Create array of all integers between two numbers, inclusive, in Javascript/jQuery

JavascriptJqueryArrays

Javascript Problem Overview


Say I have the following checkbox:

<input type="checkbox" value="1-25" />

To get the two numbers that define the boundaries of range I'm looking for, I use the following jQuery:

var value = $(this).val();
var lowEnd = Number(value.split('-')[0]);
var highEnd = Number(value.split('-')[1]);

How do I then create an array that contains all integers between lowEnd and highEnd, including lowEnd and highEnd themselves? For this specific example, obviously, the resulting array would be:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]

Javascript Solutions


Solution 1 - Javascript

var list = [];
for (var i = lowEnd; i <= highEnd; i++) {
    list.push(i);
}

Solution 2 - Javascript

ES6 :

Use Array.from (docs here):

console.log(
   Array.from({length:5},(v,k)=>k+1)
)

Solution 3 - Javascript

In JavaScript ES6:

function range(start, end) {
  return Array(end - start + 1).fill().map((_, idx) => start + idx)
}
var result = range(9, 18); // [9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
console.log(result);

For completeness, here it is with an optional step parameter.

function range(start, end, step = 1) {
  const len = Math.floor((end - start) / step) + 1
  return Array(len).fill().map((_, idx) => start + (idx * step))
}
var result = range(9, 18, 0.83);
console.log(result);

I would use range-inclusive from npm in an actual project. It even supports backwards steps, so that's cool.

Solution 4 - Javascript

I highly recommend underscore or lo-dash libraries:

http://underscorejs.org/#range

(Almost completely compatible, apparently lodash runs quicker but underscore has better doco IMHO)

_.range([start], stop, [step])

Both libraries have bunch of very useful utilities.

Solution 5 - Javascript

My version of the loop ;)

var lowEnd = 1;
var highEnd = 25;
var arr = [];
while(lowEnd <= highEnd){
   arr.push(lowEnd++);
}

Solution 6 - Javascript

fastest way

  1. while-- is faster on most browsers
  2. direct setting a variable is faster than push

function:

var x=function(a,b,c,d){d=[];c=b-a+1;while(c--){d[c]=b--}return d},

theArray=x(lowEnd,highEnd);

or

var arr=[],c=highEnd-lowEnd+1;
while(c--){arr[c]=highEnd--}

EDIT

readable version

var arr = [],
c = highEnd - lowEnd + 1;
while ( c-- ) {
 arr[c] = highEnd--
}

Demo

http://jsfiddle.net/W3CUn/

FOR THE DOWNVOTERS

performance

http://jsperf.com/for-push-while-set/2

faster in ie and 3x faster in firefox

only on aipad air the for loop is a little faster.

tested on win8, osx10.8, ubuntu14.04, ipad, ipad air, ipod;

with chrome,ff,ie,safari,mobile safari.

i would like to see the performance on older ie browsers where the for loop isn't that optimized!

Solution 7 - Javascript

function range(j, k) { 
    return Array
        .apply(null, Array((k - j) + 1))
        .map(function(_, n){ return n + j; }); 
}

this is roughly equivalent to

function range(j, k) { 
    var targetLength = (k - j) + 1;
    var a = Array(targetLength);
    var b = Array.apply(null, a);
    var c = b.map(function(_, n){ return n + j; });
    return c;
}

breaking it down:

var targetLength = (k - j) + 1;

var a = Array(targetLength);

this creates a sparse matrix of the correct nominal length. Now the problem with a sparse matrix is that although it has the correct nominal length, it has no actual elements, so, for

j = 7, k = 13

console.log(a);

gives us

Array [ <7 empty slots> ]

Then

var b = Array.apply(null, a);

passes the sparse matrix as an argument list to the Array constructor, which produces a dense matrix of (actual) length targetLength, where all elements have undefined value. The first argument is the 'this' value for the the array constructor function execution context, and plays no role here, and so is null.

So now,

 console.log(b);

yields

 Array [ undefined, undefined, undefined, undefined, undefined, undefined, undefined ]

finally

var c = b.map(function(_, n){ return n + j; });

makes use of the fact that the Array.map function passes: 1. the value of the current element and 2. the index of the current element, to the map delegate/callback. The first argument is discarded, while the second can then be used to set the correct sequence value, after adjusting for the start offset.

So then

console.log(c);

yields

 Array [ 7, 8, 9, 10, 11, 12, 13 ]

Solution 8 - Javascript

My five cents:

Both direction array of integers function.

When range(0, 5) become [0, 1, 2, 3, 4, 5].

And range(5, 0) become [5, 4, 3, 2, 1, 0].

Based on this answer.

function range(start, end) {
  const isReverse = (start > end);
  const targetLength = isReverse ? (start - end) + 1 : (end - start ) + 1;
  const arr = new Array(targetLength);
  const b = Array.apply(null, arr);
  const result = b.map((discard, n) => {
    return (isReverse) ? n + end : n + start;
  });

  return (isReverse) ? result.reverse() : result;
}

P.S. For use in real life you should also check args for isFinite() and isNaN().

Solution 9 - Javascript

function createNumberArray(lowEnd, highEnd) {
    var start = lowEnd;
    var array = [start];
    while (start < highEnd) {
        array.push(start);
        start++;
    }
} 

Solution 10 - Javascript

If the start is always less than the end, we can do:

function range(start, end) {
  var myArray = [];
  for (var i = start; i <= end; i += 1) {
    myArray.push(i);
  }
  return myArray;
};
console.log(range(4, 12));                 // → [4, 5, 6, 7, 8, 9, 10, 11, 12]

If we want to be able to take a third argument to be able to modify the step used to build the array, and to make it work even though the start is greater than the end:

function otherRange(start, end, step) {
  otherArray = [];
  if (step == undefined) {
    step = 1;
  };
  if (step > 0) {
    for (var i = start; i <= end; i += step) {
      otherArray.push(i);
    }
  } else {
    for (var i = start; i >= end; i += step) {
      otherArray.push(i);
    }
  };
  return otherArray;
};
console.log(otherRange(10, 0, -2));        // → [10, 8, 6, 4, 2, 0]
console.log(otherRange(10, 15));           // → [10, 11, 12, 13, 14, 15]
console.log(otherRange(10, 20, 2));        // → [10, 12, 14, 16, 18, 20]

This way the function accepts positive and negative steps and if no step is given, it defaults to 1.

Solution 11 - Javascript

var values = $(this).val().split('-'),
    i = +values[0],
    l = +values[1],
    range = [];

while (i < l) {
    range[range.length] = i;
    i += 1;
}

range[range.length] = l;

There's probably a DRYer way to do the loop, but that's the basic idea.

Solution 12 - Javascript

You can design a range method that increments a 'from' number by a desired amount until it reaches a 'to' number. This example will 'count' up or down, depending on whether from is larger or smaller than to.

Array.range= function(from, to, step){
	if(typeof from== 'number'){
		var A= [from];
		step= typeof step== 'number'? Math.abs(step):1;
		if(from> to){
			while((from -= step)>= to) A.push(from);
		}
		else{
			while((from += step)<= to) A.push(from);
		}
		return A;
	}	
}

If you ever want to step by a decimal amount : Array.range(0,1,.01) you will need to truncate the values of any floating point imprecision. Otherwise you will return numbers like 0.060000000000000005 instead of .06.

This adds a little overhead to the other version, but works correctly for integer or decimal steps.

Array.range= function(from, to, step, prec){
	if(typeof from== 'number'){
		var A= [from];
		step= typeof step== 'number'? Math.abs(step):1;
		if(!prec){
			prec= (from+step)%1? String((from+step)%1).length+1:0;
		}
		if(from> to){
			while(+(from -= step).toFixed(prec)>= to) A.push(+from.toFixed(prec));
		}
		else{
			while(+(from += step).toFixed(prec)<= to) A.push(+from.toFixed(prec));
		}
		return A;
	}	
}

Solution 13 - Javascript

Adding http://minifiedjs.com/ to the list of answers :)

Code is similar to underscore and others:

var l123 = _.range(1, 4);      // same as _(1, 2, 3)
var l0123 = _.range(3);        // same as _(0, 1, 2)
var neg123 = _.range(-3, 0);   // same as _(-3, -2, -1)
var empty = _.range(2,1);      // same as _()

Docs here: http://minifiedjs.com/api/range.html

I use minified.js because it solves all my problems with low footprint and easy to understand syntax. For me, it is a replacement for jQuery, MustacheJS and Underscore/SugarJS in one framework.

Of course, it is not that popular as underscore. This might be a concern for some.

Minified was made available by Tim Jansen using the CC-0 (public domain) license.

Solution 14 - Javascript

const range = (start, stop, step) => Array.from({ length: (stop - start) / step + 1}, (_, i) => start + (i * step));

source: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from

Solution 15 - Javascript

Solution with pure ES6

Inspired by m59's answer above, but without the dependency on fill:

const range = (start, stop) => Array.from({ length: stop - start + 1 }, (_, i) => start + i)

So you can use it like:

range(3,5)
=> [3, 4, 5]

Solution 16 - Javascript

_Array = (length) => Object.keys(Array.from({length}))

//_Array = [0, 1, 2, 3, 4]

Solution 17 - Javascript

Solving in underscore

data = [];
_.times( highEnd, function( n ){ data.push( lowEnd ++ ) } );

Solution 18 - Javascript

        function getRange(a,b)
        {
            ar = new Array();
            var y = a - b > 0 ? a - b : b - a;
            for (i=1;i<y;i++)
            {
                ar.push(i+b);
            }
            return ar;
        }

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Question40 Degree DayView Question on Stackoverflow
Solution 1 - JavascriptBenView Answer on Stackoverflow
Solution 2 - JavascriptAbdennour TOUMIView Answer on Stackoverflow
Solution 3 - Javascriptm59View Answer on Stackoverflow
Solution 4 - JavascriptErichBSchulzView Answer on Stackoverflow
Solution 5 - JavascriptPavel PodlipenskyView Answer on Stackoverflow
Solution 6 - JavascriptcoccoView Answer on Stackoverflow
Solution 7 - Javascriptdavid.barkhuizenView Answer on Stackoverflow
Solution 8 - JavascriptSergei PanfilovView Answer on Stackoverflow
Solution 9 - JavascriptIgorView Answer on Stackoverflow
Solution 10 - JavascriptdrjorgepolancoView Answer on Stackoverflow
Solution 11 - JavascriptsdleihssirhcView Answer on Stackoverflow
Solution 12 - JavascriptkennebecView Answer on Stackoverflow
Solution 13 - JavascriptChristianView Answer on Stackoverflow
Solution 14 - JavascriptSaberView Answer on Stackoverflow
Solution 15 - JavascriptFlavio WuenscheView Answer on Stackoverflow
Solution 16 - JavascriptHeraldView Answer on Stackoverflow
Solution 17 - JavascriptvladiimView Answer on Stackoverflow
Solution 18 - JavascriptMamdouhView Answer on Stackoverflow