Question

I want to shuffle two lists, such that they are shuffled in the same way (provided that I have a method Shuffle(List list) that shuffles one list.

List<ObjX> listA = new List<ObjX>() { A, B, C, D };
List<ObjX> listB = new List<ObjX>() { W, X, Y, Z };
ShuffleTwoLists(listA , listB )

Result:
A: C, B, D, A
B: Y, X, Z, W

Is there a quick way to implement ShuffleTwoLists(listA , listB) in C#?

Answer 1

Option 1: zip, shuffle, unzip

To expand on Marcelo’s comment, and assuming you don’t mind creating new lists instead of shuffling the existing ones:

var zipped = listA.Zip(listB, (a, b) => new { a, b } ).ToList();
Shuffle(zipped);
var newListA = zipped.Select(pair => pair.a).ToList();
var newListB = zipped.Select(pair => pair.b).ToList();

Option 2: shuffle the indexes

To expand on MAK’s answer with code:

var indexes = Enumerable.Range(0, listA.Count).ToList();
Shuffle(indexes);
var newListA = indexes.Select(index => listA[index]).ToList();
var newListB = indexes.Select(index => listB[index]).ToList();

Of course both approaches could mutate the original lists, with a bit more work.

Option 3: shuffle both lists with the same random seed

I personally like to pass a Random (or whatever) into methods/classes which need them instead of creating new ones. So I’d give my Shuffle a Random parameter. It avoids various problems, and expresses the dependency nicely. You can use this to your advantage, by creating two Random instances with the same seed:

int seed = existingRandom.Next();
Shuffle(listA, new Random(seed));
Shuffle(listB, new Random(seed));

Assuming Shuffle does the same thing when given the same sequence of random numbers, this will shuffle both lists in the same way.