

      foreach ( $matchWords as $key=>$item ) {
          if ( $item == '' || in_array(strtolower($item), $stopWords) || strlen($item) <= 3 ) {
              unset($matchWords[$key]);
          }
      }
      $wordCountArr = array();
      if ( is_array($matchWords) ) {
          foreach ( $matchWords as $key => $val ) {
              $val = strtolower($val);
              if ( !isset($wordCountArr[$val]) {
                  $wordCountArr[$val] = array();
              }
              if ( isset($wordCountArr[$val]['count']) ) {
                  $wordCountArr[$val]['count']++;
              } else {
                  $wordCountArr[$val]['count'] = 1;
              }
          }
          arsort($wordCountArr);
          $wordCountArr = array_slice($wordCountArr, 0, 10);
          foreach ( $wordCountArr as $key => $val) {
              $wordCountArr[$val]['bytotal'] = $wordCountArr[$val]['count'] / $totalWords;
          } 
      }
      return $wordCountArr;
}
This is untested code, but it should get you what you need. :)

Hey thanks for your

Hey thanks for your super-fast response! I get it quite to work (not saying the code is not right, I just didn’t figure out how to implement it). But I’ve found a way around the issue. Thank you very much!

Excellent Code Example

Excellent bit of code. I was looking for something to auto-suggest keywords for a content management system. I made the following changes : 1) removed the strtolower call in the foreach loop because the string has already been forced to lower case so all the items will be lower case anyway. if ( $item == '' || in_array($item, $stopWords) || strlen($item) <= 3 ) { 2) Added a $count parameter to the function so that when it is called you can specify how many words to return. function extractCommonWords($string,$count){ ... $wordCountArr = array_slice($wordCountArr, 0, $count); return $wordCountArr; } The hardest thing to get right is the stop word list.

Brill

Brilliant. Thanks for this piece of code. I was after doing something very similar so have used this as a base and made a few tweaks here and there so it fits my needs. 

Removing ss from words

Hi

Great code and is working a treat on my site but I have a problem with words containing 'ss'.  The double ss is being removed so a word such as 'password' is being seen as 'paword' in the keywords.

Regards

Please help me with spanish letters

 

Hi, would you please to let me know how I could include spanish characters too?.

For example I need to print: 'tamaño' instead I get: 'tamao'

Thanks in advance
Tolenca

 

philipnorton42's picture

You need to add those special

You need to add those special characters to the regular expression on line 6 of the examples above. You just need to make sure it takes alphanumeric characters as well as the spanish letters. I think that should work.

instead of$wordCountArr =

instead of

$wordCountArr = array();

      if ( is_array($matchWords) ) {
          foreach ( $matchWords as $key => $val ) {
              $val = strtolower($val);
              if ( !isset($wordCountArr[$val]) {
                  $wordCountArr[$val] = array();
              }
              if ( isset($wordCountArr[$val]['count']) ) {
                  $wordCountArr[$val]['count']++;
              } else {
                  $wordCountArr[$val]['count'] = 1;
              }
          }

I did:

 

$ignoreOccur = array(1,2);
$wordCountArr = array_diff(array_count_values(explode(" ", matchWords)), $ignoreOccur);

To get all assoc array $wordCoundArr of [word] => [occurences], ignoring words that have occurred 1 or 2 (or specified number of) times.
philipnorton42's picture

Interesting take on it. I

Interesting take on it. I like it! :)

Also remove numbers

I changed line 13 to not include 'keywords' that are just numbers using the  is_numeric() function

if ( $item == '' || in_array($item, $stopWords) || strlen($item) &lt;= 3 || is_numeric($item) ) {

 

useful code

Thanks, this is quite useful!

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