2011-07-03 90 views

回答

18

您应该通过的话字符串分割成词,然后循环并且将每一个计数器:

var wordCounts = { }; 
var words = str.split(/\b/); 

for(var i = 0; i < words.length; i++) 
    wordCounts["_" + words[i]] = (wordCounts["_" + words[i]] || 0) + 1; 

"_" +允许它来处理像constructor词已经是对象的属性。

您可能希望写入words[i].toLowerCase()以便不区分大小写。

+1

只是出于好奇 - 你有这个片段铺设在某处,或者你想出解决方案只是为了这个答案?无论哪种方式,这真棒。 :) –

+0

@ajax:我当场创建它。谢谢! – SLaks

+0

嘿,非常感谢,我只是想知道,你能解释一下/ \ b /参数吗?这是一个正则表达式是吗? –

0

从未来开始,再次询问这个问题,但是我对解决方案提前开始并将其标记为已回答。无论如何,这是SLak答案的补充。

function nthMostCommon(string, ammount) { 
    var wordsArray = string.split(/\s/); 
    var wordOccurrences = {} 
    for (var i = 0; i < wordsArray.length; i++) { 
     wordOccurrences['_'+wordsArray[i]] = (wordOccurrences['_'+wordsArray[i]] || 0) + 1; 
    } 
    var result = Object.keys(wordOccurrences).reduce(function(acc, currentKey) { 
     /* you may want to include a binary search here */ 
     for (var i = 0; i < ammount; i++) { 
      if (!acc[i]) { 
       acc[i] = { word: currentKey.slice(1, currentKey.length), occurences: wordOccurrences[currentKey] }; 
       break; 
      } else if (acc[i].occurences < wordOccurrences[currentKey]) { 
       acc.splice(i, 0, { word: currentKey.slice(1, currentKey.length), occurences: wordOccurrences[currentKey] }); 
       if (acc.length > ammount) 
        acc.pop(); 
       break; 
      } 
     } 
     return acc; 
    }, []); 
    return result; 
} 
相关问题