Creating array of length n with random numbers in

问题:

Following up on this answer for creating an array of specified length, I executed the below to get a corresponding result but filled with random numbers, instead of zeros.

var randoms = Array(4).fill(Math.floor(Math.random() * 9));

Well, mathematically speaking it's random, all right. But I'd like the randomness to be visible within the array and not only between runs, of course. Stupid computer... Don't do what I say. Do what I want!

I can iterate and put it my random (and varying) values. But I wonder, of pure curiosity, if it's possible to get the right result with a one-liner, like above, MatLab-style. Do I need to call eval(function()...)? I've heard a lot of bad things about eval...

Note that the above produces code like this:

[7, 7, 7, 7]
[3, 3, 3, 3]

etc. while I want something like

[1, 2, 3, 4]
[4, 3, 8, 4]

回答1:

What does Array#fill do?

According to MDN

The fill() method fills all the elements of an array from a start index to an end index with a static value.

You can use Function#apply, Array#map, Math.floor(), Math.random().

In ES6, Array#from and Arrow function can be used.

Array.from({length: 6}, () => Math.floor(Math.random() * 9));

Array.apply(null, Array(6)).map(() => Math.floor(Math.random() * 9));

var randomArr = Array.from({length: 6}, () => Math.floor(Math.random() * 9));

document.getElementById('result').innerHTML = JSON.stringify(randomArr, 0, 4); // For demo only

<pre id="result"></pre>

In ES5:

Array.apply(null, Array(6)).map(function(item, index){
    return Math.floor(Math.random() * 9);
});

var randomArr = Array.apply(null, Array(6)).map(function(item, index){
    return Math.floor(Math.random() * 9)
});

document.getElementById('result').innerHTML = JSON.stringify(randomArr, 0, 4);

<pre id="result"></pre>

What is Array.apply(null, Array(n))? Can new Array(n) used here?

Both the above code create new array of six elements, each element having value as undefined. However, when used new syntax, the created array is not iterable. To make the array iterable, Array.apply(null, Array(6)) syntax is used.

---

If you have lodash included on page, it's really easy.

_.times(6, _.random.bind(0, 100))
        ^                        - Number of elements in array
                         ^       - Random number range min
                            ^^^  - Random number range max

Note: This answer is inspired from Colin Toh's blog

回答2:

var randoms = Array(4).fill(Math.floor(Math.random() * 9));

This line of code will create a list of 4 of the same number because fill takes a single value and repeats it for the length of the list. What you want to do is run the random number generator each time:

var makeARandomNumber = function(){
    return Math.floor(Math.random() * 9);
}
var randoms = Array(5).fill(0).map(makeARandomNumber);
console.log(randoms)
// => [4, 4, 3, 2, 6]

https://jsfiddle.net/t4jtjcde/

回答3:

Short and simple ES6 approach -

// randomly generated n = 4 length array 0 <= array[n] <= 9
var randoms = Array.from({length: 4}, () => Math.floor(Math.random() * 10));

Enjoy!

回答4:

I wonder if it's possible to get the right result with a one-liner...

var randoms = [...Array(4)].map(虚 => Math.floor(Math.random() * 9));

document.body.innerText = randoms;

回答5:

`const t = Array.from({length: n}, mapArrowFx);

1) const t10 = Array.from({length: 10}, (v, k) => k); [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

2) const tEven = Array.from({length: 10}, (v, k) => 2*k); [0, 2, 4, 6, 8, 10, 12, 14, 16, 18]

........

3)

let n=100; const tRandom= Array.from({length: n}, (v, k) => Math.floor(Math.random()*n));

...