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嗨，我一直在寻找一种在 javascript 中开发简单的词袋类型模型的方法，并花时间查看了一些示例，但是大多数都需要从我所看到的中安装 jnode 或 browserify。我试图简单地阅读文本，将其拆分，并获取文本中最常用的单词，但是我在使用 javascript 的数组对象返回文本值时遇到问题，到目前为止我只能返回编号索引：

function bagOfWords(text){
text=text.toLowerCase(); //make everything lower case
var bag = text.split(" "); //remove blanks

//count duplicates 
var map = bag.reduce(function(prev, cur) {
  prev[cur] = (prev[cur] || 0) + 1;
  return prev;
}, {});


var arr = Object.keys( map ).map(function ( key ) { return map[key]; }); //index based on values to find top 10 possible tags
arr=arr.sort(sortNumber); //sort the numbered array

var top10 = new Array(); //the final array storing the top 10 elements
for (i = arr.length; top10.length < 10; i--) { 
if(top10.length<10){
top10.push(arr[i]);}

}

}

有没有更简单的方法使用 reduce 方法来查找、计算和搜索前 10 个单词，而无需迭代索引并引用原始文本输入（无需创建新的排序数组）？

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4 回答 4

2

我不知道这个问题是否有很好的reduce解决方案，但我想出了一个算法：

  1. 对所有单词进行排序，然后克隆这个数组。
  2. 使用克隆数组上的lastIndexOf()和indexOf()以相反的出现顺序对排序的单词列表进行排序。
  3. filter()删除重复的新数组。
  4. slice()过滤后的数组以将其限制为前 10 个单词。

片段：

function bagOfWords(text) {
  var bag = text.
              toLowerCase().
              split(' ').
              sort(),
      clone = bag.slice();  //needed because sort changes the array in place
          
  return bag.
           sort(function(a, b) { //sort in reverse order of occurrence
    	     return (clone.lastIndexOf(b) - clone.indexOf(b) + 1) -
     	            (clone.lastIndexOf(a) - clone.indexOf(a) + 1);
    	   }).
           filter(function(word, idx) { //remove duplicates
             return bag.indexOf(word) === idx;
           }).
           slice(0, 10);  //first 10 elements
} //bagOfWords

console.log(bagOfWords('four eleven two eleven ten nine one six seven eleven nine ten seven four seven six eleven nine five ten seven six eleven nine seven three five ten eleven six nine two five seven ten eleven nine six three eight eight eleven nine ten eight three eight five eleven eight ten nine four four eight eleven ten five eight six seven eight nine ten ten eleven '));

console.log(bagOfWords('Four score and seven years ago our fathers brought forth on this continent a new nation conceived in Liberty and dedicated to the proposition that all men are created equal Now we are engaged in a great civil war testing whether that nation or any nation so conceived and so dedicated can long endure We are met on a great battle-field of that war We have come to dedicate a portion of that field as a final resting place for those who here gave their lives that that nation might live It is altogether fitting and proper that we should do this But in a larger sense we can not dedicate we can not consecrate we can not hallow this ground The brave men living and dead who struggled here have consecrated it far above our poor power to add or detract The world will little note nor long remember what we say here but it can never forget what they did here It is for us the living rather to be dedicated here to the unfinished work which they who fought here have thus far so nobly advanced It is rather for us to be here dedicated to the great task remaining before us that from these honored dead we take increased devotion to that cause for which they gave the last full measure of devotion that we here highly resolve that these dead shall not have died in vain that this nation under God shall have a new birth of freedom and that government of the people by the people for the people shall not perish from the earth'));

于 2016-06-30T22:56:38.017 回答
2

您可以使用String.prototype.match(),Array.prototype.some()从结果中排除重复的对象，Array.protototype.slice()并使用参数0, 10返回数组中出现最多相同单词的前十个项目。

var text = document.querySelector("div").textContent;

var res = text.match(/[a-z]+/ig).reduce((arr, word) => {
    return !arr.some(w => w.word === word) 
           ? [...arr, {
              word: word,
              len: text.match(new RegExp("\\b(" + word + ")\\b", "g")).length
             }] 
           : arr
}, [])
.sort((a, b) => {
    return b.len - a.len
});

console.log(res.slice(0, 10));
<div>
    Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nullam et ipsum eget purus maximus suscipit. Aliquam fringilla eros at lorem venenatis, et hendrerit neque ultrices. Suspendisse blandit, nulla eu hendrerit mattis, elit nibh blandit nibh, non scelerisque leo tellus placerat est. Phasellus dignissim velit metus. Sed quis urna et nunc hendrerit tempus quis eu neque. Vestibulum placerat massa eget sapien viverra fermentum. Aenean ac feugiat nibh, eu dignissim ligula. In hac habitasse platea dictumst. Nunc ipsum dolor, consectetur at justo eget, venenatis vulputate dui. Nulla facilisi. Suspendisse consequat pellentesque tincidunt. Nam aliquam mi a risus suscipit rutrum.

Donec porta enim at lectus scelerisque, non tristique ex interdum. Nam vehicula consequat feugiat. In dictum metus a porttitor efficitur. Praesent varius elit porta consectetur ornare. Mauris euismod ullamcorper arcu. Vivamus ante enim, mollis eget auctor quis, tristique blandit velit. Aliquam ut erat eu erat vehicula sodales. Vestibulum et lectus at neque sodales congue ut id nibh. Etiam congue ornare felis eget dictum. Donec quis nisl non arcu tincidunt iaculis.

Donec rutrum quam sit amet interdum mattis. Morbi eget fermentum dui. Morbi pulvinar nunc sed viverra sollicitudin. Praesent facilisis, quam ut malesuada lobortis, elit urna luctus nulla, sed condimentum dolor arcu id metus. Donec sit amet tempus massa. Nulla facilisi. Suspendisse egestas sollicitudin tempus. Fusce rutrum vel diam quis accumsan.

Etiam purus arcu, suscipit et fermentum vel, commodo a leo. Vestibulum varius purus felis, fringilla blandit lacus luctus varius. In tempus imperdiet risus ut imperdiet. Ut ut faucibus nunc. Vivamus augue orci, lobortis at enim non, faucibus pharetra est. Pellentesque ante arcu, rhoncus eu lectus nec, ornare molestie lorem. Suspendisse at cursus erat. Vivamus quis lacinia neque. Donec euismod neque eget purus faucibus hendrerit.

Fusce in ante placerat, aliquam mauris et, condimentum ligula. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Mauris hendrerit egestas risus, at consequat metus interdum et. Proin ut tellus quis lorem auctor tempor. Mauris viverra ligula et finibus iaculis. Mauris quis enim a lorem bibendum cursus nec nec enim. Etiam porttitor ligula et erat sagittis vulputate. Fusce ornare mi quis ante faucibus mattis. Aliquam tristique libero sed magna dapibus, vitae sollicitudin lorem malesuada. Praesent dignissim malesuada tellus vitae facilisis. Nullam diam augue, tincidunt ut maximus non, convallis vel felis.
</div>

于 2016-06-30T23:07:07.157 回答
1

这是另一个算法：

function myCounter(bagWords) {  

    // Create an array of bag words from string
    var bagMap = bagWords.toLowerCase().split(' ');

    // Count duplicates
    var bagCount = bagMap.reduce( (countWords, word) => {
        countWords[word] = ++countWords[word] || 1;
        return countWords;
    }, {});

    // Create a sorted array
    var bagResult = []; 
    return bagResult = Object.keys(bagCount).map(function(el, i) {
        bagResult.push(bagCount[el]);
        return { 
            word: el, 
            count: bagCount[el]
        };
    }).sort((a,b) => {      
        return b.count-a.count;
    }).slice(0,10); 

}

var bagWords = "Pat cat hat Tat bat rat pat Cat Hat tat Bat Rat test pat pat bat bat tap tac eat bat Ope ope Asd Eat dsa";
console.log(myCounter(bagWords));

也许它可以帮助某人。

于 2016-09-09T12:18:20.690 回答
1

是否有必须使用 Array.prototype.reduce() 的要求？此方法用于将整个元素数组减少为一个值，这听起来不符合您的用例。如果您想简单地计算单词的出现次数，我喜欢使用字典。

function bagOfWords(text, topCnt) {
  text= text.toLowerCase(); //make everything lower case
  var bag = text.split(" "); //remove blanks
  //Remove "." and possibly other punctuation?

  //build the word dictionary;
  var wordDict = {};
  for(idx in bag) {
    if(bag[idx] in wordDict) {
      wordDict[bag[idx]]++;
    }
    else {
      wordDict[bag[idx]] = 1;
    }
  }
  //get the duplicate free array
  var dupFree = [];
  for(word in wordDict) {
    dupFree.push(word);
  }
  //find the top topCnt;
  //Custom sort method to sort the array based on the dict we created.
  var sorted = dupFree.sort(function(a, b) {
    if (wordDict[a] > wordDict[b]) {
      return -1;
    }
    if (wordDict[a] < wordDict[b]) {
      return 1;
    }
    return 0;
  });
  
  //Now we can just return back the spliced array.  
  //NOTE - if there is a tie, it would not return all the ties.
  //  For instance, if there were twenty words with each having the same occurance, this would not return back all 20 of them.  To do that, you would need another loop.
  return sorted.slice(0,topCnt);
}

    var lorem = "Lorem ipsum dolor sit amet consectetur adipiscing elit Duis gravida, lectus vel semper porttitor nulla nulla semper tortor et maximus quam orci a nibh Duis vel aliquet est Aliquam at elit libero Duis molestie nisi et lectus fringilla vulputate Integer in faucibus dolor Vivamus leo magna, interdum sit amet arcu et vulputate aliquam elit Pellentesque vel imperdiet nisi maximus malesuada eros Aenean sit amet turpis lorem Pellentesque in scelerisque ante Nunc sed dignissim ex Quisque neque risus feugiat a felis vitae blandit tristique mauris Etiam pharetra eleifend felis ac cursus Pellentesque ac magna nec lectus interdum lacinia Fusce imperdiet libero accumsan dolor consectetur, sed finibus justo ornare. Vivamus vehicula ornare metus quis fermentum sapien ullamcorper non Cras non odio interdum facilisis elit sit amet facilisis risus";
 console.log(bagOfWords(lorem,10));

正如我在评论中提到的，肯定可以进行一些改进。这至少应该让你开始。这里的神奇之处在于使用字典删除重复项并计算出现次数，然后使用自定义排序函数按您想要的顺序排列数组。

查看 MDN 了解您的所有 javascript 功能需求。该网站是一个了不起的资源。 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

于 2016-06-30T22:52:54.790 回答