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I need to capture maximum 5 elements of a string. However if there are less than 5, then I just need how many there are there.
var y = '1,2,3,4,5,6,7,8,9,10'
//desired result:
var out = '1,2,3,4,5' // please note there is no trailing comma
var y = '1,2,3'
//desired result:
var out = '1,2,3'
My code:
for (var i = 0; i < 5; i++) {
x += y;
x = x + ",";
}
Write(x);
A simple method will do. The below splits the string by , then takes either n elements or the total length if it is less than n and then rejoins the values with a comma
const getNValues = (str, n) => {
const values = str.split(",");
const res = values.slice(0, Math.min(values.length,n))
return res.join(",");
}
console.log(getNValues("1,2,3,4,5,6,7,8,9",5));
console.log(getNValues("1,2,3",5));
Try this simple function to do that
function getMaxLen(str) {
if(typeof str != 'string') return null;
str = str.split(',');
return str.slice(0, 5).join(',');
}
console.log(getMaxLen('1,2,3,4,5,6,7,8,9,10'))
console.log(getMaxLen('1,2,3'))
var string = '1, 2, 3, 4, 5, 6, 7, 8, 9, 10';
var out = (string.match(/^([0-9],? ?){0,5}/)[0] || '').replace(/, ?$/, '');
console.log(out)
[EDIT] Explanation
.match(^([0-9],? ?){0,5}/g):
start at the begging ^
match numbers [0-9]
then a comma if any and a space if any ,? ?.
match this expression 0 to 5 times {0, 5}
I have seen questions and answers for using variables with RegEx like this:
var re = new RegExp(yyy, 'g');
xxx.match(re);
How do I add the caret and the dollar sign for an exact match?
I am iterating through an array: [-10, 0, 10] as search values. So, I am trying to do something like this:
var re = new RegExp("^" + data[i] + "$", 'g');
If I am searching an array of [-10, 0, 10] and if I search for 0, I get all three. I need to search for only 0 with this: /^0$/
Here is my entire code with test cases.
I am trying to find unique items using RegEx / match.
function nonUniqueElements(data){
var arrForDeletion = [];
var strData = data.join(",");
for(var i = 0; i < data.length; i++){
var re = new RegExp(data[i], 'g');
console.log("re: " + re);
var matches = strData.match(re);
if(matches.length === 1){
arrForDeletion.push(data.indexOf(data[i]));
}
}
if(arrForDeletion.length > 0){ // If there are unique items in the input array, do not include them in the returned array.
var arrReturn = [];
data.forEach(function(item,idx){
if(arrForDeletion.indexOf(idx) === -1){
arrReturn.push(item);
}
});
console.log("arr to be returned:");
console.log(arrReturn);
return arrReturn;
}
else{ // There were no unique items in the input array.
console.log("no unique items:");
console.log(data);
return data;
}
}
/*
nonUniqueElements([1, 2, 3, 1, 3]); // == [1, 3, 1, 3]
nonUniqueElements([1, 2, 3, 4, 5]); // == []
nonUniqueElements([5, 5, 5, 5, 5]); // == [5, 5, 5, 5, 5]
nonUniqueElements([10, 9, 10, 10, 9, 8]); // == [10, 9, 10, 10, 9]
*/
nonUniqueElements([-10,10,0]); // []
You're getting all 3 because for each entry in your array you are searching for exactly that value. Which will of course return all 3.
If you only want to match values of exactly 0 you can simply do:
const regex = new RegExp('^0$'); // Only need to compile it once
for(let i = 0; i < arr.length; i++){
arr[i].match(re);
}
Or if you are only checking for 0 values why not just do
arr[i] === 0 or arr[i] === '0' depending on if your input is a string or a number.
Oh, wait...Geez...
I am searching the entire string of the joined array:
"-10,10,0"
So, of course it will find 0 three times.
Geez...
(See what lack of sleep will result in?)
I think my whole approach to this project was wrong. But, I wanted to use RegEx and match.
I apologize.
Thanks for looking at this.
Sorry. I am unable to delete it.
I have very long array containing numbers. I need to remove trailing zeros from that array.
if my array will look like this:
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
I want to remove everything except [1, 2, 0, 1, 0, 1].
I have created function that is doing what is expected, but I'm wondering if there is a build in function I could use.
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
for(i=arr.length-1;i>=0;i--)
{
if(arr[i]==0)
{
arr.pop();
} else {
break;
}
}
console.log(arr);
Can this be done better/faster?
Assuming:
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
You can use this shorter code:
while(arr[arr.length-1] === 0){ // While the last element is a 0,
arr.pop(); // Remove that last element
}
Result:
arr == [1,2,0,1,0,1]
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
var copy = arr.slice(0);
var len = Number(copy.reverse().join('')).toString().length;
arr.length = len;
arr -> [1, 2, 0, 1, 0, 1]
how it works
copy.reverse().join('') becomes "00000000000000000101021"
when you convert a numerical string to number all the preceding zeroes are kicked off
var len = Number(copy.reverse().join('')) becomes 101021
now by just counting the number i know from where i have to remove the trailing zeroes and the fastest way to delete traling elements is by resetting the length of the array.
arr.length = len;
DEMO
const arr = [0,0,0,1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
My solution is:
arr.join('').replace(/0+$/g,'').split('').map(Number);
It will remove trailing zeros in the given array.
Result is [0,0,0,1,2,0,1,0,1];
If you also needed to remove leading zeros, you can adjust the regex like this:
arr.join('').replace(/^0+|0+$/g,'').split('').map(Number);
Now It will remove not only trailing zeros, but leading zeros too.
Result is [1,2,0,1,0,1];
Accepted answer is perfectly well. Just for fun here is a reducing approach.
var a = [1,0,1,0,1,0,1,2,3,4,5,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0];
f = a => { var b = true;
return a.reduceRight((r,n,i) => ( b ? n && ( b = false
, r[i] = n
)
: r[i] = n
, r
)
, []
);
};
console.log(f(a));
Here's a one-liner representing javascript at both its best and worst at the same time.
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
arr.slice(0,arr.reduceRight(([d,l],c) => [d||c,l-!(d||c)], [false,arr.length])[1])
Output:
[ 1, 2, 0, 1, 0, 1 ]
A simpler immutable approach using reduceRight
const arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
const prunedArray = arr.reduceRight((acc, item) => {
if(item === 0 && acc.length === 0) {
return acc;
}
return acc.concat(item);
}, []);
console.log(prunedArray); // [1, 0, 1, 0, 2, 1]
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
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));