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Is there a simple way to split a number into an array of digits without converting it to a string and back?

I was working on a Javascript exercise that required a number to be converted into an array of single digits and the only way I can seem to achieve this is by converting the number into a string and then converting it back to a number.
let numbers = 12345;
Array.from(numbers.toString(10), Number) // [1, 2, 3, 4, 5]
Basically, I'm wondering if this is the best way to achieve a split like this on a number or if there is a more efficient method that doesn't require such a conversion.

You can always get the smallest digit with n % 10. You can remove this digit with subtraction and division by 10 (or divide and floor). This makes for a pretty simple loop:
function digits(numbers){
if (numbers == 0) return [numbers]
let res = []
while (numbers){
let n = numbers % 10
res.push(n)
numbers = (numbers - n) / 10
}
return res.reverse()
}
console.log(digits(1279020))
This takes the numbers in reverse order so you either have to unshift the results on to the array or push and reverse at the end.
One of the nice things about this, is that you can find the digits of different bases by swapping out 10 for a the base of your choice:
function digits(numbers, base){
if (numbers == 0) return [numbers]
let res = []
while (numbers){
let n = numbers % base
res.push(n)
numbers = (numbers - n) / base
}
return res.reverse()
}
// binary
console.log(digits(20509, 2).join(''))
console.log((20509).toString(2))
// octal
console.log(digits(20509, 8).join(''))
console.log((20509).toString(8))
Although once your base is larger than 10 you will have to map those digits to the appropriate letters.

One approach would be to iterate through the number of digits and calculate the difference of each modulo by base, and then populate the output list from the result of each iteration.
A quick way to identify the number of digits in your base 10 input would be the following:
Math.floor(Math.log(input) / Math.LN10 + 1) // 5 for input of 12349
Next, iterate through this range and for each iteration, calculate the base of the current and previous iterations, and perform module of the input against these. The digit for the current iteration is then derived from the difference of the modulo calculations like this:
function arrayFromInput(input) {
const output = [];
for (let i = 0; i < Math.floor(Math.log(input) / Math.LN10 + 1); i++) {
const lastBase = Math.pow(10, i);
const nextBase = Math.pow(10, i + 1);
const lastMod = input % lastBase;
const nextMod = input % nextBase;
const digit = (nextMod - lastMod) / lastBase;
output.unshift(digit);
}
return output;
}
console.log(arrayFromInput(12345), '= [1,2,3,4,5]');
console.log(arrayFromInput(12), '= [1,2]');
console.log(arrayFromInput(120), '= [1,2 0]');
console.log(arrayFromInput(9), '= [9]');
console.log(arrayFromInput(100), '= [1,0,0]');

How to add zeros if less than 4 decimal and remove if more?

Can someone show me how to write a function that adds 0's after the decimal if less than four digits appear after the decimal until 4 decimal spots long and trim digits from far right of decimal in excess of 4. These are strings. Don't want any rounding. For display only, not calculations. So for example:
719.843797 should remove last two digits to be 719.8437
21.947 should add one 0 to be 21.9470
1.3456 no change

Using toFixed or any sort of multiplication may result in rounding problems. Treating number as a string allows to avoid them. The function below passes all your given usecases. But this is kinda hacky. Though I'm not sure if there is a better way to fit your requirements.
function truncate4(x) {
var parts = x.toString().split('.'); // this is not i18n proof :(
var integral = parts[0];
var decimal = parts[1] || ''; // might be integer
return integral + '.' + (decimal + '0000').substr(0, 4)
}
console.log(truncate4(719.843797));
console.log(truncate4(21.947));
console.log(truncate4(1.3456));

This function should match
function arrondir(num){
var str = num.toString();
var index = str.indexOf(".");
var decimal = str.substr(index+1);
var integer = str.substr(0, index);
if (decimal.length < 4)
for (var i = 0; i < 5 - decimal.length; i++)
decimal = decimal.concat("0");
else decimal = decimal.substr(0,4);
return integer.concat(".").concat(decimal);
}
console.log(arrondir(719.84)); // 719.8400
console.log(arrondir(719.84657963657)); // 719.8465

How do i return a number excluding the last digit?

So I have a number like 5467. I want my code to return 546.
I tried taking the last number and subtracting it from the original number but I get 5460 instead of 546.

Combine / with %:
(5467 - (5467 % 10)) / 10
564

Sounds like you also need to divide my 10. You could do something like this:
var number = 5467;
number = number - (number % 10); // This will subtract off the last digit.
number = number / 10;
console.log(number); // 546
We first use the modulo operator % to get the last digit, and we subtract it from number. That reduces the number from 5467 to 5460. Now to chop off the last digit (which is guaranteed to be a 0) we divide by 10 and get 546.
Written more concisely you could do:
number = (number - ( number % 10)) / 10;

There's a few things you can do the most concise being:
Math.floor(num / 10);
Or, convert to a string, remove the last character and convert back to number.
parseInt(num.toString().slice(0, -1));

If string representation would be fine for you then one other way is
var num = 5467,
cut = (num/10).toFixed(); // <-'547'
Well... warning..! i have to say toFixed() method rounds if necessary. So in this particular example it doesn't work.

I dont mind some of the other answers, but i feel that this maybe too fixed on it being a number.
Which it is, but you want to remove the last digit/char, regardless of the number, so why not substr?
http://www.w3schools.com/jsref/jsref_substr.asp
var s = 5467;
s = s.toString().substr(0, s.toString().length - 1);
console.log(s)
or even easier:
var s = (5467).toString();
s = s.substr(0, s.length - 1);
console.log(s)
These dont take into account single digit numbers, so passing in 1 would return blank. To answer that you could simply do a check like:
var s = (1).toString();
if(s.length > 1)
s = s.substr(0, s.length - 1);
console.log(s)
Also, similar question to:
Remove last digits from an int
Remove the last digits of a number (not string)
Removing the last digits in string

To truncate digits from the right hand side until the number is less than 30, keep dividing by 10 and rounding down until a suitable value is reached:
var n = 12341235;
while (n > 30) n = n/10|0;
document.write(n);
The greater than and division operations will coerce n to a number, so it can be a number or string. If ToNumber(n) results in NaN (e.g. n = 'foo'), then the value of n is not modified.

You can simply divide the number by 10 and parseInt()
var num = 5467;
num = parseInt(num/10);
Update :
To repeat the process until the answer is less than 30, use while loop as
var num = 5467;
while(num >= 30) {
num = parseInt(num/10);
}
document.write(num);

Spliting the binary string in half

I am trying to split binary number in half and then just add 4 zeroes.
For example for 10111101 I want to end up with only the first half of the number and make the rest of the number zeroes. What I want to end up would be 10110000.
Can you help me with this?

Use substring to split and then looping to pad
var str = '10111101';
var output = str.substring( 0, str.length/2 );
for ( var counter = 0; counter < str.length/2; counter++ )
{
output += "0";
}
alert(output)

try this (one-liner)
var binary_str = '10111101';
var padded_binary = binary_str.slice(0, binary_str.length/2) + new Array(binary_str.length/2+1).join('0');
console.log([binary_str,padded_binary]);
sample output
['10111101','10110000']

I guess you are using JavaScript...
"10111101".substr(0, 4) + "0000";

It's a bit unclear if you are trying to operate on numbers or strings. The answers already given do a good job of showing how to operate on a strings. If you want to operate with numbers only, you can do something like:
// count the number of leading 0s in a 32-bit word
function nlz32 (word) {
var count;
for (count = 0; count < 32; count ++) {
if (word & (1 << (31 - count))) {
break;
}
}
return count;
}
function zeroBottomHalf (num) {
var digits = 32 - nlz32(num); // count # of digits in num
var half = Math.floor(digits / 2);// how many to set to 0
var lowerMask = (1 << half) - 1; //mask for lower bits: 0b00001111
var upperMask = ~lowerMask //mask for upper bits: 0b11110000
return num & upperMask;
}
var before = 0b10111101;
var after = zeroBottomHalf(before);
console.log('before = ', before.toString(2)); // outputs: 10111101
console.log('after = ', after.toString(2)); // outputs: 10110000
In practice, it is probably simplest to covert your number to a string with num.toString(2), then operate on it like a string as in one of the other answers. At the end you can convert back to a number with parseInt(str, 2)

If you have a real number, not string, then just use binary arithmetic. Assuming your number is always 8 binary digits long - your question is kinda vague on that - it'd be simply:
console.log((0b10111101 & 0b11110000).toString(2))
// 10110000

JavaScript - Convert 24 digit hexadecimal number to decimal, add 1 and then convert back?

For an ObjectId in MongoDB, I work with a 24 digit hexadecimal number. Because I need to keep track of a second collection, I need to add 1 to this hexadecimal number.
In my case, here's my value
var value = "55a98f19b27585d81922ba0b"
What I'm looking for is
var newValue = "55a98f19b25785d81922ba0c"
I tried to create a function for this
function hexPlusOne(hex) {
var num = (("0x" + hex) / 1) + 1;
return num.toString(16);
}
This works with smaller hex numbers
hexPlusOne("eeefab")
=> "eeefac"
but it fails miserably for my hash
hexPlusOne(value)
=> "55a98f19b275840000000000"
Is there a better way to solve this?

This version will return a string as long as the input string, so the overflow is ignored in case the input is something like "ffffffff".
function hexIncrement(str) {
var hex = str.match(/[0-9a-f]/gi);
var digit = hex.length;
var carry = 1;
while (digit-- && carry) {
var dec = parseInt(hex[digit], 16) + carry;
carry = Math.floor(dec / 16);
dec %= 16;
hex[digit] = dec.toString(16);
}
return(hex.join(""));
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("ffffffffffffffffffffffff"));
This version may return a string which is 1 character longer than the input string, because input like "ffffffff" carries over to become "100000000".
function hexIncrement(str) {
var hex = str.match(/[0-9a-f]/gi);
var digit = hex.length;
var carry = 1;
while (digit-- && carry) {
var dec = parseInt(hex[digit], 16) + carry;
carry = Math.floor(dec / 16);
dec %= 16;
hex[digit] = dec.toString(16);
}
if (carry) hex.unshift("1");
return(hex.join(""));
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("ffffffffffffffffffffffff"));
I was curious te see whether user2864740's suggestion of working with 12-digit chunks would offer any advantage. To my surprise, even though the code looks more complicated, it's actually around twice as fast. But the first version runs 500,000 times per second too, so it's not like you're going to notice in the real world.
function hexIncrement(str) {
var result = "";
var carry = 1;
while (str.length && carry) {
var hex = str.slice(-12);
if (/^f*$/i.test(hex)) {
result = hex.replace(/f/gi, "0") + result;
carry = 1;
} else {
result = ("00000000000" + (parseInt(hex, 16) + carry).toString(16)).slice(-hex.length) + result;
carry = 0;
}
str = str.slice(0,-12);
}
return(str.toLowerCase() + (carry ? "1" : "") + result);
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("000000000000ffffffffffff") + "<BR>");
document.write(hexIncrement("0123456789abcdef000000000000ffffffffffff"));

The error comes from attempting to covert the entire 24-digit hex value to a number first because it won't fit in the range of integers JavaScript can represent distinctly2. In doing such a conversion to a JavaScript number some accuracy is lost.
However, it can be processed as multiple (eg. two) parts: do the math on the right part and then the left part, if needed due to overflow1. (It could also be processed one digit at a time with the entire addition done manually.)
Each chunk can be 12 hex digits in size, which makes it an easy split-in-half.
1 That is, if the final num for the right part is larger than 0xffffffffffff, simply carry over (adding) one to the left part. If there is no overflow then the left part remains untouched.
2 See What is JavaScript's highest integer value that a Number can go to without losing precision?
The range is 2^53, but the incoming value is 16^24 ~ (2^4)^24 ~ 2^(4*24) ~ 2^96; still a valid number, but outside the range of integers that can be distinctly represented.
Also, use parseInt(str, 16) instead of using "0x" + str in a numeric context to force the conversion, as it makes the intent arguably more clear.