Populating a 2D array in Javascript with random numbers

Question

I'm trying to populate a 2D array in javascript with random numbers. Although each column in the array is random, each row is identical which is not what I want (see image below). I want both rows and columns to be random.

http://eeldesigns.com/image.jpg

cols = 5;
rows = 10;

front = new Array(cols).fill(new Array(rows));

// Loop through Initial array to randomly place cells
for(var x = 0; x < cols; x++){
  for(var y = 0; y < rows; y++){
    front[x][y] = Math.floor(Math.random()*5);
  }
}
console.table(front) ;

Possible duplicate of Array.prototype.fill() with object passes reference and not new instance — Sebastian Simon
– Sebastian Simon, Commented Feb 4, 2019 at 20:04
.fill(new Array(rows)) creates a single array instance and assigns it to each column. So when you think you modify only one you actually modify all of them. It's as if you were doing this: jsfiddle.net/ejqc4z8x — Alexey Lebedev
– Alexey Lebedev, Commented Feb 4, 2019 at 20:18

Code Maniac · Accepted Answer · 2019-02-04 20:27:15Z

2

One way of doing this using map

let op = new Array(10)
         .fill(0)
         .map(e=>(new Array(5)
         .fill(0)
         .map(e=> Math.floor(Math.random() * 5))))

console.log(op)

edited Feb 4, 2019 at 20:27

answered Feb 4, 2019 at 20:08

Code Maniac

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2 Comments

Ismail Over a year ago

thanks for the tip - I wasn't familiar with maps and their usage.

Code Maniac Over a year ago

@Ismail always happy to help :) you can read about map on given link in answer.

Scott Sauyet · Accepted Answer · 2019-02-04 20:33:56Z

1

The trouble is that you're not initializing the row. It's easily fixed:

cols = 5;
rows = 10;

front = new Array(cols)// .fill(new Array(rows));

// Loop through Initial array to randomly place cells
for(var x = 0; x < cols; x++){
  front[x] = [];  // ***** Added this line *****
  for(var y = 0; y < rows; y++){
    front[x][y] = Math.floor(Math.random()*5);
  }
}
console.table(front) ; // browser console only, not StackOverflow's

Update

This is a cleaner version, somewhat similar to the one from Code Maniac, but simplified a bit:

const randomTable = (rows, cols) => Array.from(
  {length: rows}, 
  () => Array.from({length: cols}, () => Math.floor(Math.random() * 5))
)

console.table(randomTable(10, 5)) // browser console only, not StackOverflow's

edited Feb 4, 2019 at 20:33

answered Feb 4, 2019 at 20:15

Scott Sauyet

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5 Comments

Ismail Over a year ago

Thanks this was exactly what I was looking for.

Ismail Over a year ago

Just a quick question - does not the 'new Array()' initalise the row?

Scott Sauyet Over a year ago

Note that although this shows where you went wrong, the answer from Code Maniac is a better solution.

Scott Sauyet Over a year ago

No, and I should have commented that out before. You're passing one value to fill, which is used to fill all the rows with the same empty array. That's one reason the other answer is cleaner.

Scott Sauyet Over a year ago

Added a cleaner version.

Alex · Accepted Answer · 2020-07-05 22:15:24Z

0

This can be accomplished using a combination of Array.prototype.fill() and Array.prototype.map():

new Array(rows).fill([]).map(x => Array(columns).fill(0).map(x => x + Math.floor(Math.random() * (max - min)) + min));

For example, we can create a 100 by 964 column array full of random numbers between 900 and 1000 using the following:

new Array(100).fill([]).map(x => Array(964).fill(0).map(x => x + Math.floor(Math.random() * (1000 - 900)) + 900));

answered Jul 5, 2020 at 22:15

Alex

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