How do you easily create empty matrices javascript?
JavascriptJavascript Problem Overview
In python, you can do this:
[([None] * 9) for x in range(9)]
and you'll get this:
[[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None],
[None, None, None, None, None, None, None, None, None]]
How can I do the equivalent in javascript?
Javascript Solutions
Solution 1 - Javascript
Array.fill
Consider using fill:
Array(9).fill().map(()=>Array(9).fill())
The idea here is that fill() will fill out the items with undefined, which is enough to get map to work on them.
You could also fill directly:
Array(9).fill(Array(9))
Alternatives to Array(9).fill() include
Array(...Array(9))
[].push(...Array(9))
[].concat(Array(9))
Array.from(Array(9))
We can rewrite the solution a bit more semantically as:
function array9() { return Array(9).fill(); }
array9().map(array9)
or
function array(n) { return Array(n).fill(); }
array(9).map(() => array(9))
Array.from provides us with an optional second mapping argument, so we have the alternative of writing
Array.from(Array(9), () => Array.from(Array(9));
or, if you prefer
function array9(map) { return Array.from(Array(9), map); }
array9(array9);
For verbose description and examples, see Mozilla's Docs on Array.prototype.fill() here.
and for Array.from(), here.
Note that neither Array.prototype.fill() nor Array.from() has support in Internet Explorer. A polyfill for IE is available at the above MDN links.
Partitioning
partition(Array(81), 9)
if you have a partition utility handy. Here's a quick recursive one:
function partition(a, n) {
return a.length ? [a.splice(0, n)].concat(partition(a, n)) : [];
}
Looping
We can loop a bit more efficiently with
var a = [], b;
while (a.push(b = []) < 9) while (b.push(null) < 9);
Taking advantage of the fact that push returns the new array length.
Solution 2 - Javascript
var matrix = [];
for(var i=0; i<9; i++) {
matrix[i] = new Array(9);
}
... or:
var matrix = [];
for(var i=0; i<9; i++) {
matrix[i] = [];
for(var j=0; j<9; j++) {
matrix[i][j] = undefined;
}
}
Solution 3 - Javascript
There is something about Array.fill I need to mention.
If you just use below method to create a 3x3 matrix.
Array(3).fill(Array(3).fill(0));
You will find that the values in the matrix is a reference.
Optimized solution (prevent passing by reference):
If you want to pass by value rather than reference, you can leverage Array.map to create it.
Array(3).fill(null).map(() => Array(3).fill(0));
Solution 4 - Javascript
// initializing depending on i,j:
var M=Array.from({length:9}, (_,i) => Array.from({length:9}, (_,j) => i+'x'+j))
// Print it:
console.table(M)
// M.forEach(r => console.log(r))
document.body.innerHTML = `<pre>${M.map(r => r.join('\t')).join('\n')}</pre>`
// JSON.stringify(M, null, 2) // bad for matrices
Beware that doing this below, is wrong:
// var M=Array(9).fill([]) // since arrays are sparse
// or Array(9).fill(Array(9).fill(0))// initialization
// M[4][4] = 1
// M[3][4] is now 1 too!
Because it creates the same reference of Array 9 times, so modifying an item modifies also items at the same index of other rows (since it's the same reference), so you need an additional call to .slice or .map on the rows to copy them (cf torazaburo's answer which fell in this trap)
note: It may look like this in the future, with Number.range() proposal
const M = [...Number.range(1,10)].map(i => [...Number.range(1,10)].map(j => i+'x'+j))
Solution 5 - Javascript
If you really like one-liners and there is a use for underscore.js in your project (which is a great library) you can do write-only things like:
_.range(9).map(function(n) {
return _.range(9).map(function(n) {
return null;
});
});
But I would go with standard for-cycle version mentioned above.
Solution 6 - Javascript
This is an exact fix to your problem, but I would advise against initializing the matrix with a default value that represents '0' or 'undefined', as Arrays in javascript are just regular objects, so you wind up wasting effort. If you want to default the cells to some meaningful value, then this snippet will work well, but if you want an uninitialized matrix, don't use this version:
/**
* Generates a matrix (ie: 2-D Array) with:
* 'm' columns,
* 'n' rows,
* every cell defaulting to 'd';
*/
function Matrix(m, n, d){
var mat = Array.apply(null, new Array(m)).map(
Array.prototype.valueOf,
Array.apply(null, new Array(n)).map(
function() {
return d;
}
)
);
return mat;
}
Usage:
< Matrix(3,2,'dobon');
> Array [ Array['dobon', 'dobon'], Array['dobon', 'dobon'], Array['dobon', 'dobon'] ]
If you would rather just create an uninitialized 2-D Array, then this will be more efficient than unnecessarily initializing every entry:
/**
* Generates a matrix (ie: 2-D Array) with:
* 'm' columns,
* 'n' rows,
* every cell remains 'undefined';
*/
function Matrix(m, n){
var mat = Array.apply(null, new Array(m)).map(
Array.prototype.valueOf,
new Array(n)
);
return mat;
}
Usage:
< Matrix(3,2);
> Array [ Array[2], Array[2], Array[2] ]
Solution 7 - Javascript
The question is slightly ambiguous, since None can translate into either undefined or null. null is a better choice:
var a = [], b;
var i, j;
for (i = 0; i < 9; i++) {
for (j = 0, b = []; j < 9; j++) {
b.push(null);
}
a.push(b);
}
If undefined, you can be sloppy and just don't bother, everything is undefined anyway. :)
Solution 8 - Javascript
For a 2-d matrix I'd do the following
var data = Array(9 * 9).fill(0);
var index = (i,j) => 9*i + j;
//any reference to an index, eg. (3,4) can be done as follows
data[index(3,4)];
You can replace 9 with any generic ROWS and COLUMNS constants.
Solution 9 - Javascript
Well, you can create an empty 1-D array using the explicit Array constructor:
a = new Array(9)
To create an array of arrays, I think that you'll have to write a nested loop as Marc described.
Solution 10 - Javascript
Coffeescript to the rescue!
[1..9].map -> [1..9].map -> null
Solution 11 - Javascript
Here's one, no looping:
(Math.pow(10, 20)+'').replace((/0/g),'1').split('').map(parseFloat);
Fill the '20' for length, use the (optional) regexp for handy transforms and map to ensure datatype. I added a function to the Array prototype to easily pull the parameters of 'map' into your functions.. bit risky, some people strongly oppose touching native prototypes, but it does come in handy..
Array.prototype.$args = function(idx) {
idx || (idx = 0);
return function() {
return arguments.length > idx ? arguments[idx] : null;
};
};
// Keys
(Math.pow(10, 20)+'').replace((/0/g),'1').split('').map(this.$args(1));
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
// Matrix
(Math.pow(10, 9)+'').replace((/0/g),'1').split('').map(this.$args(1)).map(this.$args(2))
Solution 12 - Javascript
better. that exactly will work.
let mx = Matrix(9, 9);
function Matrix(w, h){
let mx = Array(w);
for(let i of mx.keys())
mx[i] = Array(h);
return mx;
}
what was shown
Array(9).fill(Array(9)); // Not correctly working
It does not work, because all cells are fill with one array
Solution 13 - Javascript
JavaScript doesn’t have a built-in 2D array concept, but you can certainly create an array of arrays.
function createMatrix(row, column, isEmpty) {
let matrix = []
let array = []
let rowColumn = row * column
for (let i = 1; i <= rowColumn; i++) {
isEmpty ? array.push([]) : array.push(i)
if (i % column === 0) {
matrix.push(array)
array = []
}
}
return matrix
}
createMatrix(5, 3, true)
or
function createMatrix(row, column, from) {
let [matrix, array] = [[], []],
total = row * column
for (let element = from || 1; element <= total; element++) {
array.push(element)
if (element % column === 0) {
matrix.push(array)
array = []
}
}
return matrix
}
createMatrix(5, 6, 1)
Solution 14 - Javascript
Use this function or some like that. :)
function createMatrix(line, col, defaultValue = 0){
return new Array(line).fill(defaultValue).map((x)=>{ return new Array(col).fill(defaultValue); return x; });
}
var myMatrix = createMatrix(9,9);
Solution 15 - Javascript
I'll give it my shot as well
var c = Array;
for( var i = 0, a = c(9); i < 9; a[i] = c(9), i++ );
console.log( a.join(",") );
//",,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,"
Readable and maintainable !
Solution 16 - Javascript
Array.from(new Array(row), () => new Array(col).fill(0));
Solution 17 - Javascript
You can add functionality to an Array by extending its prototype object.
Array.prototype.nullify = function( n ) {
n = n >>> 0;
for( var i = 0; i < n; ++i ) {
this[ i ] = null;
}
return this;
};
Then:
var arr = [].nullify(9);
or:
var arr = [].nullify(9).map(function() { return [].nullify(9); });