Estrarre i keyphrase dal testo (1-4 word ngrams)

Question

Qual è il modo migliore per estrarre i keyphras da un blocco di testo? Sto scrivendo uno strumento per fare l'estrazione di parole chiave: something like this. Ho trovato alcune librerie per Python e Perl per estrarre n-grammi, ma sto scrivendo questo in Nodo quindi ho bisogno di una soluzione JavaScript. Se non ci sono librerie JavaScript esistenti, qualcuno potrebbe spiegare come farlo, quindi posso semplicemente scriverlo da solo?

Answer 1

mi piace l'idea, così ho implementato: vedi qui sotto (commenti descrittivi sono inclusi).
Anteprima: http://fiddle.jshell.net/WsKMx/

/*@author Rob W, created on 16-17 September 2011, on request for Stackoverflow (http://stackoverflow.com/q/7085454/938089) 
* Modified on 17 juli 2012, fixed IE bug by replacing [,] with [null] 
* This script will calculate words. For the simplicity and efficiency, 
* there's only one loop through a block of text. 
* A 100% accuracy requires much more computing power, which is usually unnecessary 
**/ 


var text = "A quick brown fox jumps over the lazy old bartender who said 'Hi!' as a response to the visitor who presumably assaulted the maid's brother, because he didn't pay his debts in time. In time in time does really mean in time. Too late is too early? Nonsense! 'Too late is too early' does not make any sense."; 

var atLeast = 2;  // Show results with at least .. occurrences 
var numWords = 5;  // Show statistics for one to .. words 
var ignoreCase = true; // Case-sensitivity 
var REallowedChars = /[^a-zA-Z'\-]+/g; 
// RE pattern to select valid characters. Invalid characters are replaced with a whitespace 

var i, j, k, textlen, len, s; 
// Prepare key hash 
var keys = [null]; //"keys[0] = null", a word boundary with length zero is empty 
var results = []; 
numWords++; //for human logic, we start counting at 1 instead of 0 
for (i=1; i<=numWords; i++) { 
    keys.push({}); 
} 

// Remove all irrelevant characters 
text = text.replace(REallowedChars, " ").replace(/^\s+/,"").replace(/\s+$/,""); 

// Create a hash 
if (ignoreCase) text = text.toLowerCase(); 
text = text.split(/\s+/); 
for (i=0, textlen=text.length; i<textlen; i++) { 
    s = text[i]; 
    keys[1][s] = (keys[1][s] || 0) + 1; 
    for (j=2; j<=numWords; j++) { 
     if(i+j <= textlen) { 
      s += " " + text[i+j-1]; 
      keys[j][s] = (keys[j][s] || 0) + 1; 
     } else break; 
    } 
} 

// Prepares results for advanced analysis 
for (var k=1; k<=numWords; k++) { 
    results[k] = []; 
    var key = keys[k]; 
    for (var i in key) { 
     if(key[i] >= atLeast) results[k].push({"word":i, "count":key[i]}); 
    } 
} 

// Result parsing 
var outputHTML = []; // Buffer data. This data is used to create a table using `.innerHTML` 

var f_sortAscending = function(x,y) {return y.count - x.count;}; 
for (k=1; k<numWords; k++) { 
    results[k].sort(f_sortAscending);//sorts results 

    // Customize your output. For example: 
    var words = results[k]; 
    if (words.length) outputHTML.push('<td colSpan="3" class="num-words-header">'+k+' word'+(k==1?"":"s")+'</td>'); 
    for (i=0,len=words.length; i<len; i++) { 

     //Characters have been validated. No fear for XSS 
     outputHTML.push("<td>" + words[i].word + "</td><td>" + 
      words[i].count + "</td><td>" + 
      Math.round(words[i].count/textlen*10000)/100 + "%</td>"); 
      // textlen defined at the top 
      // The relative occurence has a precision of 2 digits. 
    } 
} 
outputHTML = '<table id="wordAnalysis"><thead><tr>' + 
       '<td>Phrase</td><td>Count</td><td>Relativity</td></tr>' + 
       '</thead><tbody><tr>' +outputHTML.join("</tr><tr>")+ 
       "</tr></tbody></table>"; 
document.getElementById("RobW-sample").innerHTML = outputHTML; 
/* 
CSS: 
#wordAnalysis td{padding:1px 3px 1px 5px} 
.num-words-header{font-weight:bold;border-top:1px solid #000} 

HTML: 
<div id="#RobW-sample"></div> 
*/ 

Answer 2

non so un tale libreria in JavaScript, ma la logica è di testo

1. divisa in serie
2. quindi selezionare e contare

alternativa

1. suddivisa in serie
2. creare un array secondario
3. attraversa ogni elemento della prima matrice
4. controllo se l'articolo corrente esiste in matrice secondaria
5. se non esiste spinta come chiave
6. altro incremento di valore di un elemento avente una chiave = all'elemento ricercato. HTH

Ivo Stoykov