Snake Eyes Snake Eyes - 1 month ago 9
C# Question

How to combine more than two generic lists in C# Zip?

I have three (it's possible to have more than 3-4 generic list, but in this example let 3) generic lists.

List<string> list1

List<string> list2

List<string> list3

all lists have same number of elements (same counts).

I used that for combining two lists with ZIP :

var result = list1.Zip(list2, (a, b) => new {
test1 = f,
test2 = b

I used that for
statement, to avoid
each List, like

foreach(var item in result){
Console.WriteLine(item.test1 + " " + item.test2);

How to use simmilary with Zip for three lists ?



I want like:

List<string> list1 = new List<string>{"test", "otherTest"};

List<string> list2 = new List<string>{"item", "otherItem"};

List<string> list3 = new List<string>{"value", "otherValue"};

after ZIP (I don't know method), I want to result (in VS2010 debug mode)

[0] { a = {"test"},
b = {"item"},
c = {"value"}

[1] { a = {"otherTest"},
b = {"otherItem"},
c = {"otherValue"}

How to do that ?

Answer Source

The most obvious way for me would be to use Zip twice.

For example,

var results = l1.Zip(l2, (x, y) => x + y).Zip(l3, (x, y) => x + y);

would combine (add) the elements of three List<int> objects.


You could define a new extension method that acts like a Zip with three IEnumerables, like so:

public static class MyFunkyExtensions
    public static IEnumerable<TResult> ZipThree<T1, T2, T3, TResult>(
        this IEnumerable<T1> source,
        IEnumerable<T2> second,
        IEnumerable<T3> third,
        Func<T1, T2, T3, TResult> func)
        using (var e1 = source.GetEnumerator())
        using (var e2 = second.GetEnumerator())
        using (var e3 = third.GetEnumerator())
            while (e1.MoveNext() && e2.MoveNext() && e3.MoveNext())
                yield return func(e1.Current, e2.Current, e3.Current);

The usage (in the same context as above) now becomes:

var results = l1.ZipThree(l2, l3, (x, y, z) => x + y + z);

Similarly, you three lists can now be combined with:

var results = list1.ZipThree(list2, list3, (a, b, c) => new { a, b, c });