Mixing LINQ with async (getting payload from a seed)

Question

I have a collection of seeds

var seeds = new [] {1, 2, 3, 4};

From each seed I want to run an async method that performs some calculation with the seed:

async Task<int> Calculation(int seed);

My goal is to perform a select like this:

var results = from seed in seeds
              let calculation = await Calculation(seed)
              select new { seed, calculation };

Unfortunately, this syntax isn't allowed using LINQ.

How can I make the "results" variable contain both the seed and the calculation?

(I would appreciate any answer, but specially if it's using System.Reactive's Observable)

http://stackoverflow.com/questions/21868087/how-to-await-a-list-of-tasks-asynchronously-using-linq Here answer by the Stephen Cleary, about it. — mybirthname, Nov 03 '16 at 16:10

score 3 · Accepted Answer · answered Nov 03 '16 at 19:57

3

Here's an Rx solution:

var seeds = new [] {1, 2, 3, 4};
var results = Observable.ToObservable(seeds)
    .SelectMany(async i => new { seed = i, calculation = await Calculation(i)})
    .ToEnumerable();

answered Nov 03 '16 at 19:57

Shlomo

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ocuenca · Answer 2 · 2016-11-03T20:17:49.310

2

You could do the following using WhenAll static method:

var r= await Task.WhenAll(seeds.Select(async seed =>new {
                                                          Seed= seed,
                                                          Result = await Calculation(seed) 
                                                        }
                                      )
                         );

edited Nov 03 '16 at 20:17

answered Nov 03 '16 at 16:17

ocuenca

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2

There's no reason for `Task.Run` to be used here. – Servy Nov 03 '16 at 19:58
Hi @Servy, thanks for pointed out, is it correct with my last update? – ocuenca Nov 03 '16 at 20:12
You could also replace the statement lambda with an expression lambda and just inline the calculation call, simplifying the code a fair bit. The formatting is also...not ideal. – Servy Nov 03 '16 at 20:16
you caught me doing it ;) – ocuenca Nov 03 '16 at 20:16

score 1 · Answer 3 · edited May 23 '17 at 10:30

1

Change your async function to return both the calculated number and the given seed:

public static async Task<Output> Calculation(int seed)
{
    return new Output { Seed = seed, Result = 0 };
}

public class Output
{
    public int Seed { get; set; }
    public int Result { get; set; }
}

Then use the linq to return a Task[] on which you can WaitAll or WhenAll: (WaitAll vs WhenAll)

var seeds = new[] { 1, 2, 3, 4 };

var tasks = seeds.Select(Calculation);
var results = await Task.WhenAll(tasks);

foreach (var item in results)
    Console.WriteLine($"seed: {item.Seed}, result: {item.Result}");

edited May 23 '17 at 10:30

Community

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answered Nov 03 '16 at 16:09

Gilad Green

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`WhenAlll` has a overload that takes in a `Enumerable>`, no need to call ToArray() – Scott Chamberlain Nov 03 '16 at 20:04
@ScottChamberlain - you are correct :) Thanks :) Fixed it. Originally I used `WaitAll` and had to pass an array.. but forgot to update that when changed – Gilad Green Nov 03 '16 at 20:06

Mixing LINQ with async (getting payload from a seed)

3 Answers3