Frequent Word Challenge question answer

challengequestion.js

  Challenge Question Answer
function getTokens(rawString) {
  // NB: `.filter(Boolean)` removes any falsy items from an array
  return rawString.toLowerCase().split(/[ ,!.";:-]+/).filter(Boolean).sort();
}
/returning a lowercase string which will be 
/.split = split takes this expression and splits array based on character in 
  parenthesis 
if character in parenthesis is not in original array, then array will return
original array 
/[regular expression]/
/.filter(function) = scan the array, call function on each element in array and remove element where function returns false
/.sort = sorts arrays in place and returns elements in alphabetical order

//mostFrequentWord is an object--first object 
//the object has a for loop loops over words(array) to build another object
//when looping over array, if word is already in object, increment word by 1 
//if its not in the object it will add it to the obj

// 2nd object keeping in count of frequency of words (currentMaxCount)
//loops over frequency(mostFrequentWord) object to figure out which one of words appears  the mos t frequently 
//returns word that appears the most (currentMaxKey) 


function mostFrequentWord(text) {
  var words = getTokens(text);//array that has been sorted
 // words= array of strings
  var wordFrequencies = {};//empty object
  for (var i = 0; i <= words.length; i++) {//for loop
    if (words[i] in wordFrequencies) { //array in empty object?
      wordFrequencies[words[i]]++;
      **//string is indexing in the ith position; will increase by 1 a certain 
      //amount of times depending on array length. 
      // 
    }
    else {
      wordFrequencies[words[i]]=1;
      //words[i] is in the ith position; set the value associated with the key to 1 
    }
  }
  var currentMaxKey = Object.keys(wordFrequencies)[0];
  **//currentMaxKey = all of keys of object wordFrequencies has value of 0 
  //currentMaxKey becomes a new object? 
  var currentMaxCount = wordFrequencies[currentMaxKey];
  **//currentMaxCount = adding currentMaxKey object to wordFrequencies object 
  //using bracket notation ; create new object currentMaxCount to hold all 
  //key/value pairs from previous 2 objects 
  
  for (var word in wordFrequencies) 
    if (wordFrequencies[word] > currentMaxCount) {
      //if object[with newly added array] is greater than currentMaxCount,
      currentMaxKey = word;
            //currentMaxKey is equal to word array
      currentMaxCount = wordFrequencies[word];
      //object is equal to object[newly added array]
    }
  }
  return currentMaxKey;
  **//return new object with updated keys/values 
}