yield return in the lambda expression?

Question

The follow code will break the list into sub-lists which begins with "[" and end with "]". How to convert it to use yield return so it can handle very huge stream input lazily? --Or how to implement it in F# with lazily enumerating?-- (never mind, I think f#implementation should be trivial)

var list = new List<string> { "[", "1", "2", "3", "]", "[", "2", "2", "]", "[", "3", "]" };
IEnumerable<IEnumerable<string>> result = Split(list);

static IEnumerable<IEnumerable<string>> Split(List<string> list)
{
    return list.Aggregate(new List<List<string>>(), // yield return?
    (sum, current) =>
    {
        if (current == "[")
            sum.Add(new List<string>());
        else if (current == "]")
            return sum; // Convert to yield return?
        else
            sum.Last().Add(current);
        return sum; // Convert to yield return?
    });
}

Feels like aggregate is not quite the right thing to use here. Write your own version of it, then superimpose the logic on top. — leppie, Feb 18 '16 at 05:45
@leppie The problem with the `Aggregate` is that he's not actually aggregating anything. The lambda is only ever used for its side effects, so it's noting more than a much harder to read `foreach` loop. — Servy, Feb 18 '16 at 05:59

Servy · Accepted Answer · 2016-02-18T05:55:33.480

12

C# doesn't support anonymous iterator blocks, so you'll simply need to use a named method instead of an anonymous method.

public static IEnumerable<IEnumerable<string>> Split(IEnumerable<string> tokens)
{
    using(var iterator = tokens.GetEnumerator())
        while(iterator.MoveNext())
            if(iterator.Current == "[")
                yield return SplitGroup(iterator);
}

public static IEnumerable<string> SplitGroup(
    IEnumerator<string> iterator)
{
    while(iterator.MoveNext() && iterator.Current != "]")
        yield return iterator.Current;
}

edited Feb 18 '16 at 05:55

answered Feb 18 '16 at 05:52

Servy

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See https://stackoverflow.com/questions/1217729/in-c-why-cant-an-anonymous-method-contain-a-yield-statement – StayOnTarget Feb 01 '18 at 20:13

score 1 · Answer 2 · answered Sep 13 '17 at 04:32

While C# not allow yield in lambda I think we could workaround like this

static IEnumerable<IEnumerable<string>> Split(List<string> list)
{
    var tmp = list as IEnumerable<string>;
    while(tmp.FirstOrDefault() == "[")
    {
        yield return tmp.TakeWhile((current) => current != "]");
        tmp = tmp.SkipWhile((current) => current != "]");
    }
}

score -3 · Answer 3 · answered Feb 18 '16 at 06:16

-3

I found another way of doing this, using Linq. Not sure on efficience, but it works.

int index=0;
list.Select(item=> new { item=item, index= item=="]"? ++index : index })
                  .Where(c => !(c.item == "[" || c.item =="]"))
                  .GroupBy(g=>g.index)
                  .Select(e=> e.Select(c=>c.item));

Working Demo

answered Feb 18 '16 at 06:16

Hari Prasad

16,716
4
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That's not lazily generating the sequence. `Groupby` is going to need to process the *entire* sequence before it can yield any results, which the OP has said they don't want to do. – Servy Feb 18 '16 at 14:40

yield return in the lambda expression?

3 Answers3