Related
As the title suggests, I want to create a function the counts the number of values in my array between two values that have been entered by the user. So for example, if the array was [1, 4, 6, 7, 8, 6] and the user entered 5 as their first value and 7 as their second value, they would be greeted with an alert that said
"total number of values = 3".
You can create an extremely clean solution to this problem by utilizing the second property of Array#filter (which sets the this binding given to your callback of choice):
var array = [1, 4, 6, 7, 8, 6]
function inRange (x) {
return this[0] <= x && x <= this[1]
}
var result = array.filter(inRange, [5, 7]).length
console.log('Total number of values:', result)
All you need is a simple for loop.
var total = 0;
var num1 = 5;
var num2 = 7;
var array = [1,4,6,7,8,6];
for(var a = 0; a < array.length; a++) {
if(array[a] >= num1 && array[a] <= num2) {
total++;
}
}
alert("Total numbers of values = " + total);
This will loop through the array, detect nums within the range, tally up a value, and output it in a alert.
You can use Array.prototype.filter(), RegExp.prototype.test() with RegExp constructor with class from-to, where from is 5, to is 7, get .length of resulting array
var from = 5;
var to = 7;
var len = arr.filter(RegExp.prototype.test.bind(new RegExp(`[${from}-${to}]`))).length;
You can alternatively use .toString(), .match()
var arr = [1,4,6,7,8,6];
var from = 5;
var to = 7;
var res = arr.toString().match(new RegExp(`[${from}-${to}]`, "g"));
var len = res.length;
console.log(res.length);
You may do as follows;
var arr = [1,4,6,7,8,6],
input = [5,7],
result = arr.reduce((r,n) => n >= input[0] && n <= input[1] ? ++r : r, 0);
console.log(result);
var array = [1, 4, 6, 7, 8, 6];
function countNumber(arr,a,b){
let count = 0;
for (const number of arr){
if (number >= a && number <= b){
count ++;
}
}
return count;
}
console.log(countNumber(array, 5, 7));
I have a list with numbers:
2 3 11 17 21 6
at the same time i have another list of numbers with '-' separator:
-4-7-11-15-9-6-
I need to find a regular expression in javascript that match the first occurrence (in this example):
11
// returns index of first occurence of element in arr
// otherwise, returns -1
var getFirstOccurence = function (arr, str) {
// split the str by dashes and map to an array of numbers
var t = str.split("-").map(Number);
for (var i = 0; i < arr.length; i++) {
// arr[i] is in t, return its index
if (t.indexOf(arr[i]) > -1) return i;
}
// otherwise, return -1 (not in list)
return -1;
}
var str = "-4-7-11-15-9-6-",
list = [2, 3, 11, 17, 21, 6];
alert(getFirstOccurence(list, str));
// alerts 2; arr[2] is 11, which is in the list
modify the second list to (4|7|11|15|9|6) and use it as your pattern
Demo
You can
var string = '-4-7-11-15-9-6-',
array = [2, 3, 11, 17, 21, 6];
var regex = new RegExp('-(' + array.join('|') + ')-');
var match = string.match(regex);
var value = match ? match[0] : undefined;
console.log()
/(?:^|-)11(?:-|$)/
this regex will match 11 in front, in between and at the end
How do we split a string every 3 characters from the back using JavaScript?
Say, I have this:
str = 9139328238
after the desired function, it would become:
parts = ['9','139','328','238']
How do we do this elegantly?
var myString = String( 9139328238 );
console.log( myString.split( /(?=(?:...)*$)/ ) );
// ["9", "139", "328", "238"]
I can't make any performance guarantees. For smallish strings it should be fine.
Here's a loop implementation:
function funkyStringSplit( s )
{
var i = s.length % 3;
var parts = i ? [ s.substr( 0, i ) ] : [];
for( ; i < s.length ; i += 3 )
{
parts.push( s.substr( i, 3 ) );
}
return parts;
}
I know this is an old question, but I would like to provide my own one-line version to solve the problem :)
"12345678".split('').reverse().join('').match(/.{1,3}/g).map(function(x){
return x.split('').reverse().join('')
}).reverse()
This basically reverses the string, captures the groups of 3 elements, reverses each group and then reverses the whole string.
The steps are:
"12345678" -> [1, 2, 3, 4, 5, 6, 7, 8] //.split('')
[1, 2, 3, 4, 5, 6, 7, 8] -> [8, 7, 6, 5, 4, 3, 2, 1] //.reverse()
[8, 7, 6, 5, 4, 3, 2, 1] -> "87654321" //.join('')
"87654321" -> [876, 543, 21] //.match(...)
[876, 543, 21] -> [678, 345, 12] //.map(function(x){...})
[678, 345, 12] -> [12, 345, 678] //.reverse()
You can then join the array with a character (e.g. the ',' for thousands separator)
[12, 345, 678].join(',') -> "12,345,678"
There are a lot of complicated answers here.
function group(value) {
return value.match(/\d{1,3}(?=(\d{3})*$)/g);
}
console.log(group('1'));
console.log(group('123'));
console.log(group('1234'));
console.log(group('12345'));
console.log(group('123456'));
console.log(group('1234567'));
console.log(group('12345678'));
console.log(group('123456789'));
Not as elegant, but shows you a while loop
function commaSeparateNumber (val) {
val = val.toString();
while (/(\d+)(\d{3})/.test(val)){
val = val.replace(/(\d+)(\d{3})/, '$1'+','+'$2');
}
return val;
}
var str = "9139328238";
var splitStr = commaSeparateNumber(str).split(",");
console.log(splitStr);
Try this:
var str = 9139328238 + ''; //convert int to string
var reqArr = []; // required array
var len = str.length; //maintaining length
while (len > 0) {
len -= 3;
reqArr.unshift(str.slice(len)); //inserting value to required array
str = str.slice(0, len); //updating string
}
Hope it helps..
Since regex operations are not liked by everyone for various reasons: here is a regular function using a regular loop to split any regular string every X characters from back. Nothing fancy but it works:
function splitStringFromEnd(customString, every) {
var result = [], counter = every;
// loop that captures substring chungs of "every" length e.g.: 1000.00 -> ["000", ".00"]
for (var i = counter; counter <= customString.length; counter += every) {
result.unshift(customString.substr(customString.length - counter, every))
}
// check if there is a remainder and grabs it.
// Using our 1000.00 example: diff = 9 - 7; remainder = 3 - 2; -> ["1", "000", ".00"]
var diff = counter - customString.length;
var remainder = every - diff;
if(remainder > 0) { result.unshift(customString.substr(0, remainder)) }
return result;
}
for your example it would be:
splitStringFromEnd("9139328238", 3);
// :returns => ["9", "139", "328", "238"]
Enjoy :)
const price_format = (price) => {
let result = [];
let new_str = [...price].reverse().join("");
let rightSplit = new_str.match(/.{1,3}/g).reverse();
for (let item of rightSplit) {
result.push([...item].reverse().join(""));
}
return result.join(",");
}
let price = "2560000000";
console.log(price_format(price));
// output : 2,560,000,000
"12345678".split('').reverse().reduce((a, s) => (a[0].length<3?a[0]=s+a[0]:a.unshift(s),a), ['']);
Finally it seems good. This is what I have got till now without using any loops
function breakAt3(x)
{
if(x.length < 3){ var parts = [x]; return parts; }
var startPos = (x.length % 3);
var newStr = x.substr(startPos);
var remainingStr = x.substr(0,startPos);
var parts = newStr.match(/.{1,3}/g);
if(remainingStr != ''){ var length = parts.unshift(remainingStr); }
return parts;
}
var str = '92183213081';
var result = breakAt3(str); // 92,183,213,081
This is my code so far:
var n = 123456789;
var d = n.toString().length;
var digits = [];
var squaredDigits = [];
for (i = d; i >= 1; i--) {
var j = k / 10;
var r = (n % k / j) - 0.5;
var k = Math.pow(10, i);
var result = r.toFixed();
digits.push(result);
}
console.log(digits);
But when I run my code I get this: [9, 1, 2, 3, 4, 5, 6, 7, 8]
If anyone can see the problem or find a better solution I would very much appreciate it!
Why not just do this?
var n = 123456789;
var digits = (""+n).split("");
What about:
const n = 123456;
Array.from(n.toString()).map(Number);
// [1, 2, 3, 4, 5, 6]
(123456789).toString(10).split("")
^^ this will return an array of strings
(123456789).toString(10).split("").map(function(t){return parseInt(t)})
^^ this will return an array of ints
I realize this was asked several months ago, but I have an addition to samccone's answer which is more succinct but I don't have the rep to add as a comment!
Instead of:
(123456789).toString(10).split("").map(function(t){return parseInt(t)})
Consider:
(123456789).toString(10).split("").map(Number)
Modified the above answer a little bit. We don't really have to call the 'map' method explicitly, because it is already built-in into the 'Array.from' as a second argument.
As of MDN.
Array.from(arrayLike[, mapFn[, thisArg]])
let num = 1234;
let arr = Array.from(String(num), Number);
console.log(arr); // [1, 2, 3, 4]
const toIntArray = (n) => ([...n + ""].map(v => +v))
It is pretty short using Array destructuring and String templates:
const n = 12345678;
const digits = [...`${n}`];
console.log(digits);
Assuming the value n:
const n = 123456789
A minimal ES6 version if you'd like:
String(n).split("").map(Number)
An even shorter but less readable version:
[...String(n)].map(Number)
Want to go even shorter (but less readable)?
[...`${n}`].map(Number)
Shorter you say (and basically illegible)!?
[...""+n].map(Number)
Now you're a real programmer, congrats!
Side note
These aren't really efficient (as most in this thread) since you're allocating 2 arrays instead of 1. Want to be more efficient? Try this which only allocates one array:
var arr = []
var str = String(n)
for (var i = 0; i < str.length; i++) {
arr.push(Number(str[i]))
}
Oldschool but more efficient, huzzah!
It's very simple, first convert the number to string using the toString() method in JavaScript and then use split() method to convert the string to an array of individual characters.
For example, the number is num, then
const numberDigits = num.toString().split('');
This will work for a number greater than 0. You don't need to convert the number into string:
function convertNumberToDigitArray(number) {
const arr = [];
while (number > 0) {
let lastDigit = number % 10;
arr.push(lastDigit);
number = Math.floor(number / 10);
}
return arr;
}
You can get a list of string from your number, by converting it to a string, and then splitting it with an empty string. The result will be an array of strings, each containing a digit:
const num = 124124124
const strArr = `${num}`.split("")
OR to build on this, map each string digit and convert them to a Number:
const intArr = `${num}`.split("").map(x => Number(x))
Here's an alternative to Nicolás Fantone's answer. You could argue it's maybe a little less readable. The emphasis is that Array.from() can take an optional map function as a parameter. There are some performance gains this way since no intermediate array gets created.
const n = 123456;
Array.from(n.toString(), (val) => Number(val)); // [1, 2, 3, 4, 5, 6]
const number = 1435;
number.toString().split('').map(el=>parseInt(el));
let input = 12345664
const output = []
while (input !== 0) {
const roundedInput = Math.floor(input / 10)
output.push(input - roundedInput * 10)
input = roundedInput
}
console.log(output)
Suppose,
let a = 123456
First we will convert it into string and then apply split to convert it into array of characters and then map over it to convert the array to integer.
let b = a.toString().split('').map(val=>parseInt(val))
console.log(b)
Move:
var k = Math.pow(10, i);
above
var j = k / 10;
var num = 123456789;
num = num.toString(); //'123456789'
var digits = num.split(""); //[ '1', '2', '3', '4', '5', '6', '7', '8', '9' ]
It's been a 5+ years for this question but heay always welcome to the efficient ways of coding/scripting.
var n = 123456789;
var arrayN = (`${n}`).split("").map(e => parseInt(e))
Another method here. Since number in Javascript is not splittable by default, you need to convert the number into a string first.
var n = 123;
n.toString().split('').map(Number);
I ended up solving it as follows:
const n = 123456789;
let toIntArray = (n) => ([...n + ""].map(Number));
console.log(toIntArray(n));
Update with string interpolation in ES2015.
const num = 07734;
let numStringArr = `${num}`.split('').map(el => parseInt(el)); // [0, 7, 7, 3, 4]
var n = 38679;
var digits = n.toString().split("");
console.log(digits);
Now the number n is divided to its digits and they are presented in an array, and each element of that array is in string format. To transform them to number format do this:
var digitsNum = digits.map(Number);
console.log(digitsNum);
Or get an array with all elements in number format from the beginning:
var n = 38679;
var digits = n.toString().split("").map(Number);
console.log(digits);
This is actually the cleanest solution I think.
var n = 123456789;
const digits = (`${n}`).split('')
You put it in a string literal but it is kept as numbers, and then it is split to an array and assigned to digits.
const toIntArray = (n) => ([...n + ""].map(v => +v))
I am trying to solve a math problem where I take a number e.g. 45, or 111 and then split the number into separate digits e.g. 4 5 or 1 1 1. I will then save each number to a var to run a method on. Does anyone know how to split a number into individual digitals?
For example I have a loop that runs on an array :
for (var i = 0; i < range.length; i++) {
var n = range[i];
}
For each number, I would like to split its digits and add them together?
var num = 123456;
var digits = num.toString().split('');
var realDigits = digits.map(Number)
console.log(realDigits);
var number = 12354987,
output = [],
sNumber = number.toString();
for (var i = 0, len = sNumber.length; i < len; i += 1) {
output.push(+sNumber.charAt(i));
}
console.log(output);
/* Outputs:
*
* [1, 2, 3, 5, 4, 9, 8, 7]
*/
UPDATE: Calculating a sum
for (var i = 0, sum = 0; i < output.length; sum += output[i++]);
console.log(sum);
/*
* Outputs: 39
*/
You can also do it in the "mathematical" way without treating the number as a string:
var num = 278;
var digits = [];
while (num != 0) {
digits.push(num % 10);
num = Math.trunc(num / 10);
}
digits.reverse();
console.log(digits);
One upside I can see is that you won't have to run parseInt() on every digit, you're dealing with the actual digits as numeric values.
This is the shortest I've found, though it does return the digits as strings:
let num = 12345;
[...num+''] //["1", "2", "3", "4", "5"]
Or use this to get back integers:
[...num+''].map(n=>+n) //[1, 2, 3, 4, 5]
I will provide a variation on an answer already given so you can see a different approach that preserves the numeric type all along:
var number = 12354987,
output = [];
while (number) {
output.push(number % 10);
number = Math.floor(number/10);
}
console.log(output.reverse().join(',')); // 1,2,3,5,4,9,8,7
I've used a technique such as the above to good effect when converting a number to Roman numerals, which is one of my favorite ways to begin to learn a programming language I'm not familiar with. For instance here is how I devised a way to convert numbers to Roman numerals with Tcl slightly after the turn of the century: http://code.activestate.com/recipes/68379-conversion-to-roman-numerals/
The comparable lines in my Tcl script being:
while {$arabic} {
set digit [expr {$arabic%10}]
set arabic [expr {$arabic/10}]
// Split positive integer n < 1e21 into digits:
function digits(n) {
return Array.from(String(n), Number);
}
// Example:
console.log(digits(1234)); // [1, 2, 3, 4]
You can work on strings instead of numbers to achieve this. You can do it like this
(111 + '').split('')
This will return an array of strings ['1','1','1'] on which you can iterate upon and call parseInt method.
parseInt('1') === 1
If you want the sum of individual digits, you can use the reduce function (implemented from Javascript 1.8) like this
(111 + '').split('').reduce(function(previousValue, currentValue){
return parseInt(previousValue,10) + parseInt(currentValue,10);
})
Use String, split and map :
String(number).split("").map(Number);
function splitNum(num) {
return String(num).split("").map(Number);
}
console.log(splitNum(1523)); // [1, 5, 2, 3]
console.log(splitNum(2341)); // [2, 3, 4, 1]
console.log(splitNum(325)); // [3, 2, 5]
Without converting to string:
function toDigits(number) {
var left;
var results = [];
while (true) {
left = number % 10;
results.unshift(left);
number = (number - left) / 10;
if (number === 0) {
break;
}
}
return results;
}
Using String, ... and map
const num = 7890;
const digits = [...String(num)].map(Number);
console.log(digits)
Alternatively, using ... and reduce to get digits and their sum.
const sumOfDigits = num => [...""+num].reduce((acc, dig) => acc + +dig, 0);
console.log('Sum of digits: ', sumOfDigits(7890));
Separate each 2 parametr.
function separator(str,sep) {
var output = '';
for (var i = str.length; i > 0; i-=2) {
var ii = i-1;
if(output) {
output = str.charAt(ii-1)+str.charAt(ii)+sep+output;
} else {
output = str.charAt(ii-1)+str.charAt(ii);
}
}
return output;
}
console.log(separator('123456',':')); //Will return 12:34:56
With ES6, you could use Array.from with a stringed number as iterables and Number as mapping function.
const getDigits = n => Array.from(n.toString(), Number);
console.log(getDigits(12345));
A fun introduction to recursion. This answer takes a Number and returns an array of Number digits. It does not convert the number to a string as an intermediate step.
Given n = 1234,
n % 10 will return first (right-moist) digit, 4
n / 10 will return 123 with some remainder
Using Math.floor we can chop the remainder off
Repeating these steps, we can form the entire result
Now we just have to build the recursion condition,
If the number is already a single digit (n < 10), return an array singleton of the digit
otherwise (inductive) the number is 10 or greater; recur and prepend to the first digit
const digits = (n = 0) =>
n < 10
? [ n ]
: [ ... digits (Math.floor (n / 10)), n % 10 ]
console.log (digits ()) // [ 0 ]
console.log (digits (1)) // [ 1 ]
console.log (digits (12)) // [ 1, 2 ]
console.log (digits (123)) // [ 1, 2, 3 ]
console.log (digits (11234)) // [ 1, 2, 3, 4 ]
console.log (digits (123456789012))
// [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2 ]
This also works:
var number = 12354987;
console.log(String(number).split('').map(Number));
Shadow Wizard , extended version by Orien
var num:Number = 1523;
var digits:Array = [];
var cnt:int = 0;
while (num > 0) {
var mod:int = num % 10;
digits.push(mod * Math.pow(10, cnt))
num = Math.floor(num / 10);
cnt++;
}
digits.reverse();
trace(digits);
output:1000,500,20,3
A functional approach in order to get digits from a number would be to get a string from your number, split it into an array (of characters) and map each element back into a number.
For example:
var number = 123456;
var array = number.toString()
.split('')
.map(function(item, index) {
return parseInt(item);
});
console.log(array); // returns [1, 2, 3, 4, 5, 6]
If you also need to sum all digits, you can append the reduce() method to the previous code:
var num = 123456;
var array = num.toString()
.split('')
.map(function(item, index) {
return parseInt(item);
})
.reduce(function(previousValue, currentValue, index, array) {
return previousValue + currentValue;
}, 0);
console.log(array); // returns 21
As an alternative, with ECMAScript 2015 (6th Edition), you can use arrow functions:
var number = 123456;
var array = number.toString().split('').map((item, index) => parseInt(item));
console.log(array); // returns [1, 2, 3, 4, 5, 6]
If you need to sum all digits, you can append the reduce() method to the previous code:
var num = 123456;
var result = num.toString()
.split('')
.map((item, index) => parseInt(item))
.reduce((previousValue, currentValue) => previousValue + currentValue, 0);
console.log(result); // returns 21
I used this simple way of doing it.
To split digits
var N = 69;
var arr = N.toString().split('').map(Number)
// outputs [6,9]
console.log( arr );
To add them together
console.log(arr.reduce( (a,b) => a+b )); // 15
And the easiest.... num_string.split('').map(Number)
Try below:
console.log((''+123).split('').map(Number))
To just split an integer into its individual digits in the same order, Regular Expression is what I used and prefer since it prevents the chance of loosing the identity of the numbers even after they have been converted into string.
The following line of code convert the integer into a string, uses regex to match any individual digit inside the string and return an array of those, after which that array is mapped to be converted back to numbers.
const digitize = n => String(n).match(/\d/g).map(Number);
I might be wrong, but a solution picking up bits and pieces. Perhaps, as I still learning, is that the functions does many things in the same one. Do not hesitate to correct me, please.
const totalSum = (num) => [...num + ' '].map(Number).reduce((a, b) => a + b);
So we take the parameter and convert it to and arr, adding empty spaces. We do such operation in every single element and push it into a new array with the map method. Once splited, we use reduce to sum all the elements and get the total.
As I said, don't hesitate to correct me or improve the function if you see something that I don't.
Almost forgot, just in case:
const totalSum = (num) => ( num === 0 || num < 0) ? 'I need a positive number' : [...num + ' '].map(Number).reduce((a, b) => a + b);
If negatives numbers or just plain zero go down as parameters. Happy coding to us all.
I am posting this answer to introduce the use of unshift which is a modern solution. With push, you add to the end of an array while unshift adds to the beginning. This makes the mathematical approach more powerful as you won't need to reverse anymore.
let num = 278;
let digits = [];
while (num > 0) {
digits.unshift(num % 10);
num = parseInt(num / 10);
}
console.log(digits);
var num = 111,
separateDigits = num.toString().split(""), i, l = separateDigits.length;
for( i = 0; i < l; ++i ) {
someObject.someMethod( +separateDigits[i] );
}
You can try this.
var num = 99;
num=num.toString().split("").map(value=>parseInt(value,10)); //output [9,9]
Hope this helped!
function iterateNumber(N, f) {
let n = N;
var length = Math.log(n) * Math.LOG10E + 1 | 0;
for (let i = 0; i < length; i++) {
const pow = Math.pow(10, length - i - 1)
let c = (n - (n % pow)) / pow
f(c, i)
n %= pow
}
}
('' + 123456789).split('').map( x => +x ).reduce( (a,b) => a+b ) === 45
true
or without map
('' + 123456789).split('').reduce( (a,b) => (+a)+(+b) ) === 45
true
You can do it in single line, seperate each digits than add them together :
var may = 12987;
var sep = (""+may).split("").map(n=>+n).reduce((a,b)=>a+b);
This is my short solution.. with sum of number
function sum (num) {
let sNumber = num
.toString()
.split('')
.reduce((el1, el2) => {
return Number(el1) + Number(el2)
}, 0)
return sNumber
}
console.log(sum(123))
console.log(sum(456))
javascript has a function for it and you can use it easily.
console.log(new Intl.NumberFormat().format(number));
for example :
console.log(new Intl.NumberFormat().format(2334325443534));
==> 2,334,325,443,534
Iterate through each number with for...of statement.
By adding a + sign before a String, it will be converted into a number.
const num = 143,
digits = [];
for (const digit of `${num}`) {
digits.push(+digit)
}
console.log(digits);
Inspired by #iampopov You can write it with spread syntax.
const num = 143;
const digits = [...`${num}`].map(Number);
console.log(digits);
And as a one liner.
console.log(Number.MAX_SAFE_INTEGER.toString().split('').reduce((pv, v) => Number(v) + pv, 0));