Hello! It would seem trivial, but I can't think of how to make a function that will accept as parameter the number of returned elements, and the array from which we will isolate them (sorry)!

! ! ! I wrote a implementation of this function, BUT it has a disadvantage , which I urge you to consider

The fact that this function in 1ВЫХ may return fewer , and in OF 2, every time she sees the elements of the array with 0го index, and the chance that in a variable

```
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
function getRandom( n, array ) { // n is the number of random elements from array
// . . . code which will allow you to obtain a certain number of elements from the array
};
getRandom( 3, arr ); // must succeed for example [4, 6, 7] or [1, 4, 6]
```

! ! ! I wrote a implementation of this function, BUT it has a disadvantage , which I urge you to consider

```
let arr = [1,2,3,4,5,6,7,8,9,10];
getRandomRiver function(n, a){
let result = [];
for(let k=0; k<a.length; k++){
if(!Math.the round(Math.random())) result.push(a[k]); // Math.the round(Math.random()) returns either 0 or 1, if 0 then we add this element to the array
if(result.length>=n) break;
};
console.log(result);
};
getRandomRiver(3, arr);
```

The fact that this function in 1ВЫХ may return fewer , and in OF 2, every time she sees the elements of the array with 0го index, and the chance that in a variable

`result`

will get the elements closer to the end of the array is very small, and I need a full Rand covering the full array! Thank you in advance! asked April 19th 20 at 12:48

2 answers

answered on

Solution

```
const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
getRandomRiver function(n, arr) {
const indexes = new Set();
const limit = arr.length;
n = Math.min(n, limit);
while (indexes.size < n) {
const index = Math.floor(limit * Math.random());
indexes.add(index);
}
const result = [...indexes].map(index => arr[index]);
return result;
};
const randomSet = getRandomRiver(3, arr);
console.log(randomSet);
```

answered on

Solution

`const getRandomRiver = (n, arr) => arr.sort(() => 0.5 - Math.random()).slice(0, n);`

or

```
const getRandomRiver = (n, arr) => {
return arr
.map(x => ({ x, r: Math.random() }))
.sort((a, b) => a.r - b.r)
.map(a => a.x)
.slice(0, n);
}
```

Hmm, cool, peretasoval array and return the first three elements are, huh! God, I feel stupid! The second example I have not yet dismantled, I think in my case and the first will be limited! Damn well, how is it then, well, it's damn easy !=// commented on April 19th 20 at 12:55

@haven_Corke, this is a VERY suboptimal solution commented on April 19th 20 at 12:58

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`[indexes...]`

, this is equivalent to`Array.from(indexes)`

.If you talk, why the solution @kaylie sub-optimal, then all the matter in the average number of operations that will be produced during operation of the algorithm.

Let the initial array will have

`N`

elements. Then when sorting occursat least`N · log(N)`

operations. It is still possible to add`N`

operations on every call`.map()`

.In the end, if an array of 1000 elements it is necessary to choose 3 items, it will be absolutely approximately 12,000 operations.

In the proposed embodiment, the number of operations depends not on the number of elements in the array, and the number of elements that have to choose. And if you re to consider an array of 1000 elements from which to choose 3 elements, the algorithm, proposed by me, will spend a total of 12 operations. - malika.Boyer92 commented on April 19th 20 at 12:56