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Discussion on: How to shuffle an array in JavaScript

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Replies for: Is the first method really flawed? Also, rather than Math.random(), I can always use crypto.getRandomValues() developer.mozilla.org/en-US/docs/W...

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codebubb profile image
James Bubb •

Yeah, apparently - I think it boils down to a question of sort() having a strong bias for keeping elements in their initial positions (there's a bit of a discussion about it on one of the answers here: stackoverflow.com/questions/245095...).

Thanks for the tip regarding .getRandomValues(). How would that work with strings or objects?

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patarapolw profile image
Pacharapol Withayasakpunt • • Edited

I read the solution part of javascript.info/array-methods#shuf..., but not really understand the logic, or is it the fault of JS Engine?

For .getRandomValues(), there are two more methods of shuffling, but of course, not performant, even if your first method is not faulty.

function shuffle1(arr: any[]): any[] {
    return Array(arr.length).fill(null)
        .map((_, i) => [Math.random(), i])
        .sort(([a], [b]) => a - b)
        .map(([, i]) => arr[i])
}

function shuffle2(arr: any[]): any[] {
    const ord = new Uint8Array(arr.length)
    crypto.getRandomValues(ord)

    return ord
        .map((a, i) => [a, i])
        .sort(([a], [b]) => a - b)
        .map(([, i]) => arr[i])
}
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codebubb profile image
James Bubb •

Haha! I like this from that site: But due to the utter randomness of the comparison the black box goes mad, and how exactly it goes mad depends on the concrete implementation that differs between engines.

So yes, I think it's down to the JavaScript runtime itself where the problem lies.

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tvogel profile image
Tilman Vogel •

Well, the generic Array.sort() algorithm puts several requirements on the comparison function, see developer.mozilla.org/en-US/docs/W... which in turn it may use to implement the sorting algorithm. In the "Custom sort" approach above, in particular the requirement of the comparison to be "stable" is violated.
So, you are breaking the Array.sort() contract and may receive any result. In the worst case, the algorithm might not even terminate.