Word frequency for javascript key / value array

Question

I am trying to implement a piece of javascript code to parse a word / frequency on a given line. My goal is to return the array as follows:

[{text: firstword, size:3 },{text:secondword , size:5 },{text: nword, size: 1},...]

I implemented the following code, but I ran out of memory, so I don’t know if it is good or not.

function wordFrequency(txt){
    var wordArray = txt.split(/[ .?!,*'"]/);
    var newArray = [];
    $.each(wordArray, function (ix, word) {
        if (newArray.length >= 1){
            newArray.some(function (w){
                if (w.text === word){
                    w.size++;
                } else {
                    newArray.push({text: word, size: 1});
                }
            });
        } else {
            newArray.push({text: word, size: 1});
        }
    });
    return newArray;
}

Answer 1

Array.prototype.some expects this callback to return true or false and return true as soon as your callback returns true for this element, otherwise it returns false.

some , , , . , some .

, , , newArray , , , .

, newArray :

[{word:"test"},{word:"another"},{word:"one"},{word:"more"}]

even :

[{word:"test"},{word:"another"},{word:"one"},{word:"more"},{word:"even"},{word:"even"},{word:"even"},{word:"even"}]

Array.prototype.filter , , , $.each Array.prototype.forEach:

function wordFrequency(txt){
  var wordArray = txt.split(/[ .?!,*'"]/);
  var newArray = [], wordObj;
  wordArray.forEach(function (word) {
    wordObj = newArray.filter(function (w){
      return w.text == word;
    });
    if (wordObj.length) {
      wordObj[0].size += 1;
    } else {
      newArray.push({text: word, size: 1});
    }
  });
  return newArray;
}
document.write(JSON.stringify(wordFrequency("count everything, count all the words, count all the words!").sort(function(a,b){return a.size<b.size})).split("},").join("}<br/>"));

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Answer 2

. words :

var freq = words.reduce(function(p, c) {
    p[c] = (p[c] || 0) + 1;
    return p;
}, {});

:

var array = Object.keys(freq).map(function(key) {
   return { text: key, size: freq[key] };
});

Answer 3

, , -. , some each, newArray , .

JavaScript. , , .

:

function wordFrequency(txt){
    var wordArray = txt.split(/[ .?!,*'"]/);
    var map = {};
    $.each(wordArray, function(ix, word) {
      // skip empty results
      if (!word.length) {
        return;
      }
      // add word to map
      if (!map[word]) {
        map[word] = 0;
      } 
      map[word]++;
    });
    return map;
}

:

var text = "hello!world*hello foo  'bar'foo";
var result = wordFrequency(text);

// iterate over results
Object.keys(result).forEach(function(w) {
  console.log(w + ": " + result[w]);
});

// or use for...in
for (var w in result) {
  console.log(w + ": " + result[w]);
}

, :

var mappedResult = Object.keys(result).map(function(w) {
  return { text: w, size: result[w] };
});
console.log(mappedResult);

, forEach jQuery $.each, , Object.keys.

JSBin.

Answer 4

, . Array.prototype , . ... ( ).

function wordFrequency(txt) {
    var words = txt.split(/[ \.\?!,\*'"]+/),
        seen = [];
    for (var i = 0; i < words.length; i++) {
        var w = words[i],
            found = false;
        for (var j = 0; j < seen.length; j++) {
            if (w === seen[j].text) {
                seen[j].size++;
                found = true;
                break;
            }
        }
        if (!found) seen.push( { text: w, size: 1 } );
    }
    return seen;
}

( , for , , for . , , , .)

Array.prototype.forEach() Array.prototype.indexOf(), . , . ( , Array.prototype.findIndex(), .)

function wordFrequency2(txt) {
    var words = txt.split(/[ \.\?!,\*'"]+/),
        seen = [],
        freq = [];
    // get frequencies
    words.forEach(function (w) {
        var idx = seen.indexOf(w);
        if (idx >= 0) {
            freq[idx]++;
        }
        else {
            seen.push(w);
            freq.push(1);
        }
    });
    // produce the results array
    var r = [];
    seen.forEach(function (w, idx) {
        r.push( { text: w, size: freq[idx] } );
    });
    return r;
}

, , , ...

Answer 5

var words = (function(){

var sWords = document.body.innerText.toLowerCase().trim().replace(/[,;.]/g,'').split(/[\s\/]+/g).sort();
var iWordsCount = sWords.length; // count w/ duplicates

// array of words to ignore
var ignore = ['and','the','to','a','of','for','as','i','with','it','is','on','that','this','can','in','be','has','if'];
ignore = (function(){
var o = {}; // object prop checking > in array checking
var iCount = ignore.length;
for (var i=0;i<iCount;i++){
o[ignore[i]] = true;
}
return o;
}());

var counts = {}; // object for math
for (var i=0; i<iWordsCount; i++) {
var sWord = sWords[i];
if (!ignore[sWord]) {
counts[sWord] = counts[sWord] || 0;
counts[sWord]++;
}
}

var arr = []; // an array of objects to return
for (sWord in counts) {
arr.push({
text: sWord,
frequency: counts[sWord]
});
}

// sort array by descending frequency | http://stackoverflow.com/a/8837505
return arr.sort(function(a,b){
return (a.frequency > b.frequency) ? -1 : ((a.frequency < b.frequency) ? 1 : 0);
});

}());

(function(){
var iWordsCount = words.length; // count w/o duplicates
for (var i=0; i<iWordsCount; i++) {
var word = words[i];
console.log(word.frequency, word.text);
}
}());