使用组合器lamba在C#中合并2个列表

Question

我需要以特定方式合并两个列表,如下面的函数所述.这个实现使用递归和工作但似乎kludgy.有没有人知道用LINQ做更好的方法,似乎应该有类似的东西SelectMany可以参考外部(unflattened)元素,但我找不到任何东西

/// <summary>
/// Function merges two list by combining members in order with combiningFunction
/// For example   (1,1,1,1,1,1,1) with 
///               (2,2,2,2)       and a function that simply adds
/// will produce  (3,3,3,3,1,1,1)
/// </summary>
public static IEnumerable<T> MergeList<T>(this IEnumerable<T> first, 
                                          IEnumerable<T> second, 
                                          Func<T, T, T> combiningFunction)
{
    if (!first.Any())
        return second;

    if (!second.Any())
        return first;

    var result = new List<T> {combiningFunction(first.First(), second.First())};
    result.AddRange(MergeList<T>(first.Skip(1), second.Skip(1), combiningFunction));

    return result;
}

Answer 1

Enumerable.Zip 正是你想要的.

var resultList = Enumerable.Zip(first, second,
// or, used as an extension method:  first.Zip(second,
    (f, s) => new
              {
                  FirstItem = f,
                  SecondItem = s,
                  Sum = f + s
              });

编辑:似乎我没有说明即使一个列表完成,拉链的"外部"风格仍在继续.这是一个解决方案:

public static IEnumerable<TResult> OuterZip<TFirst, TSecond, TResult>(
    this IEnumerable<TFirst> first, IEnumerable<TSecond> second,
    Func<TFirst, TSecond, TResult> resultSelector)
{
    using (IEnumerator<TFirst> firstEnumerator = first.GetEnumerator())
    using (IEnumerator<TSecond> secondEnumerator = second.GetEnumerator())
    {
        bool firstHasCurrent = firstEnumerator.MoveNext();
        bool secondHasCurrent = secondEnumerator.MoveNext();

        while (firstHasCurrent || secondHasCurrent)
        {
            TFirst firstValue = firstHasCurrent
                ? firstEnumerator.Current
                : default(TFirst);

            TSecond secondValue = secondHasCurrent
                ? secondEnumerator.Current
                : default(TSecond);

            yield return resultSelector(firstValue, secondValue);

            firstHasCurrent = firstEnumerator.MoveNext();
            secondHasCurrent = secondEnumerator.MoveNext();
        }
    }
}

可以很容易地修改此函数以将布尔值传递给结果选择器函数,以表示是否存在第一个或第二个元素,如果需要显式检查(而不是使用default(TFirst)或default(TSecond)在lambda中).