Develop Reference

Create a function that will format a string to always show two significant digits after the decimal, without methods

I'm trying to format a string to show two sig figs after the decimal. I want to do this without using built-in methods like 'Math' and such. I have something where I can do that, but the issue comes when the number is just '4.5' for example, I need to add a 0 onto it, but I'm having trouble. Here's my code so far.
function toFixed(num){
let elem = "";
// let x = "";
for (let i = 0; i < num.length; i++) {
if(num[i - 2] === '.'){ // num[i] happens to be the third 8
break;
}
// elem += num[i];
elem += num[i];
elem += "0";
}
return elem
}
console.log(toFixed("7.88888888")) // "7.88"
console.log(toFixed("77645345.987654")) //77645345.88
console.log(toFixed("1")) // "1.00"
console.log(toFixed("4.5")) // "4.50"
Right now I'm getting 0's after each number it seems.

Without Math:
function toFixed(num) {
let [a, b = ""] = num.split(".");
b += "00";
return `${a}.${b.slice(0, 2)}`;
}
console.log(toFixed("7.88888888")) // "7.88"
console.log(toFixed("77645345.987654")) // "77645345.98"
console.log(toFixed("1")) // "1.00"
console.log(toFixed("4.5")) // "4.50"
Without any builtin methods:
function toFixed(num) {
const len = num.length;
let result = "";
for (let i = 0; i < len; i++) {
const c = num[i];
if (c === ".") {
return `${result}.${num[i + 1] || 0}${num[i + 2] || 0}`;
}
result += c;
}
return `${result}.00`;
}
console.log(toFixed("7.88888888")) // "7.88"
console.log(toFixed("77645345.987654")) // "77645345.98"
console.log(toFixed("1")) // "1.00"
console.log(toFixed("4.5")) // "4.50"

function toFixedNumber(num) {
let elem = +num;
return elem.toFixed(2)
}
console.log(toFixedNumber("7.88888888")) // "7.88"
console.log(toFixedNumber("77645345.987654")) //77645345.88
console.log(toFixedNumber("1")) // "1.00"
console.log(toFixedNumber("4.5")) // "4.50"

Javascript: function to describe array is skipping most of the actual function and returns undefined?

I am trying to write a single javascript function that will take in a list of numbers as arguments and will output the number of odd number, number of negative numbers, will average the numbers, and will output the median. I believe I basically have completed all of the code, but am either confusing my syntax or am incorrectly returning.
Code:
var arrayAnalyze = function(numbers){
var oddNum = []; //Array of odd numbers
var negNum = []; //Array of negative numbers
var numSum = 0; // Sum of all numbers
var avgNum = 0; //Average of all numbers
var midNum = []; //Median number
//Return odd numbers to oddNum array
for (i = 0; i < numbers.length; i++){
if (numbers[i] % 2 !== 0){
oddNum.push(numbers[i]);
}
}
//Return negative numbers to negNum array
for (i = 0; i < numbers.length; i++){
if (Math.abs(numbers[i]) + numbers[i] === 0){
negNum.push(numbers[i]);
}
}
//Return sum of all numbers to numSum variable
for (i = 0; i < numbers.length; i++){
numSum += i;
}
//Return average of numbers to avgNum variable
avgNum = numSum / numbers.length;
//Return median of numbers to midNum array
numbers.sort(function(a,b){return a - b;});
var evenSplit = Math.floor(numbers.length / 2);
if(numbers.length % 2){
midNum = numbers[evenSplit];
}else{
midNum = (numbers[evenSplit - 1] + numbers[evenSplit]) / 2.0; }
midNum.push();
return "Odds: " + oddNum.length, "Negatives: " + negNum.length, "Average: " + avgNum.toFixed(2), "Median: " + midNum[0];
};
console.log(arrayAnalyze(7, -3, 0, 12, 44, -5, 3));
Output:
TypeError: numbers.sort is not a function

There's a number of errors that you'd want to correct - comments inline
var arrayAnalyze = function (numbers) {
var oddNum = []; //Array of odd numbers
var negNum = []; //Array of negative numbers
var numSum = 0; // Sum of all numbers
var avgNum = 0; //Average of all numbers
var midNum = []; //Median number
//Return odd numbers to oddNum array
for (i = 0; i < numbers.length; i++) {
// always check the element at index, i is just the index
if (numbers[i] % 2 !== 0) {
// return exits the currently running function! (not the block)
oddNum.push(numbers[i]);
}
}
//Return negative numbers to negNum array
for (i = 0; i < numbers.length; i++) {
// exclude 0 here
if (Math.abs(numbers[i]) + numbers[i] === 0 && numbers[i]) {
negNum.push(numbers[i]);
}
}
//Return sum of all numbers to numSum variable
for (i = 0; i < numbers.length; i++) {
numSum += numbers[i];
}
//Return average of numbers to avgNum variable
avgNum = numSum / numbers.length;
//Return median of numbers to midNum array
// if you are using a function you need to invoke it to get it's value
midNum.push((function median(numbers) {
// note that this will actually sort the elements of the array you pass in in-place
numbers.sort(function (a, b) { return a - b; });
var evenSplit = Math.floor(numbers.length / 2);
if (numbers.length % 2) {
return numbers[evenSplit];
} else {
return (numbers[evenSplit - 1] + numbers[evenSplit]) / 2.0;
}
})(numbers));
// use + to concatenate the strings, otherwise it just becomes a bunch of comma separated expressions
return "Odds: " + oddNum.length + ",Negatives: " + negNum.length + ",Average: " + avgNum.toFixed(2) + ",Median: " + midNum[0];
};
// an array is passed in using square brackets
console.log(arrayAnalyze([7, -3, 0, 12, 44, -5, 3]));

when a function hits a return statement it will exit out meaning the any code beneath the return will not get executed.

In your odd number sorter you use the %/Modulus on the counter/index rather than numbers[i] to use it on each element of the numbers array parameter. This also needs fixed when you push to the appropriate results array. I have spotted this same concept being done multiple times throughout the function so I would go back and correct that as it would break a couple things.
Also to give you a tip in the right direction in terms of learning return like other users are saying, let's take a look at this part of your code:
for (i = 0; i < numbers.length; i++){
return numSum += i;
}
You do not need to return numSum as your are returning its value later at the end. Just updated the variable you initialized at the beginning by doing the following (also updated in regards to my suggestion above):
for (i = 0; i < numbers.length; i++){
numSum += numbers[i];
}

You cleary are pretty confused about how the return keyword works; I suggest you to check out some documentation here and here.
As an example, you need to change that piece of code
if (numbers % 2 !== 0){
return oddNum.push(numbers);
}else{
return false;
}
into
if (numbers % 2 !== 0){
oddNum.push(numbers);
}
All the others if structures have the same error.

I believe you have a fundamental misunderstand of what return does. MDN has a page on it: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/return#Examples
return interrupts code execution, exits the innermost function and returns a value to the caller.
function myFunction() {
return "foo";
alert("This will never be reached!");
}
alert(myFunction()) // Equivalent to alert("foo").
For example, in your code:
//Return odd numbers to oddNum array
for (i = 0; i < numbers.length; i++){
if (i % 2 !== 0){
return oddNum.push(i); // <- code execution will stop here
}else{
return false; // <- or here, whichever is reached first.
}
}
Which means your loop will never execute for more than one iteration. So when you call
console.log(arrayAnalyze(7, -3, 0, 12, 44, -5, 3));
The first value is odd, so the function will stop at return oddNum.push(i);. And since oddNum.push(i) itself returns nothing (undefined), arrayAnalyze will return undefined too, and the log will be equivalent to
console.log(undefined);
Which is what you are seeing.
In this case the returns are completely unnecessary and the loop should read:
//Return odd numbers to oddNum array
for (i = 0; i < numbers.length; i++){
if (i % 2 !== 0){
oddNum.push(i); // <- now code execution is not interrupted!
}
}
And so on, through the rest of the code.
Also, at the end you declare a function called median:
//Return median of numbers to midNum array
function median(numbers) {
[...]
}
But you never invoke it (calling median(someValue)), which means the code inside it will never be executed either. I haven't checked the code for mathematical correctness, but I believe you should just remove the median declaration and leave its body inside the main arrayAnalyze function.

Your code should look like this :
var arrayAnalyze = function (numbers) {
var oddNum = []; //Array of odd numbers
var negNum = []; //Array of negative numbers
var numSum = 0; // Sum of all numbers
var avgNum = 0; //Average of all numbers
var midNum = []; //Median number
//Return odd numbers to oddNum array
for (i = 0; i < numbers.length; i++) {
if (i % 2 !== 0) {
oddNum.push(numbers[i]);
}
}
//Return negative numbers to negNum array
for (i = 0; i < numbers.length; i++) {
if (Math.abs(numbers[i]) + numbers[i] === 0) {
negNum.push(numbers[i]);
}
}
//Return sum of all numbers to numSum variable
for (numbers[i] = 0; numbers[i] < numbers.length; numbers[i]++) {
numSum += numbers[i];
}
//Return average of numbers to avgNum variable
avgNum = numSum / numbers.length;
//Return median of numbers to midNum array
var newArrayOfNumber = numbers;
newArrayOfNumber.sort();
var evenSplit = Math.floor(newArrayOfNumber.length / 2);
if (newArrayOfNumber.length % 2) {
midNum = newArrayOfNumber[evenSplit];
} else {
midNum = (newArrayOfNumber[evenSplit - 1] + newArrayOfNumber[evenSplit]) / 2.0;
}
return "Odds: " + oddNum.length + ", Negatives: " + negNum.length +", Average: " + avgNum.toFixed(2) +", Median: " + midNum;
};
When you call the function you should pass a array to it, so just add [] to your numbers like this : arrayAnalyze([7, -3, 0, 12, 44, -5, 3])
it should return :"Odds: 3, Negatives: 3, Average: 3.50, Median: 7.5"
When you want to add some number to a array in a for block or a if block, dont use return, just do the operation and if you want to break the for loop, use the line :
Break;
and when you want to access the number in your for loop, you should call the array and not only the i, like this : number[i]
espering that it help you
sorry for my mistakes in writting in english

A while loop to add the digits of a multi-digit number together? (Javascript)

I need to add the digits of a number together (e.g. 21 is 2+1) so that the number is reduced to only one digit (3). I figured out how to do that part.
However,
1) I may need to call the function more than once on the same variable (e.g. 99 is 9+9 = 18, which is still >= 10) and
2) I need to exclude the numbers 11 and 22 from this function's ambit.
Where am I going wrong below?
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
var proc = num.toString().split("");
var total = 0;
for (var i=0; i<proc.length; i++) {
total += +proc[i];
};
};
while(x > 9 && x != 11 && x != 22) {
numberMagic(x);
};
} else {
xResult = x;
};
console.log(xResult);
//repeat while loop for y and z

Here are the problems with your code
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
var proc = num.toString().split("");
var total = 0;
for (var i=0; i<proc.length; i++) {
total += +proc[i]; // indentation want awry
}; // don't need this ; - not a show stopper
// you're not returning anything!!!!
};
while(x > 9 && x != 11 && x != 22) {
numberMagic(x);
}; // ; not needed
// because x never changes, the above while loop would go on forever
} else { // this else has no if
xResult = x; // even if code was right, x remains unchanged
};
console.log(xResult);
Hope that helps in some way
Now - here's a solution that works
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
while (num > 9) {
if (num == 11 || num == 22) {
return num;
}
var proc = num.toString().split("");
num = proc.reduce(function(previousInt, thisValueString) {
return previousInt + parseInt(thisValueString);
}, 0);
}
return num;
}
console.log(numberMagic(x));
console.log(numberMagic(y));
console.log(numberMagic(z));

I'm not sure to understand what you want..
with this function you reduce any number to one single digit
while(num > 9){
if(num == 11 || num == 22) return;
var proc = num.toString();
var sum = 0;
for(var i=0; i<proc.length; i++) {
sum += parseInt(proc[i]);
}
num = sum;
}
is it what you are looking at?

I wrote an example at Jsfiddle that you can turn any given number into a single digit:
Example input: 551
array of [5, 5, 1] - add last 2 digits
array of [5, 6] - add last 2 digits
array of [1, 1] - add last 2 digits
array of [2] - output
Here is the actual code:
var number = 1768;
var newNumber = convertToOneDigit(number);
console.log("New Number: " + newNumber);
function convertToOneDigit(number) {
var stringNumber = number.toString();
var stringNumberArray = stringNumber.split("");
var stringNumberLength = stringNumberArray.length;
var tmp;
var tmp2;
var tmp3;
console.log("Array: " + stringNumberArray);
console.log("Array Length: " + stringNumberLength);
while (stringNumberLength > 1) {
tmp = parseInt(stringNumberArray[stringNumberLength - 1]) + parseInt(stringNumberArray[stringNumberLength - 2]);
stringNumberArray.pop();
stringNumberArray.pop();
tmp2 = tmp.toString();
if (tmp2.length > 1) {
tmp3 = tmp2.split("");
for (var i = 0; i < tmp3.length; i++) {
stringNumberArray.push(tmp3[i]);
}
} else {
stringNumberArray.push(tmp2);
}
stringNumberLength = stringNumberArray.length;
console.log("Array: " + stringNumberArray);
console.log("Array Length: " + stringNumberLength);
}
return stringNumberArray[0];
}

function addDigits(n) {
let str = n.toString().split('');
let len = str.length;
let add,
acc = 0;
for (i=0; i<=len-1; i++) {
acc += Number(str[i]);
}
return acc;
}
console.log( addDigits(123456789) ); //Output: 45
Just make it a While loop, remember a While loops it's just the same as a For loop, only you add the counter variable at the end of the code, the same way you can do with a Do{code}while(condition) Only need to add a counter variable at the end and its gonna be the same. Only that the variable its global to the loop, I mean comes from the outside.
Ej.
let i = 0; //it's global to the loop, ( wider scope )
while (i<=x) {
//Code line;
//Code line;
//Code line;
//Code line;
i++
}
Now this is working with an outside variable and it's NOT recommended.. unless that var its local to a Function.

Please look at the this solution also
var x = 123;
var y = 456;
var z = 789;
var numberMagic = function (num) {
var total = 0;
while (num != 0) {
total += num % 10;
num = parseInt(num / 10);
}
console.log(total);
if (total > 9)
numberMagic(total);
else
return total;
}
//Call first time function
numberMagic(z);

JavaScript Function Arrays

How would I use a function that returns the sum of a given array while getting the sum of the even numbers and sum the odd numbers? I'm not understanding how that is done. Can someone please explain a little more in depth?
Here is my entire code:
function main()
{
var evenNum = 0;
//need a total Even count
var oddNum = 0;
//need a total Odd count
var counter = 1;
var num = 0;
function isOdd(x) {
if ((num % 2) == 0)
{
return false;
}
else
{
return true;
}
}
function isEven(x) {
if ((num % 2) == 0)
{
return false;
}
else
{
return true;
}
}
for (counter = 1; counter <= 100; counter++)
{
num = Math.floor(1 + Math.random() * (100-1));
var total = 0;
for(var j = 0; j < length; j++)
total += a[j];//Array?
console.log(num);
console.log("The count of even number is " + evenNum);
console.log("The count of odd number is " + oddNum);
return 0;
}
main()

If I understand your question correctly, you need a function that returns two values, one for the sum of even numbers and one for the sum of odd numbers. It's not clear if you use even/odd referring to the index of the array or the values in array.
In both cases you can return an object that contains both values:
function sum(array) {
var evenSum = 0;
var oddSum = 0;
...calculate...
var res = {};
res.evenSum = evenSum;
res.oddSum = oddSum;
return res;
}
Hope this will help

Javascript Happy Numbers not working

Here I have a function that should take a number n into the disHappy(n) to check if all
n in [n-0) are happy.
Happy Numbers wikipedia
If I only run happyChecker(n), I can tell that 7 is happy, but disHappy(n) doesn't show it. It is as if it doesn't receive the true. I have used console.log()'s all over the place and happyChecker(n) shows a number that SHOULD return true. When I placed a console.log() above the return true; for if(newNum===1), it showed that it branched into that branch but it just didn't seem to return the true.
function happyChecker(n) {
var arr = [];
var newNum = 0;
//here I split a number into a string then into an array of strings//
num = n.toString().split("");
for (var i = 0; i < num.length; i++) {
arr[i] = parseInt(num[i], 10);
}
//here I square each number then add it to newNum//
for (var i = 0; i < arr.length; i++) {
newNum += Math.pow(arr[i], 2);
}
//here I noticed that all unhappy numbers eventually came into one of these three//
//( and more) numbers, so I chose them to shorten the checking. A temporary solution for sure//
if (newNum === 58 || newNum === 4 || newNum == 37) {
return false;
}
if (newNum === 1) {
return true;
} else {
happyChecker(newNum);
}
}
function disHappy(num) {
for (j = num; j > 0; j--) {
if (happyChecker(j)) {
console.log(j + " is a Happy Number. It's so happy!!!.");
}
}
}

When you recurse, you need to return the value returned:
if (newNum === 1) {
return true;
} else {
return happyChecker(newNum);
}
You also should declare "num" with var.
I'm ordinarily not a "code golfer", but this is a good example of how the (new-ish) iterator utility methods on the Array prototype can clean up code. You can use the .reduce() function to traverse the array of digit characters and do the work of squaring and summing all at once:
var newNum = n.toString()
.split('')
.reduce(function(sum, digit) {
return sum + (+digit * +digit);
}, 0);
The call to .toString() returns a string, then .split('') gives you an array. Then .reduce() starts with an initial sum of 0 and for each element of the array (each digit), it adds to it the square of that digit. (Instead of parseInt() I just used the + unary operator; we know for sure that each string will be a valid number and an integer.)

You need to add return to the happyChecker call.
return happyChecker(newNum);
see:
http://jsfiddle.net/YjgL8/2/

here is my implementation
var getSum = function (n) {
if (!n >= 0) return -1;
var digits = n.toString().split("");
var sum = 0;
for (var i = 0; i < digits.length; i++) {
var digit = parseInt(digits[i], 10);
sum += digit * digit;
}
return sum;
}
/**
* #param {number} n
* #return {boolean}
*/
var isHappy = function(n, visited) {
if (n < 0) return false;
if (n === 1) return true;
if (typeof visited === 'undefined') visited = {};
sum = getSum(n);
if (visited[sum]) return false; // cycle
visited[sum] = true;
return isHappy(sum, visited);
};

Complete Example of finding happy numbers in range of custom number.
function happyNumbers() {
var result = document.getElementById("happy-result")
var inputy = parseInt(document.getElementById("happyValue").value)
result.innerHTML=""
for (i = 1; i < inputy; i++) {
(happy(i, i))
}
}
function happy(value,value2) {
var result = document.getElementById("happy-result")
var lengthNum = value.toString().length;
var resultNumbers = 0
for (var b = 0 ; b < lengthNum; b++) {
resultNumbers = resultNumbers + parseInt(value.toString().charAt(b)) * parseInt(value.toString().charAt(b))
}
if (resultNumbers == 4) {
return false
} else if (resultNumbers == 1) {
result.innerHTML += "<br> happy number " + i
return true
}else{
happy(resultNumbers, value2);
}
}
window.onload=happyNumbers()
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<div class="panel panel-default">
<div class="panel-heading">happy numbers</div>
<div class="panel-body">
<label>Enter the number that you want ot have see happy numbers uo to it</label>
<input id="happyValue" oninput="happyNumbers()" value="100" class="form-control" />
<div id="happy-result"></div>
</div>
</div>