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JavaScript: Rearrange an array in order – largest, smallest, 2nd largest, 2nd smallest, 3rd largest, 3rd smallest,

Given an array of integers, where the values should be sorted in the following order:
if we have an array
[1, -1, -3, 9, -2, -5, 4, 8,]
we must rearrange it this way: largest number, smallest number, 2nd largest number, 2nd smallest number, ...
[9, -5, 8, -3, 4, -2, 1, -1 ]
I get the first largest and smallest numbers, but can't figure out how to make it dynamic for all values in the array.
I know that I must take two variables, say firstSmallest and firstLargest and point them to the first and last index of the array respectively, run a loop, which I do already in the code below, and store value into new array by incrementing firstSmallest and decrementing firstLargest, but couldn't implement into code.
let unsortedArr = [1, 5, 8 , 7, 6, -1, -5, 4, 9, 5]
let output = [];
function meanderArray(unsorted){
let sorted = unsorted.sort((a, b) => a-b);
let firstSmallest = sorted[0];
let firstLargest = sorted[unsorted.length-1];
for(let i = 0; i <= sorted.length; i++){
//I should increment firstSmallest and decrement firstLargest numbers and store in output
}
return output;
}
meanderArray(unsortedArr);
console.log(output);

You could take a toggle object which takes the property of either the first item or last from an array and iterate until no more items are available.
function meanderArray([...array]) {
const
result = [],
toggle = { shift: 'pop', pop: 'shift' };
let fn = 'shift';
array.sort((a, b) => a - b);
while (array.length) result.push(array[fn = toggle[fn]]());
return result;
}
console.log(...meanderArray([1, 5, 8, 7, 6, -1, -5, 4, 9, 5]));

You can sort an array by descending, then logic is the following: take first from start and first from end, then second from start-second from end, etc.
let unsortedArr = [1, 5, 8 , 7, 6, -1, -5, 4, 9, 5]
let output = [];
function meanderArray(unsorted){
let sorted = unsorted.sort((a, b) => b-a);
let output = []
for(let i = 0; i < sorted.length/2; i++){
output.push(sorted[i])
if(i !== sorted.length - 1 - i){
output.push(sorted[sorted.length - 1 - i])
}
}
return output;
}
let result = meanderArray(unsortedArr);
console.log(result);

You can sort, then loop and extract the last number with pop() and extract the first number with shift().
let unsortedArr = [1, -1, -3, 9, -2, -5, 4, 8,]
let output = [];
function meanderArray(unsorted){
let sorted = unsorted.sort((a, b) => a - b);
for(let i = 0; i < unsortedArr.length + 2; i++){
output.push(sorted.pop());
output.push(sorted.shift());
}
console.log(output);
return output;
}
meanderArray(unsortedArr);

Fastest Meandering Array method among all solutions mentioned above.
According to the JSBench.me, this solution is the fastest and for your reference i have attached a screenshot below.
I got a different approach, but i found that was very close to one of above answers from elvira.genkel.
In my solution for Meandering Array, First I sorted the given array and then i tried to find the middle of the array. After that i divided sorted array in to two arrays, which are indices from 0 to middle index and other one is from middle index to full length of sorted array.
We need to make sure that first half of array's length is greater than the second array. Other wise when applying for() loop as next step newly created array will contains some undefined values. For avoiding this issue i have incremented first array length by one.
So, always it should be firstArr.length > secondArr.length.
And planned to create new array with values in meandering order. As next step I created for() loop and try to push values from beginning of the first array and from end of the second array. Make sure that dynamically created index of second array will receive only zero or positive index. Other wise you can find undefined values inside newly created Meandering Array.
Hope this solution will be helpful for everyone, who loves to do high performance coding :)
Your comments and suggestions are welcome.
const unsorted = [1, 5, 8, 7, 6, -1, -5, 4, 9, 5];
const sorted = unsorted.sort((a,b)=>a-b).reverse();
const half = Math.round(Math.floor(sorted.length/2)) + 1;
const leftArr = sorted.slice(0, half);
const rightArr = sorted.slice(half, sorted.length);
const newArr = [];
for(let i=0; i<leftArr.length; i++) {
newArr.push(leftArr[i]);
if (rightArr.length-1-i >= 0) {
newArr.push(rightArr[rightArr.length-1-i]);
}
}

How to ignore a negative number in Javascript?

I understand that Math.abs is used to convert negative numbers to absolute in JavaScript, but say I want to filter out negative numbers in an array for example, how do I do that? Is there a function for that?

This should solve your issue [1, 2, -3 , 4, -5].filter(val => val > 0);

If you want to eliminate the values, #Antoni 's approach will do.
However, if you want to retain elements but bound the values (also known as clamping) I usually use Math.max(0, number); to set a lower numerical bound - if number is <0, it returns 0.
You can apply this to an array using map:
var array = [1, -2, 3];
array = array.map(number => Math.max(0, number));
//Sets array == [1, 0, 3]
Similarly for an upper bound you can use Math.min(upperLimit, number); and of course you can apply them both for upper and lower bounds:
var array = [1, -2, 3];
const lowerLimit = 0;
const upperLimit = 2;
array = array.map(number => Math.max(lowerLimit, Math.min(upperLimit, number)));
//Sets array == [1, 0, 2]

Javascript reduce function

Hello I have an array as follows:
[[1,1],[2,8],[3,12],[4,13],[5,23],[6,30],[7,41],[8,44],[9,50]]
So lets say it is the format of [[x,y]]
As you can see the above array, the first element inside inner arrays goes from 1,2...6 which is x, and the second element of inner arrays are 1,8,12,13,23 and 30 which is y.
This is a cumulative array of y and I want to convert it into non-cumulative but with a twist.
I want to convert y to non-cumulative array by getting the difference from the value of y at last 3rd value of x
Therefore I want the final result to be as follows:
[[1,1],[2,8],[3,0],[4,1],[5,11],[6,0],[7,11],[8,14],[9,0]]
I have tried a fiddle here:
https://jsfiddle.net/gxytx6ka/1/
So far I have been able to get the difference between two array elements by some help of Dekel (stackoverflow guy) as you can see in the above fiddle:
$(document).ready(function() {
var old_arr = [ [1, 1], [2, 8], [3, 12], [4, 13], [5, 23], [6, 30], [7, 41], [8, 44], [9, 50] ];
var new_arr = old_arr.reduce(function(a, b, c, d) {
if (a.length) {
a.push([d[c][0], d[c][1] - d[c - 1][1]])
} else {
a = [b]
}
return a;
}, []);
console.log(new_arr);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
which outputs:
[[1,1],[2,7],[3,4],[4,1],[5,10],[6,7],[7,11],[8,3],[9,6]]
but I want:
[[1,1],[2,8],[3,0],[4,1],[5,11],[6,0],[7,11],[8,14],[9,0]]
which I am not able to get.
I tried using a%3==0 in if condition, but not working...
Basically I want the final output in an efficient way.

Why reduce? A simple loop does the job perfectly fine:
let a = [[1,1],[2,8],[3,12],[4,13],[5,23],[6,30],[7,41],[8,44],[9,50]];
let out = [],
n = 0,
lastY = 0;
for (let [x, y] of a) {
if (++n % 3 == 0)
lastY = y;
out.push([x, y - lastY]);
}
console.log(JSON.stringify(out))

You could use Array#map, which returns simply an array with the result.
var array = [[1,1], [2,8], [3,12], [4,13], [5,23], [6,30], [7,41], [8,44], [9,50]],
result = array.map(function (a, i) {
(i + 1) % 3 || (this.last = a[1]);
return [a[0], a[1] - this.last];
}, { last: 0 });
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
ES6
var array = [[1,1], [2,8], [3,12], [4,13], [5,23], [6,30], [7,41], [8,44], [9,50]],
result = array.map((l => ([x, y], i) => (++i % 3 || (l = y), [x, y - l]))(0));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

It is possible to do this with reduce — You just need to figure out when to zero-out every third value and when to find the difference between the current value and it's previous (third) value.
Naming parameters with more intelligible names also helps out.
var old_arr = [ [1, 1], [2, 8], [3, 12], [4, 13], [5, 23], [6, 30], [7, 41], [8, 44], [9, 50] ];
var new_arr = old_arr.reduce(function(result, value, index, arr) {
if (index % 3 == 2) {
return result.concat([[value[0], 0]]); // Set every third value to 0.
} else {
var prevIndex = index - (index % 3) - 1;
var prevVal = prevIndex > -1 ? arr[prevIndex][1] : 0;
return result.concat([[value[0], Math.abs(prevVal - value[1])]]);
}
}, []);
console.log(JSON.stringify(new_arr)); // [[1,1],[2,8],[3,0],[4,1],[5,10],[6,0],[7,11],[8,3],[9,0]]

With reduce you can do it like this (also using ES6 arrow function and spread operator):
var arr = [[1,1], [2,8], [3,12], [4,13], [5,23], [6,30], [7,41], [8,44], [9,50]];
var result = arr.reduce( ([arr, v], [x, y], i) =>
[[...arr,[x, i%3==2 ? 0 : y-v]], i%3==2 ? y : v], [[],0])[0];
console.log(result);
Explanation
reduce is called with a double initial value (a pair):
[[],0]
The first element is the array that will accumulate into the final result, and the second value is the current value to which subsequent y values will be offset.
These two values are captured by the argument [arr, v] which is a destructuring assignment (ES6). The reduce callback will always return a new pair like that, with extended arr and (potentially) updated v.
The callback function also takes the current [x, y] pair from the original array. Again this is a destructuring assignment, so you have direct access to x and y variables. The third argument i is the current, zero-based index in the original array.
As said, the callback returns a pair. The first element in that pair is constructed as follows:
[...arr, [x, i%3==2 ? 0 : y-v]]
The ...arr notation spreads the previously collected result elements as a list, and one pair is added to that list: [x, i%3==2 ? 0 : y-v]. The x is just reproduced from the original [x, y], but the y follows the requested logic with the ternary operator: if we are at an index that has a remainder of 2 when divided by 3, then the y value should be 0. Otherwise, it should be offset against the previously set value v.
The second element in the pair, must be the new value for v:
i%3==2 ? y : v
Again, according to the requested logic, v remains unchanged when the remainder is not 2, but otherwise it is set to the current value of y.
So reduce will thus accumulate/update these two values in each iteration, and finally return a pair, of which only the first has our interest, which is why there is [0] at the end.
Notes
As you seemed to be looking for a reduce implementation, that is what I went with, but whenever your output array has just as many elements as your input array (like is the case here) you can also consider to use map as a good alternative (See answer that NinaScholz posted).
If you are open for a less functional programming way, you can also choose to use a for loop and maybe gain a bit of performance.

The right way to create an array with N range? [duplicate]

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})()]

Find the number in an array that is closest to a given number

I have an Array of integers in javascript, [5,10,15,20,25,30,35]
when given a number x, how can I find the element in the array that is closest to that number?
If the number is over a value, but less than halfway to the next number, I would choose the smaller value, if it were over halfway to the next number, I would choose the higher number.
For example 7 would return 5, but 8 would return 10. How can I accomplish this? Any help or tips would be appreciated. I have searched and cannot find a solution. I'm sure this is sort of common.

Probably the easiest thing to do is sort based on distance from the reference value x, and then take the first item.
The built-in Array.prototype.sort() can take a comparison function which will be called for pairs of values from the array. Then the key is simply to pass in a comparison function which compares the two values based on their distance from the reference value x.
let x = 8;
let array = [5, 10, 15, 20, 25, 30, 35];
let closest = array.sort( (a, b) => Math.abs(x - a) - Math.abs(x - b) )[0];
See this simple demo.

function getClosest(array, target) {
var tuples = _.map(array, function(val) {
return [val, Math.abs(val - target)];
});
return _.reduce(tuples, function(memo, val) {
return (memo[1] < val[1]) ? memo : val;
}, [-1, 999])[0];
}
If using a functional approach is applicable then you can map the set to tuples of (value, distance) then reduce that set of tuples to the tuple with the smallest distance. We return the value in that tuple.
To explain the useage of _.map. You map all the values in your array to new values and the function will return the array of new values. In this case an array of tuples.
To explain the useage of _.reduce. You reduce the array to a single value. You pass in an array and a memo. The memo is your "running counter" as you move through the array. In this case we check whether the current tuple is closer then the memo and if so make it the memo. We then return the memo at the end.
The code snippet above relies on underscore.js to remove the nitty gritty of functional style javascript

Your example list is sorted. If this is always the case, then binary search for your number. If you don't find the exact number, make the binary search end off by checking the two numbers around where the number would be and return the closest. Be careful with edge cases where all numbers are greater or are all smaller than the target number
If the list isn't always sorted, then go through the list keeping track of the largest number <= the target number and the smallest number >= the target number. Return the one that's closest to the target.
In either solution, you'll need to decide which side to favour if for example you're searching for 2 in [1, 3].

Create a temporary array of the same size as your original array, and populate it with the differences between your x and the array element.
For example, let the temporary array be temp[], and your original array be a[]:
temp[i]=Math.abs(x-a[i]);
Then, return the index of the minimum value in temp[] to the user.

Assuming the array is sorted, step through each adjacent pair of integers in the array. For each pair (say "5 and 10" or "20 and 25"), test if x is in between them, and if so, return whichever one is closer to x (with a bias towards the lower one).
You would also need a special case for when x is less than the first number (return the first number) or greater than the last number (return the last number).
If the array is not sorted, sort it first.

I created my own function since i could not find any that meets my requeriments.
function closest_number(quantities, number, closest_factor)
{
if (closest_factor == 'ceil')
{
quantities.sort(function(a, b)
{
return a - b
}
);
for (var i = 0; i < quantities.length; i++)
{
if (quantities[i] >= number)
{
return quantities[i];
}
last_value = quantities[i];
}
return last_value;
}
else if (closest_factor == 'floor')
{
quantities.sort(function(a, b)
{
return a - b
}
);
min_value = quantities[0];
for (var i = 0; i < quantities.length; i++)
{
if (number == quantities[i])
{
return number;
}
else if (quantities[i] < number)
{
min_value = quantities[i];
}
else if(quantities[i] > number)
{
return min_value;
}
}
return min_value;
}
else
{
return false;
}
};

Since Array.reduce is a reality for so long (even IE9 supports it), the problem is easily solvable with it. This way, no need to sort the array first (no array mutation at all):
var numbers = [20, 25, 30, 35, 5, 10, 15], x = 7;
var output = numbers.reduce(function (prev, curr) {
return Math.abs(curr - x) < Math.abs(prev - x) ? curr : prev
});
console.log(output);
You can go further and solve it with only one line of ES6 (ECMAScript 2015) syntax, by using an arrow function (but with no IE support in this case):
const numbers = [20, 25, 30, 35, 5, 10, 15], x = 7;
const output = numbers.reduce((prev, curr) => Math.abs(curr - x) < Math.abs(prev - x) ? curr : prev);
console.log(output);
Of course, for flexibility and reusability, it's easy to make it as a function:
const closest = (array, goal) => array.reduce((prev, curr) => Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
console.log(closest([20, 25, 30, 35, 5, 10, 15], 7));
console.log(closest([20, 25, 30, 35, 5, 10, 15], 8));
console.log(closest([1, 5, 7], -5));
console.log(closest([1, 5, 7], 4));
console.log(closest([1, 5, 7], 20));