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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T10:56:31+00:00 2026-05-12T10:56:31+00:00

I am using WCF REST Preview 2 to test some REST services. The package

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I am using WCF REST Preview 2 to test some REST services. The package has an extension to IEnumerable as ToDictionary(Func(TSource, TKey) keySelctor. Not sure how to define a lambda function to return keySelector?

Here is one example:

 var items = from x in entity.Instances // a customized Entity class with list instances of MyClass
             select new { x.Name, x};
 Dictionary<string, MyClass> dic = items.ToDictionary<string, MyClass>(
      (x, y) => ... // what should be here. I tried x.Name, x all not working

Not sure what should be the lambda Func should be to return a KeySelector?

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  1. Editorial Team
    Editorial Team
    2026-05-12T10:56:31+00:00Added an answer on May 12, 2026 at 10:56 am

    Since items is of type IEnumerable<MyClass>, you should be able to do the following:

    items.ToDictionary(x => x.Name)
    

    You could even have done:

    entity.Instances.ToDictionary(x => x.Name)
    

    You don’t need to specify the type parameters, since they can be correctly inferred.

    Edit:

    items.ToDictionary(x => x.Name) is actually incorrect, because items is not of type IEnumerable<MyClass>. It is actually an IEnumerable of the anonymouse type, that has 2 properties (Name, which contains the myClass.Name property, and x, which is of type MyClass).

    In that case, assuming you can do:

    var items = from instance in entity.Instances 
                select new { Name = instance.Name, // don't have to specify name as the parameter
                             Instance = instance
                            };
    
    var dict = items.ToDictionary(item => item.Name, 
                                  item => item.Instance);
    

    The second example is a bit easier to use in this case. Essentially, you don’t get any value from the linq query to get items, if all you’re trying to do is generate a dictionary.

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