Get previous element in IObservable without re-evaluating the sequence

Question

In an IObservable sequence (in Reactive Extensions for .NET), I'd like to get the value of the previous and current elements so that I can compare them. I found an example online similar to below which accomplishes the task:

sequence.Zip(sequence.Skip(1), (prev, cur) => new { Previous = prev, Current = cur })

It works fine except that it evaluates the sequence twice, which I would like to avoid. You can see that it is being evaluated twice with this code:

var debugSequence = sequence.Do(item => Debug.WriteLine("Retrieved an element from sequence"));
debugSequence.Zip(debugSequence.Skip(1), (prev, cur) => new { Previous = prev, Current = cur }).Subscribe();

The output shows twice as many of the debug lines as there are elements in the sequence.

I understand why this happens, but so far I haven't found an alternative that doesn't evaluate the sequence twice. How can I combine the previous and current with only one sequence evaluation?

Answer 1

There's a better solution to this I think, that uses Observable.Scan and avoids the double subscription:

public static IObservable<Tuple<TSource, TSource>>
    PairWithPrevious<TSource>(this IObservable<TSource> source)
{
    return source.Scan(
        Tuple.Create(default(TSource), default(TSource)),
        (acc, current) => Tuple.Create(acc.Item2, current));
}

I've written this up on my blog here: http://www.zerobugbuild.com/?p=213

Addendum

A further modification allows you to work with arbitrary types more cleanly by using a result selector:

public static IObservable<TResult> CombineWithPrevious<TSource,TResult>(
    this IObservable<TSource> source,
    Func<TSource, TSource, TResult> resultSelector)
{
    return source.Scan(
        Tuple.Create(default(TSource), default(TSource)),
        (previous, current) => Tuple.Create(previous.Item2, current))
        .Select(t => resultSelector(t.Item1, t.Item2));
}

Answer 2

@James World addendum looks great to me, if not for Tuple<>, which I almost always dislike: "Was .Item1 the previous? Or was it the current one? I can't remember. And what's the first argument to the selector, was it the previous item?".

For that part I liked @dcstraw definition of a dedicated ItemWithPrevious<T>. So there you go, putting the two together (hopefully I did not mix up previous with current) with some renaming and facilities:

public static class ObservableExtensions
{
    public static IObservable<SortedPair<TSource>> CombineWithPrevious<TSource>(
        this IObservable<TSource> source, 
        TSource initialValue = default(TSource))
    {
        var seed = SortedPair.Create(initialValue, initialValue);

        return source.Scan(seed,
            (acc, current) => SortedPair.Create(current, acc.Current));
    }

    public static IObservable<TResult> CombineWithPrevious<TSource, TResult>(
        this IObservable<TSource> source,
        Func<SortedPair<TSource>, TResult> resultSelector,
        TSource initialValue = default(TSource))
    {
        var seed = SortedPair.Create(initialValue, initialValue);

        return source
            .Scan(seed,
                (acc, current) => SortedPair.Create(current, acc.Current))
            .Select(p => resultSelector(p));
    }
}

public class SortedPair<T>
{
    public SortedPair(T current, T previous)
    {
        Current = current;
        Previous = previous;
    }

    public SortedPair(T current) : this(current, default(T)) { }

    public SortedPair() : this(default(T), default(T)) { }

    public T Current;
    public T Previous;
}

public class SortedPair
{
    public static SortedPair<T> Create<T>(T current, T previous)
    {
        return new SortedPair<T>(current, previous);
    }

    public static SortedPair<T> Create<T>(T current)
    {
        return new SortedPair<T>(current);
    }

    public static SortedPair<T> Create<T>()
    {
        return new SortedPair<T>();
    }
}

Answer 3

Evaluating twice is an indicator of a Cold observable. You can turn it to a Hot one by using .Publish():

var pub = sequence.Publish();
pub.Zip(pub.Skip(1), (...
pub.Connect();

Answer 4

If you only need to access the previous element during subscription, this is probably the simplest thing that will work. (I'm sure there's a better way, maybe a buffer operator on IObservable? The documentation is pretty sparse at the moment, so I can't really tell you.)

    EventArgs prev = null;

    sequence.Subscribe(curr => 
    {
        if (prev != null)
        {
            // Previous and current element available here
        }

        prev = curr;                              

    });

EventArgs is just a stand-in for the type of your event's argument.

Answer 5

It turns out you can use a variable to hold the previous value and refer to it and reassign it within the chain of IObservable extensions. This even works within a helper method. With the code below I can now call CombineWithPrevious() on my IObservable to get a reference to the previous value, without re-evaluating the sequence.

public class ItemWithPrevious<T>
{
    public T Previous;
    public T Current;
}

public static class MyExtensions
{
    public static IObservable<ItemWithPrevious<T>> CombineWithPrevious<T>(this IObservable<T> source)
    {
        var previous = default(T);

        return source
            .Select(t => new ItemWithPrevious<T> { Previous = previous, Current = t })
            .Do(items => previous = items.Current);
    }
}