How to iterate through two IEnumerables simultaneo

问题:

I have two enumerables: IEnumerable<A> list1 and IEnumerable<B> list2. I would like to iterate through them simultaneously like:

foreach((a, b) in (list1, list2))
{
    // use a and b
}

If they don't contain the same number of elements, an exception should be thrown.

What is the best way to do this?

回答1:

Here's an implementation of this operation, typically called Zip:

using System;
using System.Collections.Generic;

namespace SO2721939
{
    public sealed class ZipEntry<T1, T2>
    {
        public ZipEntry(int index, T1 value1, T2 value2)
        {
            Index = index;
            Value1 = value1;
            Value2 = value2;
        }

        public int Index { get; private set; }
        public T1 Value1 { get; private set; }
        public T2 Value2 { get; private set; }
    }

    public static class EnumerableExtensions
    {
        public static IEnumerable<ZipEntry<T1, T2>> Zip<T1, T2>(
            this IEnumerable<T1> collection1, IEnumerable<T2> collection2)
        {
            if (collection1 == null)
                throw new ArgumentNullException("collection1");
            if (collection2 == null)
                throw new ArgumentNullException("collection2");

            int index = 0;
            using (IEnumerator<T1> enumerator1 = collection1.GetEnumerator())
            using (IEnumerator<T2> enumerator2 = collection2.GetEnumerator())
            {
                while (enumerator1.MoveNext() && enumerator2.MoveNext())
                {
                    yield return new ZipEntry<T1, T2>(
                        index, enumerator1.Current, enumerator2.Current);
                    index++;
                }
            }
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            int[] numbers = new[] { 1, 2, 3, 4, 5 };
            string[] names = new[] { "Bob", "Alice", "Mark", "John", "Mary" };

            foreach (var entry in numbers.Zip(names))
            {
                Console.Out.WriteLine(entry.Index + ": "
                    + entry.Value1 + "-" + entry.Value2);
            }
        }
    }
}

To make it throw an exception if just one of the sequences run out of values, change the while-loop so:

while (true)
{
    bool hasNext1 = enumerator1.MoveNext();
    bool hasNext2 = enumerator2.MoveNext();
    if (hasNext1 != hasNext2)
        throw new InvalidOperationException("One of the collections ran " +
            "out of values before the other");
    if (!hasNext1)
        break;

    yield return new ZipEntry<T1, T2>(
        index, enumerator1.Current, enumerator2.Current);
    index++;
}

回答2:

You want something like the Zip LINQ operator - but the version in .NET 4 always just truncates when either sequence finishes.

The MoreLINQ implementation has an EquiZip method which will throw an InvalidOperationException instead.

var zipped = list1.EquiZip(list2, (a, b) => new { a, b });

foreach (var element in zipped)
{
    // use element.a and element.b
}

回答3:

In short, the language offers no clean way to do this. Enumeration was designed to be done over one enumerable at a time. You can mimic what foreach does for you pretty easily:

using(IEnumerator<A> list1enum = list1.GetEnumerator())
using(IEnumerator<B> list2enum = list2.GetEnumerator())    
while(list1enum.MoveNext() && list2enum.MoveNext()) {
        // list1enum.Current and list2enum.Current point to each current item
    }

What to do if they are of different length is up to you. Perhaps find out which one still has elements after the while loop is done and keep working with that one, throw an exception if they should be the same length, etc.

回答4:

In .NET 4, you can use the .Zip extension method on IEnumerable<T>

IEnumerable<int> list1 = Enumerable.Range(0, 100);
IEnumerable<int> list2 = Enumerable.Range(100, 100);

foreach (var item in list1.Zip(list2, (a, b) => new { a, b }))
{
    // use item.a and item.b
}

It won't throw on unequal lengths, however. You can always test that, though.

回答5:

Go with IEnumerable.GetEnumerator, so you can move around the enumerable. Note that this might have some really nasty behavior, and you must be careful. If you want to get it working, go with this, if you want to have maintainable code, use two foreach.

You could create a wrapping class or use a library (as Jon Skeet suggests) to handle this functionality in a more generic way if you are going to use it more than once thru your code.

The code for what I suggest:

var firstEnum = aIEnumerable.GetEnumerator();
var secondEnum = bIEnumerable.GetEnumerator();

var firstEnumMoreItems = firstEnum.MoveNext();
var secondEnumMoreItems = secondEnum.MoveNext();    

while (firstEnumMoreItems && secondEnumMoreItems)
{
      // Do whatever.  
      firstEnumMoreItems = firstEnum.MoveNext();
      secondEnumMoreItems = secondEnum.MoveNext();   
}

if (firstEnumMoreItems || secondEnumMoreItems)
{
     Throw new Exception("One Enum is bigger");
}

// IEnumerator does not have a Dispose method, but IEnumerator<T> has.
if (firstEnum is IDisposable) { ((IDisposable)firstEnum).Dispose(); }
if (secondEnum is IDisposable) { ((IDisposable)secondEnum).Dispose(); }

回答6:

using(var enum1 = list1.GetEnumerator())
using(var enum2 = list2.GetEnumerator())
{
    while(true)
    {
        bool moveNext1 = enum1.MoveNext();
        bool moveNext2 = enum2.MoveNext();
        if (moveNext1 != moveNext2)
            throw new InvalidOperationException();
        if (!moveNext1)
            break;
        var a = enum1.Current;
        var b = enum2.Current;
        // use a and b
    }
}

回答7:

You can do something like this.

IEnumerator enuma = a.GetEnumerator();
IEnumerator enumb = b.GetEnumerator();
while (enuma.MoveNext() && enumb.MoveNext())
{
    string vala = enuma.Current as string;
    string valb = enumb.Current as string;
}

C# has no foreach that can do it how you want (that I am aware of).

回答8:

Use the Zip function like

foreach (var entry in list1.Zip(list2, (a,b)=>new {First=a, Second=b}) {
    // use entry.First und entry.Second
}

This doesn't throw an exception, though ...