2

I have an array

var numbers = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18"];

and trying to get random items from it, so:

console.log(_.sample(numbers, 5));

this will give me 5 random numbers (strings) from the array in a random order, like:

"17", "2", "3", "18", "10"

How do I get a sorted list or random items, like?

"2", "3", "10", "17", "18"

_.sample will probably not be the best choice here. I am trying to get random items from a given array and have these items picked up from left to right of the array.

How to do this in javascritp?

Thank you.

EDIT: I have an array of strings, not numbers, so I cannot sort the randomly picked items.

EDIT2: To avoid confusing, in the array are words (= strings), I used there numbers as strings to more easily demonstrate what I am trying to achieve. (sorry for possible confusions)

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  • Strings that contain numbers can easily be sorted numerically too. Or are the strings of numbers just an example? Commented Apr 24, 2016 at 0:37
  • In the example above, I used numbers there because of easier demonstration of what I want to do - in the array are strings (words). Commented Apr 24, 2016 at 0:43

7 Answers 7

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3

You can use Array.prototype.sort to sort the returned array:

ie.

_.sample(numbers, 5).sort(function(a, b) { return parseInt(a, 10) - parseInt(b, 10) })

A better random would be:

var randomChoice = numbers[~~(Math.random() * numbers.length)]

Note: the ~~ performs the same action as Math.floor() in this context. They can be interchanged.

All together:

var numbers = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18"];

var randomSample = []

for(var i=0; i < 5; i++) {
   var randomChoice = numbers[~~(Math.random() * numbers.length)]
   randomSample.push(randomChoice)
}

var sortedRandomSample = randomSample.sort(function(a, b) { return parseInt(a, 10) - parseInt(b, 10) })

Demo: https://jsbin.com/zosizefaga/edit?html,js,console,output

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5 Comments

That for loop could potentially pick the same element five times.
You're right. Thats the nature of random choice. @nnnnnn.
But that's not what the OP wants. They clearly want five different elements selected randomly.
@nnnnnn if OP asks for it, this can be changed slightly to remove chosen elements from the numbers array to ensure all the choices are different.
note: if you need to add notes to your code it isn't readable enough :D, I'd go with Math.floor instead of trying to be too clever
2

Here is a solution that doesn't make any assumption about the original order. The idea is to lookup the element's position in the original array and sort by that. However, that assumes that every element is unique.

sample.sort(function(a, b) {
    return numbers.indexOf(a) - numbers.indexOf(b);
});

This will also be quite slow for large arrays.

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1 Comment

I think the complexity would be O(n² log n). If you wanted to improve that, you could make a hash that uses the strings as keys and the indexes as values. If you do this before the sort and use the hash rather than indexOf, it makes it O(n log n) again.
0

Why don't you implement your own method of sample and after calling _.sample you call the method sort?

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1 Comment

That seems more like a comment than an answer.
0

The easiest method I can think of is as follows:

var randomSample = _.sample(numbers.map(function(v,i){ return i; }), 5)
                    .sort(function(a,b){ return a-b; })
                    .map(function(v){ return numbers[v]; });

That is, make a temporary array that holds the indices of the original array, i.e., just the numbers 0 through numbers.length - 1):

var indices = numbers.map(function(v,i){ return i; })

Take a random sample from that array:

var sampleIndices = _.sample(indices, 5)

Sort the sample:

sampleIndices.sort(function(a,b){ return a-b; })

Then use the sorted, randomly selected indices to get values out of the original array:

var randomSample = sampleIndices.map(function(v){ return numbers[v]; });

And as shown at the beginning of my answer, you can do it all in one line without using the indices and sampleIndices variables. Although if you are going to be regularly taking samples from the same numbers array it would probably make sense to keep the indices variable to save rebuilding it every time, especially if the original array is quite large.

This will work regardless of what type of values are in the original array, because those values are just selected out at the end once random indices have been selected.

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0

Try this:

function random(array, elements) {
   return array.concat().sort(function() {
     if (Math.random() < 0.5) {
       return -1;
     } else {
       return 1;
     }
   }).slice(0, elements).sort(
     function(a, b) {
       return a - b
     });
}

Here's the fiddle:

JSFiddle

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3 Comments

But...this modifies the original array, and it returns the selected elements in a random order. The OP doesn't want to do either of those things.
It doesn't modify the original array, check the fiddle. And yes it was returning a random order array, my mistake.
It does modify the original array. Add console.log(numbers); after you call your function and you'll see.
0

This is a proposal without sorting and uses an helper array random for the selected items.

First get an empty array, then fill with true until count elements are filled and the filter the original array with the random selected positions.

This solution works for any content of the given array, without sort or lookup with indexOf.

function getSortedRandom(array, count) {
    var random = array.map(function () { return false; }),
        r;

    while (count) {
        r = Math.floor(Math.random() * array.length);
        if (!random[r]) {
            random[r] = true;
            count--;
        }
    }
    return array.filter(function (_, i) {
        return random[i];
    });
}

var random = getSortedRandom(["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18"], 5);

document.write('<pre>' + JSON.stringify(random, 0, 4) + '</pre>');

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Comments

0

As of lodash 4.0.0, you can use the _.sampleSize function in conjunction with sort:

var numbers = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18"];

var randomSample = _.sampleSize(numbers, 5).sort();

console.log(randomSample);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.14.1/lodash.min.js"></script>

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