在JavaScript中使用随机数创建长度为n的数组

Question

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]

Answer 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

Answer 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/

Answer 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!

Answer 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; 

Answer 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] 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] 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));

...

Answer 6

Solution from Sized array of random unique numbers

 const uniqRandomNumbers = _.sampleSize(_.range(9), 4); console.log(uniqRandomNumbers); 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.min.js"></script>