Yield return from an indexed iteration via LINQ

Question

Leaving the performance cost of LINQ usage, I would like to know how to convert the following code into a LINQ expression

for (int i = 0; i < someArray.length(); i++)
  yield return new SomeEntity(someFunction(i));

Important: I need the use of the incremented index

Update:

Rather than someArray.length(), number should be used:

for (int i = 0; i < number; i++)
  yield return new SomeEntity(someFunction(i));

2nd update

I'm still getting the compilation error "not all code paths return value"

My code:

public static IEnumerable function()
{
    Enumerable.Range(0,5).Select(i => new Entity());
}

3rd update

Didn't think it's relevant until I found out it's the cause for this error..

public static IEnumerable function()
{
    int[] arr = { 1, 2, 3, 4 };
    foreach (int i in arr)
    {
        Enumerable.Range(0,5).Select(i => new Entity());
    }
}

If you take out the foreach 1st loop out of the equation, all replies answer to this question, but my issue is n^2.. 2 nested loops...

Any ideas?

Possible duplicate of [How do you add an index field to Linq results](http://stackoverflow.com/questions/269058/how-do-you-add-an-index-field-to-linq-results) — qxg, Sep 01 '16 at 07:38
You shouldn't edit questions to make existing answer invalid. You should instead add an update at the end of the existing question. — Enigmativity, Sep 01 '16 at 07:39

score 6 · Answer 1 · edited Sep 25 '17 at 06:03

6

Use the overload of Enumerable.Select that has an index into the collection:

someArray.Select((x, i) => new SomeEntity(someFunction(i)));

Edit

As you've modified your example and are not actually using a collection to iterate and index to, use Enumerable.Range:

Enumerable.Range(0, number).Select(i => new SomeEntity(someFunction(i)));

edited Sep 25 '17 at 06:03

Graham

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answered Sep 01 '16 at 07:30

Yuval Itzchakov

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this is an inner loop where, someArray.length() is actually an Xpath function generating a node count, so basically someArray.Length() is just a number :/ editting it to the OQ – USS-Montana Sep 01 '16 at 07:33
@USS-Montana I've edited my answer to reflect your new question. – Yuval Itzchakov Sep 01 '16 at 07:40
I'm getting an error - not all code paths return a value, tried it with this example, Enumerable.Range(0, 2).Select(i => 1); – USS-Montana Sep 01 '16 at 07:47
@USS-Montana - You need `return` in front of the code. – Enigmativity Sep 01 '16 at 07:48

score 1 · Answer 2 · answered Sep 01 '16 at 07:36

1

Use Enumerable.Range to generate the numbers:

Enumerable.Range(0,number).Select(i=>new SomeEntity(someFunction(i)));

answered Sep 01 '16 at 07:36

Damien_The_Unbeliever

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I'm getting an error - not all code paths return a value, tried it with this example, Enumerable.Range(0, 2).Select(i => 1); – USS-Montana Sep 01 '16 at 07:47
@USS-Montana - then how about showing us more than two lines of code and expecting us to divine the context and fix bugs in code we cannot see? – Damien_The_Unbeliever Sep 01 '16 at 07:48
@USS-Montana - You need `return` in front of the code. – Enigmativity Sep 01 '16 at 07:49

score 0 · Answer 3 · answered Sep 01 '16 at 07:56

0

Here's my LinqPad snippet.

void Main()
{

    var e = SomeEntity.GetEntities(new List<int> { 1, 2, 3});
    e.Dump();
}

public class SomeEntity
{
    public int m_i;
    public SomeEntity(int i)
    {
        m_i = i;
    }

    public override string ToString()
    {
        return m_i.ToString();
    }


    public static int someFunction(int i){ return i+100;}

    public static IEnumerable<SomeEntity> GetEntities(IEnumerable<int> someArray)
    {
//        for (int i = 0; i < someArray.Count();i++)
//            yield return new SomeEntity(someFunction(i));

        // *** Equivalent linq function ***    
        return someArray.Select(a => new SomeEntity(someFunction(a)));    
    }
}

answered Sep 01 '16 at 07:56

Phillip Ngan

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Finally found the problem!! – USS-Montana Sep 01 '16 at 08:34
my linq for loop is nested inside a foreach loop, and if I add this to your code you'l get a similar problem – USS-Montana Sep 01 '16 at 08:35
int[] arr = { 1, 2, 3, 4 }; foreach (int i in arr) { return Enumerable.Range(0, 7).Select(a => new SomeEntity(someFunction(a))); } – USS-Montana Sep 01 '16 at 08:35

Yield return from an indexed iteration via LINQ

3 Answers3

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