Drills Answers

drilsanswers.js

EXAMPLE PROBLEM W/MICHAEL
//var number_list = [1,3,6,6,7,8,2]

function problem(number_list){
//var number_list = [4,6,10,12]
var sum_counter = 0

for(var i = 0; i <number_list.length; i++) {
  var value = number_list[i]
  sum_counter = value + sum_counter
  
}
  return sum_counter;
}

-----------------------------------------------------------

1.)// Given a string, adds "-key" to it
// addKey('dog') => 'dog-key'
function addKey(string) {
  return string + "-key";
}

-------------------------------------------------------------------------
2.)// Changes the object to remove oldKey and add newKey with previous oldKey's value
// changeObjectKey({ dogs: 3, cats: 4 }, 'cats', 'lions') => { dogs: 3, lions: 4 }
function changeObjectKey(object, oldKey, newKey) {
  object[newKey] = object[oldKey]
  delete object[oldKey]
  return object;
}
--------------------------------------------------------
3.)// Add '-key' to a given key of the object
// addKeyToObjectKey({ dogs: 3, cats: 4 }, 'cats') => { dogs: 3, cats-key: 4 }
function addKeyToObjectKey(object, target_key) {
 return changeObjectKey(object,target_key,addKey(target_key))
}
function addKey(string) {
  return string + "-key";
}
function changeObjectKey(object, oldKey, newKey) {
  object[newKey] = object[oldKey]
  delete object[oldKey]
  return object;
}
------------------------------------------------------
4.)// Returns the number of keys present on the object
// howManyKeys({ dogs: 1 }) => 1
// howManyKeys({ answerToTheUniverse: 42, everythingElse: 42 }) => 2
function howManyKeys(object) {
  var keyList = Object.keys(object)
  return keyList.length
}
------------------------------------------------------------
5.)// Increases the counter for every value in object
// increaseMultipleCounters({ answerToTheUniverse: 42, everythingElse: 42 }) => { answerToTheUniverse: 43, everythingElse: 43 }
function increaseMultipleCounters(object) {
   var increaseResult = Object.keys(object)
   for(var i = 0; i < increaseResult.length; i ++){
    value = increaseResult[i]
    object[value] = object[value] + 1 
   }
    
    return object;
}
--------------------------------------------------------
6.) //addKeyToMultipleObjectKeys({ dogs: 3, cats: 4 }) => { dogs-key: 3, cats-key: 4 }
function addKeyToMultipleObjectKeys(object) {
  var key_list = Object.keys(object)
  for(var i = 0; i < key_list.length; i ++){
    key = key_list[i]
    changeObjectKey(object,key,addKey(key))
    
  }
  
  return object;
}
function addKey(string){
  return string + "-key";
}
function changeObjectKey(object,oldKey,newKey) {
  object[newKey] = object[oldKey]
  delete object[oldKey]
  return object;
}
-----------------------------------------------
7.)
//CORRECT ANSWER
//1 is odd
//2 is even
function increaseMultipleCounters(object) {
    var listOfKeys = Object.keys(object)
   for(var i = 0; i < listOfKeys.length; i ++){
    key = listOfKeys[i]
    if (object[key]%2 == 0){
       object[key] = object[key] + 2;
    }  
     if (object[key]%2==1) {
      object[key] = object[key]+1;
    }
      
  } 
return object;
} 

console.log(increaseMultipleCounters({dogs:2,cats:3}));
-------------------------------------------------------------
//SOLVE WITH SWITCH IN ANSWERS
//1=even
//2=odd
function increaseMultipleCounters(object) {
    var listOfKeys = Object.keys(object)
   for(var i = 0; i < listOfKeys.length; i ++){
    key = listOfKeys[i]
    if (object[key]%2 == 0){
       object[key] = object[key] + 1;
    }  
     if (object[key]%2==1) {
      object[key] = object[key]+2;
    }
      
  }   
return object;
} 
--------------------------------------
8.)//function objectFromPairs(pairs){
//} 

//Option A: .forEach method
//var buddies = [['hugs',4],['sun', 6]];
function objectFromPairs(pairs){
   var object= {};
   pairs.forEach(function(element){
                   object[element[0]]=element[1];
                  
      });
          return object;
 }

// option B: for loop
function objectPairs(pairs){
  var object = {};
  for(i = 0; i <pairs.length; i ++){
    element=pairs[i]
    object[element[0]]= element[1];
    
  }
   return object;
}


console.log("objectPairs([['hugs', 4], ['sun',8],[777,6668]])")
console.log(objectPairs([['hugs', 4], ['sun',8],[777,6668]]));

------------------------------------------

9.)// Count how many times a given word appears on a list
// singleWordCounter(['apple', 'apple', 'dog', 'cat', 'orange', 'apple'], 'dog') => 1
// singleWordCounter(['apple', 'apple', 'dog', 'cat', 'orange', 'apple'], 'apple') => 3
var words = ['an','apple','went','shopping','an','cookie'];
function singleWordCounter(list,target_word){
  var count = 0 ;
  for(var i = 0; i<list.length; i ++){
    if(list[i]===target_word){
      count ++ ; 
    }
  }
   return count; 
}
 singleWordCounter(words,'an');
 singleWordCounter(words,'apple');
----------------------------------------------------------------------------------------
  
10.) //Same as before, but returns an object with{count:number_of_occurrences}
function singleWordCounterAsObject(list,target_word){
 //var object={};
 //object['count']= 0;
  var object = { 'count' : 0 }
 for(var i = 0; i<list.length; i ++){
 if(list[i]===target_word){
      object['count'] ++ ; 
    }
}
   return object; 
 }

 console.log("singleWordCounterAO:")
 console.log(singleWordCounterAsObject(['an','kiwi','went','shopping','an','cookie'], 'cat'))
 console.log(singleWordCounterAsObject(words,'an'));
 console.log(singleWordCounterAsObject(words,'kiwi'));
-------------------------------------------------------------------------
11.) //Same as before, but returns an object with {target_word:number_of_occurrences}
function singleWordCounterAsKeyObject(list, target_word){
  var object= {};
  object[target_word] = 0;
  for(var i = 0;i <list.length; i ++){
    if(list[i]==target_word){
      object[target_word] ++;       
    }
  }

  if(object[target_word] > 1 ){
      object[target_word + "s"] = object[target_word]
      delete object[target_word]
      }

  return object;
}
  console.log("singleWordCounterAkO:")
 console.log(singleWordCounterAsKeyObject(['an','kiwi','went','shopping','an','cookie'], 'cat'))
 console.log(singleWordCounterAsKeyObject(words,'an'));
 console.log(singleWordCounterAsKeyObject(words,'kiwi'));

------------------------------------------------------------------
12.) //Same as before, but add an 's' to the key if the count is greater than 1 
function pluralSingleWordCounterAsKeyObject(list,target_word){
var object= {};
 var mostFrequentWords = Object.keys(object)
object[target_word]= 0 ;
 for(var i = 0; i <list.length; i ++){
   if(list[i]===target_word){
       return target_word + "-s";
     } if(list[i])
  }
  
 }
  //STILL INCOMPLETE? 
----------------------------------------------------------------------
 //HARDER DRILLS
  
 13.) //Replaces every value with "even" if it is even, and with "odd" otherwise
 function increaseMultipleCounters(object) {
   var listOfKeys = Object.keys(object)
  for(var i = 0; i < listOfKeys.length; i ++){
   key = listOfKeys[i]
  if (object[key]%2 == 0){
      object[key] = "even";
    }  
     if (object[key]%2==1) {
      object[key] = "odd";
   }
      
   } 
return object;
 } 

console.log(increaseMultipleCounters({dogs:2,cats:3}));
 --------------------------------------------------------------------------
   14.) // Returns the sum for every value from object
// sumValues({ dogs: 1, cats: 3, lions: 4 }) => 8
var words = ['an','apple','went','shopping','an','cookie'];
function sumValues(object){
  var count = 0;
  for(var i = 0;i<object.length;i ++){
    if(object[i])= target_word) {
      count = count + [target_word];
    }
  }
  return count; 
}
sumvalues(words,'an');
   
   
   
   ---------------------------------------------------------------------
TUTORING W/ MICHEAL 
/ Count how many times a given word appears on a list
// singleWordCounter(['apple', 'apple', 'dog', 'cat', 'orange', 'apple'], 'dog') => 1
// singleWordCounter(['apple', 'apple', 'dog', 'cat', 'orange', 'apple'], 'apple') => 3
 var words = ['an','apple','went','shopping','an','cookie'];
 function singleWordCounter(list,target_word){
   var count = 0 ;
   for(var ixx = 0; ixx<list.length; ixx ++){
     if(list[ixx]===target_word){
       count ++ ; 
     }
   }
    return count; 
 }
count = count + 1
i = i+1
count + object[target_word]

 console.log("singleWordCounter:")
 console.log(singleWordCounter(['an','apple','went','shopping','an','cookie'], 'cat'))
 console.log(singleWordCounter(words,'an'));
 console.log(singleWordCounter(words,'apple'));

// -----------------------------------------------------------------------------------
function singleWordCounterAsObject(list,target_word){
 //var object={};
 //object['count']= 0;
  var object = { 'count' : 0 }
 for(var i = 0; i<list.length; i ++){
 if(list[i]===target_word){
      object['count'] ++ ; 
    }
}
   return object; 
 }

 console.log("singleWordCounterAO:")
 console.log(singleWordCounterAsObject(['an','kiwi','went','shopping','an','cookie'], 'cat'))
 console.log(singleWordCounterAsObject(words,'an'));
 console.log(singleWordCounterAsObject(words,'kiwi'));
// ----------------------------------------------------------------------------
//same, but returns an object with {target_word:number_of_occurrences}
function singleWordCounterAsKeyObject(list, target_word){
  var object= {};
  object[target_word] = 0;
  for(var i = 0;i <list.length; i ++){
    if(list[i]==target_word){
      object[target_word] ++;       
    }
  }

  if(object[target_word] > 1 ){
      object[target_word + "s"] = object[target_word]
      delete object[target_word]
      }

  return object;
}
  console.log("singleWordCounterAkO:")
 console.log(singleWordCounterAsKeyObject(['an','kiwi','went','shopping','an','cookie'], 'cat'))
 console.log(singleWordCounterAsKeyObject(words,'an'));
 console.log(singleWordCounterAsKeyObject(words,'kiwi'));

// //---------------------------------------------------
// // 1/3 solution: function addKey(string) {
// //   return string + "-key";
// // }
// // if statement
// // for loop
// //if keys in object appear once, print key/value
// //if elements in array appear more than once, make plural, by adding -s, and print newkey
// function pluralSingleWordCounterAsKeyObject(list,target_word){
//   var object= {};
//   var mostFrequentWords = Object.keys(object)
//   //object[target_word]= 0 ;
//   for(var i = 0; i <list.length; i ++){
//     if(list[i]===target_word){
//       return target_word + "-s";
//     } if(list[i])
//   }
  
// }
-----------------------------------------------
  WHAT SOLUTION IS THIS FOR:=??
    //function objectFromPairs(pairs){
//} 
//Option A: 
//var buddies = [['hugs',4],['sun', 6]];
function objectFromPairs(pairs){
   var object= {};
   pairs.forEach(function(element){
                   object[element[0]]=element[1];
                  
      });
          return object;
 }

//for loop
function objectPairs(pairs){
  var object = {};
  for(i = 0; i <pairs.length; i ++){
    element=pairs[i]
    object[element[0]]= element[1];
    
  }
   return object;
}


console.log("objectPairs([['hugs', 4], ['sun',8],[777,6668]])")
console.log(objectPairs([['hugs', 4], ['sun',8],[777,6668]]));
  
  
  
  HW--2/8/17
  var obj1 = {
  name: 'john',
  age: 32
}

var obj2 = {
  gender: 'male',
  happy: true
}


function mergeObjects(obj1,obj2){
  Object.keys(obj1).forEach(function(key)){
   obj3[key] = obj1[key]
  }
   Object.keys(obj2).forEach(function(key)){
    obj3[key] = obj1[key] + obj2[key]                        
   }
  
  
}
//can use Object.values for other problems

// key[value] 
//   object[key]
//   obj3=object[key]
  
                             
 POSSIBLE ANSWER?
                             14.) // Returns the sum for every value from object
// sumValues({ dogs: 1, cats: 3, lions: 4 }) => 8
var words = ['an','apple','went','shopping','an','cookie'];
function sumValues(object){
  var count = 0;
  for(var i = 0;i<object.length;i ++){
    if(object[i])= target_word) {
      count = count + [target_word];
    }
  }
  return count; 
}
sumvalues(words,'an');


Object.values()
  
  ----------------------------------------
  NOTES WITH MICHAEL 2-10-2017
 //.split method example
  "xx;yy;zz.q".split(/;/)
return ["xx","yy","zz.q"]

"xx;yy;zz.q".split(/;+/)
return ["xx","yy","zz.q"]

"xx;;yy;zz.q".split(/;/)
["xx", "","yy","zz.q"]

"xx;;yy;zz.q".split(/;+/)
["xx", "yy", "zz.q"]

Regular expression special characters
+ = 1 or more
- //does not exist;regular character
* = 0 or more
? = 0 or 1 


.split method example 

"x,y,z".split(':')
["x,y,z"]--returns array exactly like it is 

"x:y,z".split(":")---return 
["x"],["y","z"]//incorrect
["x","y,z"]
["x","y","z"]//incorrect
["xy","z"]//incorrect

"x:y,z".split(",")---return 
["x:y","z"]


"xx;yy;zz.q".split(";")
["xx","yy","zz.q"]


"xx;;yy;zz.q".split(";")
["xx","","yy","zz.q"]


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 

---------
  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 
}

//OBJECT DRILLS 2 SOLUTIONS

//1.) makeStudentsReport
function makeStudentsReport(data) {//function name w/ argument
  var answers = [];//create an empty array
  for (var i=0; i<data.length; i++) {//for loop that measures length
    var reports = data[i];//create another variable that has the argument
    //inputting the for loop variable?
    answers.length(reports.name  + reports.grade);//ask the computer to 
    //print the length of the answers
  }
  return answers;//returning the array with proper information 
}

console.log['Sarah', 'C'];
console.log['David', 'A'];

//2.) enroll in Summer School
//factory function solution option 
function enrollInSummerSchool (students) {
  var student1= {
    name:'sally',
    status: 'new student',
    course: 'history'
  },
      var student2 = {
        name:'david',
        status: 'withdrawn',
        course: 'math'
      }
};

function SummerSchool (students) {
  return {
    name: 'sally',
    status: 'in summer school',
    course: 'history'
  };
  return {
    name: 'david',
    status: 'in summer school',
    course: 'math'
  }
};

//3.) Find by id problem



//4.) validateKeys problem
//function called validateKeys
//function validateKeys takes two arguments, object & expectedKeys
//argument object is an object to validate keys 
//argument expectedKeys is an array of keys found in object
//return true if object has all of the keys from 
// expectedKeys AND no additional keys
// return false if 1 or more key is missing from object OR if object 
// has extra keys not in expectedKeys

//SOLUTION SHOULD RETURN BOOLEAN VALUE

//QUESTION:when a problem mentions 'if' does that mean you HAVE to 
//use an if, if/else, or else/if statement? 


var expectedKeys = ['name','age','eyeColor','hometown'];
function validateKeys(object,expectedKeys) {
  if (object1 == expectedKeys){
    return true;
  }else(object1.keys > expectedKeys){
    return false;
  } if (object2.keys < expectedKeys){
    return false;
  }else(object2.keys > expectedKeys){
    return false;
  }
  
}


//objects to test when ready to run problem
var object1 = {
  name: 'sally',
  age: '34',
  eyeColor: 'brown',
  hometown: 'miami'
}

var object2 = {
  name: 'dave',
  age: '12',
  hometown: 'minneapolis'
}


// //possible if statements to solve problem
// if (Object.keys === expectedKeys){
//   return true
// } if (expectedKeys){
//   return true
// }else (Object.keys <(less than) expectedKeys){
//   return false
// } else (expectedKeys >(more than) Object.Keys){
//   return false 
// }
 
  
  
 //maybe this too? ---> var i = 0 ; i< expectedKeys.length; i ++ )