CODEWITHSUNDEEP
javascriptarrays
Tomkay
Tomkay

Reputation: 5160

Javascript | Set all values of an array

Code

var cool = new Array(3);
cool[setAll] = 42; //cool[setAll] is just a pseudo selector..
alert(cool);

Result

A alert message:

42,42,42

How do I change/set all values of an array to a specific value?

Upvotes: 44

Views: 148291

Answers (9)

Reza Saadati
Reza Saadati

Reputation: 5419

You can use an empty statement.

const arr = [1, 2 ,3];
for (let i = 0; i < arr.length; arr[i++] = 42);
console.log(arr);

Upvotes: 0

greenthings
greenthings

Reputation: 1

Found this while working with Epicycles - clearly works - where 'p' is invisible to my eyes.

/** Convert a set of picture points to a set of Cartesian coordinates */
function toCartesian(points, scale) {
  const x_max = Math.max(...points.map(p=>p[0])),
  y_max = Math.max(...points.map(p=>p[1])),
  x_min = Math.min(...points.map(p=>p[0])),
  y_min = Math.min(...points.map(p=>p[1])),
  signed_x_max = Math.floor((x_max - x_min + 1) / 2),
  signed_y_max = Math.floor((y_max - y_min + 1) / 2);

  return points.map(p=>
  [ -scale * (signed_x_max - p[0] + x_min),
  scale * (signed_y_max - p[1] + y_min) ] );
}

Upvotes: 0

user8608046
user8608046

Reputation:

It's 2019 and you should be using this:

let arr = [...Array(20)].fill(10)

So basically 20 is the length or a new instantiated Array.

Upvotes: 4

Сергей Язев
Сергей Язев

Reputation: 33

Actually, you can use this perfect approach:

let arr = Array.apply(null, Array(5)).map(() => 0);
// [0, 0, 0, 0, 0]

This code will create array and fill it with zeroes. Or just:

let arr = new Array(5).fill(0)

Upvotes: 3

mustafa.0x
mustafa.0x

Reputation: 1506

map is the most logical solution for this problem.

let xs = [1, 2, 3];
xs = xs.map(x => 42);
xs // -> [42, 42, 42]

However, if there is a chance that the array is sparse, you'll need to use for or, even better, for .. of.

See:

Upvotes: 29

Faris Zacina
Faris Zacina

Reputation: 14274

The ES6 approach is very clean. So first you initialize an array of x length, and then call the fill method on it.

let arr = new Array(3).fill(9)

this will create an array with 3 elements like:

[9, 9, 9]

Upvotes: 48

Alnitak
Alnitak

Reputation: 339816

There's no built-in way, you'll have to loop over all of them:

function setAll(a, v) {
    var i, n = a.length;
    for (i = 0; i < n; ++i) {
        a[i] = v;
    }
}

http://jsfiddle.net/alnitak/xG88A/

If you really want, do this:

Array.prototype.setAll = function(v) {
    var i, n = this.length;
    for (i = 0; i < n; ++i) {
        this[i] = v;
    }
};

and then you could actually do cool.setAll(42) (see http://jsfiddle.net/alnitak/ee3hb/).

Some people frown upon extending the prototype of built-in types, though.

EDIT ES5 introduced a way to safely extend both Object.prototype and Array.prototype without breaking for ... in ... enumeration:

Object.defineProperty(Array.prototype, 'setAll', {
    value: function(v) {
        ...
    }
});

EDIT 2 In ES6 draft there's also now Array.prototype.fill, usage cool.fill(42)

Upvotes: 59

RobG
RobG

Reputation: 147403

The other answers are Ok, but a while loop seems more appropriate:

function setAll(array, value) {
  var i = array.length;
  while (i--) {
    array[i] = value;
  }
}

A more creative version:

function replaceAll(array, value) {
  var re = new RegExp(value, 'g');
  return new Array(++array.length).toString().replace(/,/g, value).match(re);
}

May not work everywhere though. :-)

Upvotes: 1

Quentin
Quentin

Reputation: 943569

Use a for loop and set each one in turn.

Upvotes: 3

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