I'm looking for any alternatives to the below for creating a JavaScript array containing 1 through to N where N is only known at runtime.
var foo = [];
for (var i = 1; i <= N; i++) {
foo.push(i);
}
To me it feels like there should be a way of doing this without the loop.
In ES6 using Array from() and keys() methods.
Array.from(Array(10).keys())
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Shorter version using spread operator.
[...Array(10).keys()]
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Start from 1 by passing map function to Array from(), with an object with a length property:
Array.from({length: 10}, (_, i) => i + 1)
//=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
You can do so:
var N = 10;
Array.apply(null, {length: N}).map(Number.call, Number)
result: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
or with random values:
Array.apply(null, {length: N}).map(Function.call, Math.random)
result: [0.7082694901619107, 0.9572225909214467, 0.8586748542729765,
0.8653848143294454, 0.008339877473190427, 0.9911756622605026, 0.8133423360995948, 0.8377588465809822, 0.5577575915958732, 0.16363654541783035]
Explanation
First, note that Number.call(undefined, N) is equivalent to Number(N), which just returns N. We'll use that fact later.
Array.apply(null, [undefined, undefined, undefined]) is equivalent to Array(undefined, undefined, undefined), which produces a three-element array and assigns undefined to each element.
How can you generalize that to N elements? Consider how Array() works, which goes something like this:
function Array() {
if ( arguments.length == 1 &&
'number' === typeof arguments[0] &&
arguments[0] >= 0 && arguments &&
arguments[0] < 1 << 32 ) {
return [ … ]; // array of length arguments[0], generated by native code
}
var a = [];
for (var i = 0; i < arguments.length; i++) {
a.push(arguments[i]);
}
return a;
}
Since ECMAScript 5, Function.prototype.apply(thisArg, argsArray) also accepts a duck-typed array-like object as its second parameter. If we invoke Array.apply(null, { length: N }), then it will execute
function Array() {
var a = [];
for (var i = 0; i < /* arguments.length = */ N; i++) {
a.push(/* arguments[i] = */ undefined);
}
return a;
}
Now we have an N-element array, with each element set to undefined. When we call .map(callback, thisArg) on it, each element will be set to the result of callback.call(thisArg, element, index, array). Therefore, [undefined, undefined, …, undefined].map(Number.call, Number) would map each element to (Number.call).call(Number, undefined, index, array), which is the same as Number.call(undefined, index, array), which, as we observed earlier, evaluates to index. That completes the array whose elements are the same as their index.
Why go through the trouble of Array.apply(null, {length: N}) instead of just Array(N)? After all, both expressions would result an an N-element array of undefined elements. The difference is that in the former expression, each element is explicitly set to undefined, whereas in the latter, each element was never set. According to the documentation of .map():
callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values.
Therefore, Array(N) is insufficient; Array(N).map(Number.call, Number) would result in an uninitialized array of length N.
Compatibility
Since this technique relies on behaviour of Function.prototype.apply() specified in ECMAScript 5, it will not work in pre-ECMAScript 5 browsers such as Chrome 14 and Internet Explorer 9.
Multiple ways using ES6
Using spread operator (...) and keys method
[ ...Array(N).keys() ].map( i => i+1);
Fill/Map
Array(N).fill().map((_, i) => i+1);
Array.from
Array.from(Array(N), (_, i) => i+1)
Array.from and { length: N } hack
Array.from({ length: N }, (_, i) => i+1)
Note about generalised form
All the forms above can produce arrays initialised to pretty much any desired values by changing i+1 to expression required (e.g. i*2, -i, 1+i*2, i%2 and etc). If expression can be expressed by some function f then the first form becomes simply
[ ...Array(N).keys() ].map(f)
Examples:
Array.from({length: 5}, (v, k) => k+1);
// [1,2,3,4,5]
Since the array is initialized with undefined on each position, the value of v will be undefined
Example showcasing all the forms
let demo= (N) => {
console.log(
[ ...Array(N).keys() ].map(( i) => i+1),
Array(N).fill().map((_, i) => i+1) ,
Array.from(Array(N), (_, i) => i+1),
Array.from({ length: N }, (_, i) => i+1)
)
}
demo(5)
More generic example with custom initialiser function f i.e.
[ ...Array(N).keys() ].map((i) => f(i))
or even simpler
[ ...Array(N).keys() ].map(f)
let demo= (N,f) => {
console.log(
[ ...Array(N).keys() ].map(f),
Array(N).fill().map((_, i) => f(i)) ,
Array.from(Array(N), (_, i) => f(i)),
Array.from({ length: N }, (_, i) => f(i))
)
}
demo(5, i=>2*i+1)
If I get what you are after, you want an array of numbers 1..n that you can later loop through.
If this is all you need, can you do this instead?
var foo = new Array(45); // create an empty array with length 45
then when you want to use it... (un-optimized, just for example)
for(var i = 0; i < foo.length; i++){
document.write('Item: ' + (i + 1) + ' of ' + foo.length + '<br/>');
}
e.g. if you don't need to store anything in the array, you just need a container of the right length that you can iterate over... this might be easier.
See it in action here: http://jsfiddle.net/3kcvm/
Arrays innately manage their lengths. As they are traversed, their indexes can be held in memory and referenced at that point. If a random index needs to be known, the indexOf method can be used.
This said, for your needs you may just want to declare an array of a certain size:
var foo = new Array(N); // where N is a positive integer
/* this will create an array of size, N, primarily for memory allocation,
but does not create any defined values
foo.length // size of Array
foo[ Math.floor(foo.length/2) ] = 'value' // places value in the middle of the array
*/
ES6
Spread
Making use of the spread operator (...) and keys method, enables you to create a temporary array of size N to produce the indexes, and then a new array that can be assigned to your variable:
var foo = [ ...Array(N).keys() ];
Fill/Map
You can first create the size of the array you need, fill it with undefined and then create a new array using map, which sets each element to the index.
var foo = Array(N).fill().map((v,i)=>i);
Array.from
This should be initializing to length of size N and populating the array in one pass.
Array.from({ length: N }, (v, i) => i)
In lieu of the comments and confusion, if you really wanted to capture the values from 1..N in the above examples, there are a couple options:
if the index is available, you can simply increment it by one (e.g., ++i).
in cases where index is not used -- and possibly a more efficient way -- is to create your array but make N represent N+1, then shift off the front.
So if you desire 100 numbers:
let arr; (arr=[ ...Array(101).keys() ]).shift()
In ES6 you can do:
Array(N).fill().map((e,i)=>i+1);
http://jsbin.com/molabiluwa/edit?js,console
Edit:
Changed Array(45) to Array(N) since you've updated the question.
console.log(
Array(45).fill(0).map((e,i)=>i+1)
);
Use the very popular Underscore _.range method
// _.range([start], stop, [step])
_.range(10); // => [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
_.range(1, 11); // => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
_.range(0, 30, 5); // => [0, 5, 10, 15, 20, 25]
_.range(0, -10, -1); // => [0, -1, -2, -3, -4, -5, -6, -7, -8, -9]
_.range(0); // => []
function range(start, end) {
var foo = [];
for (var i = start; i <= end; i++) {
foo.push(i);
}
return foo;
}
Then called by
var foo = range(1, 5);
There is no built-in way to do this in Javascript, but it's a perfectly valid utility function to create if you need to do it more than once.
Edit: In my opinion, the following is a better range function. Maybe just because I'm biased by LINQ, but I think it's more useful in more cases. Your mileage may vary.
function range(start, count) {
if(arguments.length == 1) {
count = start;
start = 0;
}
var foo = [];
for (var i = 0; i < count; i++) {
foo.push(start + i);
}
return foo;
}
the fastest way to fill an Array in v8 is:
[...Array(5)].map((_,i) => i);
result will be: [0, 1, 2, 3, 4]
Performance
Today 2020.12.11 I performed tests on macOS HighSierra 10.13.6 on Chrome v87, Safari v13.1.2 and Firefox v83 for chosen solutions.
Results
For all browsers
solution O (based on while) is the fastest (except Firefox for big N - but it's fast there)
solution T is fastest on Firefox for big N
solutions M,P is fast for small N
solution V (lodash) is fast for big N
solution W,X are slow for small N
solution F is slow
Details
I perform 2 tests cases:
for small N = 10 - you can run it HERE
for big N = 1000000 - you can run it HERE
Below snippet presents all tested solutions A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
function A(N) {
return Array.from({length: N}, (_, i) => i + 1)
}
function B(N) {
return Array(N).fill().map((_, i) => i+1);
}
function C(N) {
return Array(N).join().split(',').map((_, i) => i+1 );
}
function D(N) {
return Array.from(Array(N), (_, i) => i+1)
}
function E(N) {
return Array.from({ length: N }, (_, i) => i+1)
}
function F(N) {
return Array.from({length:N}, Number.call, i => i + 1)
}
function G(N) {
return (Array(N)+'').split(',').map((_,i)=> i+1)
}
function H(N) {
return [ ...Array(N).keys() ].map( i => i+1);
}
function I(N) {
return [...Array(N).keys()].map(x => x + 1);
}
function J(N) {
return [...Array(N+1).keys()].slice(1)
}
function K(N) {
return [...Array(N).keys()].map(x => ++x);
}
function L(N) {
let arr; (arr=[ ...Array(N+1).keys() ]).shift();
return arr;
}
function M(N) {
var arr = [];
var i = 0;
while (N--) arr.push(++i);
return arr;
}
function N(N) {
var a=[],b=N;while(b--)a[b]=b+1;
return a;
}
function O(N) {
var a=Array(N),b=0;
while(b<N) a[b++]=b;
return a;
}
function P(N) {
var foo = [];
for (var i = 1; i <= N; i++) foo.push(i);
return foo;
}
function Q(N) {
for(var a=[],b=N;b--;a[b]=b+1);
return a;
}
function R(N) {
for(var i,a=[i=0];i<N;a[i++]=i);
return a;
}
function S(N) {
let foo,x;
for(foo=[x=N]; x; foo[x-1]=x--);
return foo;
}
function T(N) {
return new Uint8Array(N).map((item, i) => i + 1);
}
function U(N) {
return '_'.repeat(5).split('').map((_, i) => i + 1);
}
function V(N) {
return _.range(1, N+1);
}
function W(N) {
return [...(function*(){let i=0;while(i<N)yield ++i})()]
}
function X(N) {
function sequence(max, step = 1) {
return {
[Symbol.iterator]: function* () {
for (let i = 1; i <= max; i += step) yield i
}
}
}
return [...sequence(N)];
}
[A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X].forEach(f=> {
console.log(`${f.name} ${f(5)}`);
})
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.20/lodash.min.js" integrity="sha512-90vH1Z83AJY9DmlWa8WkjkV79yfS2n2Oxhsi2dZbIv0nC4E6m5AbH8Nh156kkM7JePmqD6tcZsfad1ueoaovww==" crossorigin="anonymous"> </script>
This snippet only presents functions used in performance tests - it does not perform tests itself!
And here are example results for chrome
This question has a lot of complicated answers, but a simple one-liner:
[...Array(255).keys()].map(x => x + 1)
Also, although the above is short (and neat) to write, I think the following is a bit faster
(for a max length of:
127, Int8,
255, Uint8,
32,767, Int16,
65,535, Uint16,
2,147,483,647, Int32,
4,294,967,295, Uint32.
(based on the max integer values), also here's more on Typed Arrays):
(new Uint8Array(255)).map(($,i) => i + 1);
Although this solution is also not so ideal, because it creates two arrays, and uses the extra variable declaration "$" (not sure any way to get around that using this method). I think the following solution is the absolute fastest possible way to do this:
for(var i = 0, arr = new Uint8Array(255); i < arr.length; i++) arr[i] = i + 1;
Anytime after this statement is made, you can simple use the variable "arr" in the current scope;
If you want to make a simple function out of it (with some basic verification):
function range(min, max) {
min = min && min.constructor == Number ? min : 0;
!(max && max.constructor == Number && max > min) && // boolean statements can also be used with void return types, like a one-line if statement.
((max = min) & (min = 0)); //if there is a "max" argument specified, then first check if its a number and if its graeter than min: if so, stay the same; if not, then consider it as if there is no "max" in the first place, and "max" becomes "min" (and min becomes 0 by default)
for(var i = 0, arr = new (
max < 128 ? Int8Array :
max < 256 ? Uint8Array :
max < 32768 ? Int16Array :
max < 65536 ? Uint16Array :
max < 2147483648 ? Int32Array :
max < 4294967296 ? Uint32Array :
Array
)(max - min); i < arr.length; i++) arr[i] = i + min;
return arr;
}
//and you can loop through it easily using array methods if you want
range(1,11).forEach(x => console.log(x));
//or if you're used to pythons `for...in` you can do a similar thing with `for...of` if you want the individual values:
for(i of range(2020,2025)) console.log(i);
//or if you really want to use `for..in`, you can, but then you will only be accessing the keys:
for(k in range(25,30)) console.log(k);
console.log(
range(1,128).constructor.name,
range(200).constructor.name,
range(400,900).constructor.name,
range(33333).constructor.name,
range(823, 100000).constructor.name,
range(10,4) // when the "min" argument is greater than the "max", then it just considers it as if there is no "max", and the new max becomes "min", and "min" becomes 0, as if "max" was never even written
);
so, with the above function, the above super-slow "simple one-liner" becomes the super-fast, even-shorter:
range(1,14000);
Using ES2015/ES6 spread operator
[...Array(10)].map((_, i) => i + 1)
console.log([...Array(10)].map((_, i) => i + 1))
You can use this:
new Array(/*any number which you want*/)
.join().split(',')
.map(function(item, index){ return ++index;})
for example
new Array(10)
.join().split(',')
.map(function(item, index){ return ++index;})
will create following array:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
If you happen to be using d3.js in your app as I am, D3 provides a helper function that does this for you.
So to get an array from 0 to 4, it's as easy as:
d3.range(5)
[0, 1, 2, 3, 4]
and to get an array from 1 to 5, as you were requesting:
d3.range(1, 5+1)
[1, 2, 3, 4, 5]
Check out this tutorial for more info.
This is probably the fastest way to generate an array of numbers
Shortest
var a=[],b=N;while(b--)a[b]=b+1;
Inline
var arr=(function(a,b){while(a--)b[a]=a;return b})(10,[]);
//arr=[0,1,2,3,4,5,6,7,8,9]
If you want to start from 1
var arr=(function(a,b){while(a--)b[a]=a+1;return b})(10,[]);
//arr=[1,2,3,4,5,6,7,8,9,10]
Want a function?
function range(a,b,c){c=[];while(a--)c[a]=a+b;return c}; //length,start,placeholder
var arr=range(10,5);
//arr=[5,6,7,8,9,10,11,12,13,14]
WHY?
while is the fastest loop
Direct setting is faster than push
[] is faster than new Array(10)
it's short... look the first code. then look at all other functions in here.
If you like can't live without for
for(var a=[],b=7;b>0;a[--b]=b+1); //a=[1,2,3,4,5,6,7]
or
for(var a=[],b=7;b--;a[b]=b+1); //a=[1,2,3,4,5,6,7]
If you are using lodash, you can use _.range:
_.range([start=0], end, [step=1])
Creates an array of numbers
(positive and/or negative) progressing from start up to, but not
including, end. A step of -1 is used if a negative start is specified
without an end or step. If end is not specified, it's set to start
with start then set to 0.
Examples:
_.range(4);
// ➜ [0, 1, 2, 3]
_.range(-4);
// ➜ [0, -1, -2, -3]
_.range(1, 5);
// ➜ [1, 2, 3, 4]
_.range(0, 20, 5);
// ➜ [0, 5, 10, 15]
_.range(0, -4, -1);
// ➜ [0, -1, -2, -3]
_.range(1, 4, 0);
// ➜ [1, 1, 1]
_.range(0);
// ➜ []
the new way to filling Array is:
const array = [...Array(5).keys()]
console.log(array)
result will be: [0, 1, 2, 3, 4]
with ES6 you can do:
// `n` is the size you want to initialize your array
// `null` is what the array will be filled with (can be any other value)
Array(n).fill(null)
Very simple and easy to generate exactly 1 - N
const [, ...result] = Array(11).keys();
console.log('Result:', result);
Final Summary report .. Drrruummm Rolll -
This is the shortest code to generate an Array of size N (here 10) without using ES6. Cocco's version above is close but not the shortest.
(function(n){for(a=[];n--;a[n]=n+1);return a})(10)
But the undisputed winner of this Code golf(competition to solve a particular problem in the fewest bytes of source code) is Niko Ruotsalainen . Using Array Constructor and ES6 spread operator . (Most of the ES6 syntax is valid typeScript, but following is not. So be judicious while using it)
[...Array(10).keys()]
https://stackoverflow.com/a/49577331/8784402
With Delta
For javascript
smallest and one-liner
[...Array(N)].map((v, i) => from + i * step);
Examples and other alternatives
Array.from(Array(10).keys()).map(i => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
[...Array(10).keys()].map(i => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]
Array(10).fill(0).map((v, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array(10).fill().map((v, i) => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]
[...Array(10)].map((v, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Range Function
const range = (from, to, step) =>
[...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);
range(0, 9, 2);
//=> [0, 2, 4, 6, 8]
// can also assign range function as static method in Array class (but not recommended )
Array.range = (from, to, step) =>
[...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);
Array.range(2, 10, 2);
//=> [2, 4, 6, 8, 10]
Array.range(0, 10, 1);
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Array.range(2, 10, -1);
//=> []
Array.range(3, 0, -1);
//=> [3, 2, 1, 0]
As Iterators
class Range {
constructor(total = 0, step = 1, from = 0) {
this[Symbol.iterator] = function* () {
for (let i = 0; i < total; yield from + i++ * step) {}
};
}
}
[...new Range(5)]; // Five Elements
//=> [0, 1, 2, 3, 4]
[...new Range(5, 2)]; // Five Elements With Step 2
//=> [0, 2, 4, 6, 8]
[...new Range(5, -2, 10)]; // Five Elements With Step -2 From 10
//=>[10, 8, 6, 4, 2]
[...new Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]
// Also works with for..of loop
for (i of new Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2
As Generators Only
const Range = function* (total = 0, step = 1, from = 0) {
for (let i = 0; i < total; yield from + i++ * step) {}
};
Array.from(Range(5, -2, -10));
//=> [-10, -12, -14, -16, -18]
[...Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]
// Also works with for..of loop
for (i of Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2
// Lazy loaded way
const number0toInf = Range(Infinity);
number0toInf.next().value;
//=> 0
number0toInf.next().value;
//=> 1
// ...
From-To with steps/delta
using iterators
class Range2 {
constructor(to = 0, step = 1, from = 0) {
this[Symbol.iterator] = function* () {
let i = 0,
length = Math.floor((to - from) / step) + 1;
while (i < length) yield from + i++ * step;
};
}
}
[...new Range2(5)]; // First 5 Whole Numbers
//=> [0, 1, 2, 3, 4, 5]
[...new Range2(5, 2)]; // From 0 to 5 with step 2
//=> [0, 2, 4]
[...new Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]
using Generators
const Range2 = function* (to = 0, step = 1, from = 0) {
let i = 0,
length = Math.floor((to - from) / step) + 1;
while (i < length) yield from + i++ * step;
};
[...Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]
let even4to10 = Range2(10, 2, 4);
even4to10.next().value;
//=> 4
even4to10.next().value;
//=> 6
even4to10.next().value;
//=> 8
even4to10.next().value;
//=> 10
even4to10.next().value;
//=> undefined
For Typescript
class _Array<T> extends Array<T> {
static range(from: number, to: number, step: number): number[] {
return Array.from(Array(Math.floor((to - from) / step) + 1)).map(
(v, k) => from + k * step
);
}
}
_Array.range(0, 9, 1);
Solution for empty array and with just number in array
const arrayOne = new Array(10);
console.log(arrayOne);
const arrayTwo = [...Array(10).keys()];
console.log(arrayTwo);
var arrayThree = Array.from(Array(10).keys());
console.log(arrayThree);
const arrayStartWithOne = Array.from(Array(10).keys(), item => item + 1);
console.log(arrayStartWithOne)
✅ Simply, this worked for me:
[...Array(5)].map(...)
There is another way in ES6, using Array.from which takes 2 arguments, the first is an arrayLike (in this case an object with length property), and the second is a mapping function (in this case we map the item to its index)
Array.from({length:10}, (v,i) => i)
this is shorter and can be used for other sequences like generating even numbers
Array.from({length:10}, (v,i) => i*2)
Also this has better performance than most other ways because it only loops once through the array.
Check the snippit for some comparisons
// open the dev console to see results
count = 100000
console.time("from object")
for (let i = 0; i<count; i++) {
range = Array.from({length:10}, (v,i) => i )
}
console.timeEnd("from object")
console.time("from keys")
for (let i =0; i<count; i++) {
range = Array.from(Array(10).keys())
}
console.timeEnd("from keys")
console.time("apply")
for (let i = 0; i<count; i++) {
range = Array.apply(null, { length: 10 }).map(function(element, index) { return index; })
}
console.timeEnd("apply")
Fast
This solution is probably fastest it is inspired by lodash _.range function (but my is simpler and faster)
let N=10, i=0, a=Array(N);
while(i<N) a[i++]=i;
console.log(a);
Performance advantages over current (2020.12.11) existing answers based on while/for
memory is allocated once at the beginning by a=Array(N)
increasing index i++ is used - which looks is about 30% faster than decreasing index i-- (probably because CPU cache memory faster in forward direction)
Speed tests with more than 20 other solutions was conducted in this answer
Using new Array methods and => function syntax from ES6 standard (only Firefox at the time of writing).
By filling holes with undefined:
Array(N).fill().map((_, i) => i + 1);
Array.from turns "holes" into undefined so Array.map works as expected:
Array.from(Array(5)).map((_, i) => i + 1)
In ES6:
Array.from({length: 1000}, (_, i) => i).slice(1);
or better yet (without the extra variable _ and without the extra slice call):
Array.from({length:1000}, Number.call, i => i + 1)
Or for slightly faster results, you can use Uint8Array, if your list is shorter than 256 results (or you can use the other Uint lists depending on how short the list is, like Uint16 for a max number of 65535, or Uint32 for a max of 4294967295 etc. Officially, these typed arrays were only added in ES6 though). For example:
Uint8Array.from({length:10}, Number.call, i => i + 1)
ES5:
Array.apply(0, {length: 1000}).map(function(){return arguments[1]+1});
Alternatively, in ES5, for the map function (like second parameter to the Array.from function in ES6 above), you can use Number.call
Array.apply(0,{length:1000}).map(Number.call,Number).slice(1)
Or, if you're against the .slice here also, you can do the ES5 equivalent of the above (from ES6), like:
Array.apply(0,{length:1000}).map(Number.call, Function("i","return i+1"))
Array(...Array(9)).map((_, i) => i);
console.log(Array(...Array(9)).map((_, i) => i))
for(var i,a=[i=0];i<10;a[i++]=i);
a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
It seems the only flavor not currently in this rather complete list of answers is one featuring a generator; so to remedy that:
const gen = N => [...(function*(){let i=0;while(i<N)yield i++})()]
which can be used thus:
gen(4) // [0,1,2,3]
The nice thing about this is you don't just have to increment... To take inspiration from the answer #igor-shubin gave, you could create an array of randoms very easily:
const gen = N => [...(function*(){let i=0;
while(i++<N) yield Math.random()
})()]
And rather than something lengthy operationally expensive like:
const slow = N => new Array(N).join().split(',').map((e,i)=>i*5)
// [0,5,10,15,...]
you could instead do:
const fast = N => [...(function*(){let i=0;while(i++<N)yield i*5})()]
I have a number array [2, 1, 3, 4, 5, 1] and want to remove the smallest number in the list. But somehow my IF statement gets skipped.
I checked and by itself "numbers[i + 1]" and "numbers[i]" do work, but "numbers[i + 1] < numbers[i]" doesn't...
function removeSmallest(numbers) {
var smallestNumberKEY = 0;
for (i = 0; i <= numbers.lenths; i++) {
if (numbers[i + 1] < numbers[i]) {
smallestNumberKEY = i + 1;
}
}
numbers.splice(smallestNumberKEY, 1);
return numbers;
}
document.write(removeSmallest([2, 1, 3, 4, 5, 1]));
You have a typo in your code, array doesn't have lenths property
function removeSmallest(numbers) {
var smallestNumberKEY = 0;
for (var i = 0; i < numbers.length - 1; i++) {
if (numbers[i + 1] < numbers[i]) {
smallestNumberKEY = i + 1;
numbers.splice(smallestNumberKEY, 1);
}
}
return numbers;
}
document.write(removeSmallest([2, 1, 3, 4, 5, 1]));
But your algorithm wont work for another array, e.g [5, 3, 1, 4, 1], it will remove a value 3 too.
You can find the min value with Math.min function and then filter an array
function removeSmallest(arr) {
var min = Math.min(...arr);
return arr.filter(e => e != min);
}
You can use Array#filter instead
function removeSmallest(arr) {
var min = Math.min.apply(null, arr);
return arr.filter((e) => {return e != min});
}
console.log(removeSmallest([2, 1, 3, 4, 5, 1]))
Short one liner. If the smallest value exist multiple times it will only remove ONE. This may or may not be what you want.
const result = [6,1,3,1].sort().filter((_,i) => i) // result = [1,3,6]
It works by sorting and then creating a new array from the items where indeces are truthy(anything but 0)
another solution with splice and indexOf:
array = [2, 1, 3, 4, 5, 1];
function replace(arr){
arr = arr.slice(); //copy the array
arr.splice( arr.indexOf(Math.min.apply(null, arr)),1)
return arr;
}
document.write( replace(array) ,'<br> original array : ', array)
edit : making a copy of the array will avoid the original array from being modified
"Short" solution using Array.forEach and Array.splice methods:
function removeSmallest(numbers) {
var min = Math.min.apply(null, numbers);
numbers.forEach((v, k, arr) => v !== min || arr.splice(k,1));
return numbers;
}
console.log(removeSmallest([2, 1, 3, 4, 5, 1])); // [2, 3, 4, 5]
This is a proposal with a single loop of Array#reduce and without Math.min.
The algorithm sets in the first loop min with the value of the element and returns an empty array, because the actual element is the smallest value and the result set should not contain the smallest value.
The next loop can have
a value smaller than min, then assign a to min and return a copy of the original array until the previous element, because a new minimum is found and all other previous elements are greater than the actual value and belongs to the result array.
a value greater then min, then the actual value is pushed to the result set.
a value equal to min, then the vaue is skipped.
'use strict';
var removeSmallest = function () {
var min;
return function (r, a, i, aa) {
if (!i || a < min) {
min = a;
return aa.slice(0, i);
}
if (a > min) {
r.push(a);
}
return r;
}
}();
document.write('<pre>' + JSON.stringify([2, 1, 3, 2, 4, 5, 1].reduce(removeSmallest, []), 0, 4) + '</pre>');
I like this oneliner: list.filter(function(n) { return n != Math.min.apply( Math, list ) })
check it out here: https://jsfiddle.net/rz2n4rsd/1/
function remove_smallest(list) {
return list.filter(function(n) { return n != Math.min.apply( Math, list ) })
}
var list = [2, 1, 0, 4, 5, 1]
console.log(list) // [2, 1, 0, 4, 5, 1]
list = remove_smallest(list)
console.log(list) // [2, 1, 4, 5, 1]
list = remove_smallest(list)
console.log(list) // [2, 4, 5]
I had to do this but I needed a solution that did not mutate the input array numbers and ran in O(n) time. If that's what you're looking for, try this one:
const removeSmallest = (numbers) => {
const minValIndex = numbers.reduce((finalIndex, currentVal, currentIndex, array) => {
return array[currentIndex] <= array[finalIndex] ? currentIndex : finalIndex
}, 0)
return numbers.slice(0, minValIndex).concat(numbers.slice(minValIndex + 1))
}
function sumOfPaiars(ints){
var array = [];
var min = Math.min(...ints)
console.log(min)
for(var i=0;i<ints.length;i++){
if(ints[i]>min){
array.push(ints[i])
}
}
return array
}
If you only wish to remove a single instance of the smallest value (which was my use-case, not clear from the op).
arr.sort().shift()
Here is a piece of code that is work properly but is not accepted from codewars:
let numbers = [5, 3, 2, 1, 4];
numbers.sort(function numbers(a, b) {
return a - b;
});
const firstElement = numbers.shift();
The following was my interview question. But I couldn't crack it and even could not think how to get this done.
var arr = [1,4,5,8,3,2,6,9,7,10];
Expected output of alternate sorting:
[10,1,9,2,8,3,7,4,6,5]
What I have tried:
I tried slicing out the Math.max.apply(null,arr) and Math.min.apply(null,arr) alternatively to push into separate empty array. But It was told that the algorithm is not optimal.
I would sort the array, and then iterate it, picking values from the begining and the end (inloop calculated offsets), in each iteration. A final check to odd arrays would complete the process.
let a = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
a.sort((a, b) => a - b);
let b =[];
let l = a.length-1; // micro optimization
let L = l/2; // micro optimization
for(var i=0; i<L; i++) b.push( a[l-i] ,a[i] );
if(a.length%2) b.push( a[i] ); // add last item in odd arrays
console.log(b);
Result :
b = [10, 1, 9, 2, 8, 3, 7, 4, 6, 5]
Algorithm bennefits:
Avoiding alterations in the original array (through pop and shift), improves the performance considerably.
Precalculating l and L before the loop , prevents the need of being calculated repeatedly in each iteration.
A single conditional cheking at the end of the procces, to handle odd arrays, slightly improves the speed.
I've prepared some PERFORMANCE TESTS, with some of the proposed algorithms :
Original Array(10 items) and Big Array(1000 items)
Here is one way to do it:
var arr = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
// Sort the source array
arr.sort((a, b) => a - b);
// This will be the final result
var result = [];
// Create two pointers
var a = 0,
b = arr.length - 1;
while (result.length < arr.length) {
// Push the elements from start and end to the result array
result.push(arr[b]);
// Avoid bug when array is odd lengthed
if (a !== b) {
result.push(arr[a]);
}
a++;
b--;
}
console.log(result);
The idea is to have two pointers (a and b) traversing the the sorted original array from both the directions and appending the elements in result.
If you assume that the array will be a set of sequential numbers (a good question to ask about the data) you can do this very quickly with no need to sort or mutate the original array(i.e O(n)):
var arr = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
let a = arr.reduce((a, c, i) => {
a[c > arr.length >> 1 ? (arr.length - c) << 1 : (c << 1) - 1] = c
return a
}, [])
console.log(a)
Here's my answer, based off the intuition that you're taking from the front then the back repeatedly from the sorted array until you're empty. The trick is avoiding "max" and "min" which evaluate the entire array, and just sorting it once.
Many of the other answers will put an undefined into the array if the original array has an odd length. I would leave a comment on those but I do not have the reputation. This is why I bounds check twice per loop.
var arr = [1,4,5,8,3,2,6,9,7,10];
// Sort numerically (not lexicographically)
arr.sort((a, b) => a - b)
// The output array
var out = []
// Take from the front, then back until original array is empty
while (true) {
if (arr.length == 0) break
out.push(arr.pop())
if (arr.length == 0) break
out.push(arr.shift())
}
// Output answer
console.log(out)
My solution for readability / no hidden magic:
// Input
var arr = [1,4,5,8,3,2,6,9,7,10];
// Sort
var arr1 = arr.sort((a,b) => (a - b));
// Compose
var arr2 = [];
for (var i = 0; i < arr1.length; i++) {
arr2.push(arr1[(i % 2) === 0
? arr1.length-1-(i/2) // get from end half
: (i-1)/2 // get from begin half
])
}
// Output
console.log(arr2); // = [10, 1, 9, 2, 8, 3, 7, 4, 6, 5]
Their interview answer "that the algorithm is not optimal." is not unexpected ofcourse. I would inquire why they say that, and ask if its really benefitial to spend dollar time on dimes here. (or tens of dollars on cents, actually)
Alternative method with only one variable to increment:
var arr = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
arr = arr.sort((a, b) => b - a);
var result = [];
var a = 0;
while (result.length < arr.length) {
result.push(arr[a]);
result.push(arr[arr.length - a - 1]);
a++;
}
console.log(result);
var a = [1,4,5,8,3,2,6,9,7,10];
var b = a.sort((a, b) => a - b);
var c = a.sort((a, b) => a - b).reverse();
var d = [];
let e = a.length-1;
let f = e/2;
for(let i=0; i<f; i++) d.push( b.pop(), c.pop() );
Replace b and c in the for loop with functions to test:
for(let i=0; i<f; i++) d.push( a.sort((a, b) => a - b).pop(), a.sort((a, b) => a - b).reverse().pop() );
sort the array and divide into two parts , now use reduce to put elements from the two arrays
//original array
var arr = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
//sorting origina array in ascending order
var m = arr.sort(function(a, b) {
return a - b;
});
// diving the sorted array in two parts
let getFirstSet = m.splice(0, arr.length / 2);
// now m containleft over items after the splice
// again sorted it in descending order to avoid back looping
let getSecondSet = m.sort(function(a, b) {
return b - a;
});
//using reduce function
let newArray = getFirstSet.reduce(function(acc, curr, index) {
// pushing element from second array containing 10,9,8,7,6
acc.push(getSecondSet[index]);
// pushing element from first array containing 1,2,3,4,5
acc.push(getFirstSet[index]);
return acc;
}, []); // [] is the initial array where elements will be pushed
console.log(newArray)
Another alternative view ... should this funky sort be done in place, like .sort is?
let input = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
input.sort((a, b) => b - a).every((n, i, a) => (a.splice((i * 2 + 1), 0, a.pop()), (i * 2) < a.length));
console.log(input);
Here is a quick solution, using ternary operators and modulo operator for toggling.
let arr = [1, 4, 5, 8, 3, 2, 6, 9, 7, 10];
let j = 0;
let k = arr.length - 1;
// sort array
arr.sort((a, b) => a - b);
let new_array = [];
for (let i in arr) {
new_array[i] = i % 2 == 0 ? arr[k--] : arr[j++];
}
// prints array
console.log(new_array);