Using Linq Aggregate to return array of values

Question

I have something like this:

long[] f = new long[4]{1,10,100,1000};

I want to divide 1000 by 100, 100 by 10 and 10 by 1

Is there a way to return results in an array with the results eg/ 10,10,10

UPDATE: This seems to confuse a few so here is another example

long[] f = new long[3]{1,2,6};

I want to divide 6 by 2 and 2 by 1 with the results in an array

Answer 1

I don't think aggregate is gonna help in that case... Zip would be better suited:

long[] f = new long[4]{1,10,100,1000};
long[] result = f.Skip(1).Zip(f, (a, b) => a / b);

---

EDIT: if you're on .NET 3.5, you can easily write a Zip extension method yourself:

    public static IEnumerable<TResult> Zip<TFirst, TSecond, TResult>(this IEnumerable<TFirst> first, IEnumerable<TSecond> second, Func<TFirst, TSecond, TResult> selector)
    {

        if (first == null)
            throw new ArgumentNullException("first");
        if (second == null)
            throw new ArgumentNullException("second");
        if (selector == null)
            throw new ArgumentNullException("selector");

        return first.ZipIterator(second, selector);
    }

    private static IEnumerable<TResult> ZipIterator<TFirst, TSecond, TResult>(this IEnumerable<TFirst> first, IEnumerable<TSecond> second, Func<TFirst, TSecond, TResult> selector)
    {
        using (var enum1 = first.GetEnumerator())
        using (var enum2 = second.GetEnumerator())
        {
            while (enum1.MoveNext() && enum2.MoveNext())
            {
                yield return selector(enum1.Current, enum2.Current);
            }
        }
    }

Answer 2

If I understand you correctly, you probably don't want to use Aggregate here but instead Pairwise:

long[] result = f.Pairwise((x, y) => y / x).ToArray();

Here is an example implementation of Pairwise:

public static IEnumerable<TResult> Pairwise<TSource, TResult>(
    this IEnumerable<TSource> source,
    Func<TSource, TSource, TResult> resultSelector)
{
    TSource previous = default(TSource);

    using (var it = source.GetEnumerator())
    {
        if (it.MoveNext())
            previous = it.Current;

        while (it.MoveNext())
            yield return resultSelector(previous, previous = it.Current);
    }
}

Source

If you want the results in reverse order then add a call to Reverse.

Answer 3

Simple solution without using Linq:

    IEnumerable<long> GetResults(long[] input)
    {
        for (int i = input.Length -1; i >= 1; --i)
            yield return input[i] / input[i - 1];
    }

You can use Linq on the return value tough and it works in .NET 3.5 :)