challenge: word frequency exercise

word frequency

function getTokens(rawString) {
  // NB: `.filter(Boolean)` removes any falsy items from an array
  return rawString.toLowerCase().split(/[ ,!.";:-]+/).filter(Boolean).sort();
}
  
function mostFrequentWord(text) {
  //this is saying we will use a function called mostFrequentWord that will have text as its only parameter
  
  var words = getTokens(text);
  //now we are making a variable called words that will basically be the getTokens function
  //declared above(puts the strings to lowercase and sorts)

  var wordFrequencies = {};
  //now, wordFrequencies is an object, but it is currently empty

  for (var i = 0; i <= words.length; i++) {
    //this is a for loop. we are saying some variable (called i) will start at the 0 spot in an array - it will continue as  
    //long as this is less than or equal to the length of the array - once we do the stuff below in the for loop, add 1

    if (words[i] in wordFrequencies) {
      //this is the beginning of an if statement - it says if the i variable in the words array is in the wordFrequency object...

      wordFrequencies[words[i]]++;
    }
    // (continued from prev line), then add 1 to wordFrequencies (to increase the counter)

    else {
      wordFrequencies[words[i]]=1;
    }
    //if the previous statement is not true, then leave the word frequency to 1
  }

  var currentMaxKey = Object.keys(wordFrequencies)[0];
//this is an Object.key function. this is being used because you need to pair words (keys) to their wordFrequencies (values)

  var currentMaxCount = wordFrequencies[currentMaxKey];
//we created a currentMaxCount variable because eventually, we want this to hold the word with the highest frequency
  
  for (var word in wordFrequencies) {
    //opening another for loop...

    if (wordFrequencies[word] > currentMaxCount) {
//part of the for loop.. if the wordFrequencies var that we made a loop for before is begger than the currentMaxCount just created...

      currentMaxKey = word;
      //currentMaxKey then equals the word (basically, stop there)

      currentMaxCount = wordFrequencies[word];
      //similar to line above
    }
  }
  return currentMaxKey;
  //the answer of the most frequently used word is now in this key, so return/display it
}