Is `OrderBy(x => r.Next())` a Smart Way to Shuffle a List?

<.> r.next ()) `a sart way to shuffle a list? " />

var r = new Random();
var shuffled = ordered.OrderBy(x => r.Next());

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This method is not an ideal shuffle method. Although it generates the only random number for each element, it is an O (N LOG N) operation, and there is a more effective O (N) algorithm. <原> Working principle

In essence, this method is distributed to elements randomly and sorted them based on these numbers. This ensures that each element appears in different positions, but the location is actually determined by the generated random number.

<替> Alternative method

It is recommended to use the Fisher-Yates shuffle algorithm of the DURSTENFELD version, which directly exchange elements. Can be implemented using such extensions like this:

<能> Performance optimization

In order to further optimize performance, the elements can be returned immediately when shuffling, thereby reducing unnecessary work: (This code is exactly the same as the previous code, repeated)

public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source, Random rng)
{
    T[] elements = source.ToArray();
    for (int i = elements.Length - 1; i >= 0; i--)
    {
        int swapIndex = rng.Next(i + 1);
        yield return elements[swapIndex];
        elements[swapIndex] = elements[i];
    }
}

<说> Important description

Must use the right Random instance to avoid generating the same digital sequence multiple times and keep thread security.

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