shuffle (rearrange randomly) a List<string>

Question

I need to rearrange my List array, it has a non-determinable number of elements in it.

Can somebody give me example of how i do this, thanks

Solution 1

List<Foo> source = ...
var rnd = new Random();
var result = source.OrderBy(item => rnd.Next());

Obviously if you want real randomness instead of pseudo-random number generator you could use RNGCryptoServiceProvider instead of Random.

Solution 2

This is an extension method that will shuffle a List<T>:

    public static void Shuffle<T>(this IList<T> list) {
        int n = list.Count;
        Random rnd = new Random();
        while (n > 1) {
            int k = (rnd.Next(0, n) % n);
            n--;
            T value = list[k];
            list[k] = list[n];
            list[n] = value;
        }
    }

Comments

[chillitom]

No.. this is a bad idea, read this: blogs.msdn.com/b/ericlippert/archive/2011/01/31/…

[Sanjeevakumar Hiremath]

Is there a real randmoness? I thought it is impossible.

[Darin Dimitrov]

@Sanjeevakumar Hiremath, no there is no real randomness indeed. But with RNGCryptoServiceProvider at least the randomness should be sufficient :-)

[jp2code]

+1 I've not tested this in my code (i.e. not sure if it works), but I like it so far! Does every element get shuffled, or is that not necessary?

[kprobst]

The whole list is shuffled, yes. Note that for larger lists you might want to integrate a more random random generator (Darin's answer below) into the solution. I use it for smaller sets (say ~300 tops) and it works well enough.

[Kevin]

You can just say int k = rnd.Next(0, n). The % is unnecessary since the dividend is always less than n.

[Jon Hanna]

Looks like a sound Fisher-Yates implementation, which is what I was going to offer.

[Jon Hanna]

@chillitom, no this doesn't have the problem in that article. In Eric's article, the randomness is poorly seeded, and also the comparison is breaking the total ordering requirement of comparison methods. In Darin's example here, each item is assigned a random number once, and then ordered according to it. I think kprobst's solution is still better (well-known efficient shuffling algorithm), but Darin's is correct too.

[o0'.]

Bad idea, see comments here.

[Piotr Kula]

Doesnt seem to work. Always get a list in the same order? :(

[Veverke]

same as @ppumkin here.

[ErikE]

There is a problem with this implementation! Initializing a new Random on each call will yield predictable shuffling results. Instead, place private static readonly Random rnd = new Random(); into the static class this method sits in. Finally, it would have been nice if you had given your source for this answer credit, because then the flaw in your own post might have been discovered sooner before people went off and used it without any hint it needed fixing.

[Joshua Grosso Reinstate CMs]

@ppumpkin @Veverke This doesn't work as-is, because rnd is set to new Random() each time (i.e. with the same seed). Basically, set rnd at the beginning of your program and then never assign to it again (make it static or something). Then, the only line you actually need at shuffle-time is the first and the third.

[Veverke]

@ BalinKingOfMoria: thanks, but I must have tried it like you say, since I am used to Random and know that I need to give it a seed and use the same instance. I do not remember what I tried then, ... sounds strange I tried it as it is. Thanks, anyway.

[TaW]

I am used to Random and know that I need to give it a seed yes, give it a seed to create always the same sequence. don't give it a seed to create uniques sequences.

[nishantvodoo]

stackoverflow.com/questions/273313/randomize-a-listt

[tolache]

What is item in this example? Please forgive noobness.