Guerrilla
Guerrilla

Reputation: 14906

Lambda to turn IDictionary<string,object> values into object[]

I am trying to create a Lambda expressions that creates an object array from an IDictionary based on given named input.

This is what I have so far:

var dic = new Dictionary<string, object>() 
        { 
            {"some", "value"}, 
            {"dateTime", DateTime.Now},
            {"someNum", 52},
            {"other", "values"}
        };

string[] inputArray = { "input", "dateTime", "someNume" };

var processedArray = inputArray.Select(i => dic.Where(d => d.Key == i).Select(kvp => kvp.Value)).ToArray();

This doesn't work, it returns an IEnumerable<object> but all the members are type KeyValuePair<string,object> just, it appears, without the key data.

I think there are a few things wrong with this. I don't think I should be using dic.Select() as key entries in dictionary are unique and I am obviously referencing the value wrong or missing a way to cast each result.

I am not sure how to achieve this using a single Lambda. I need to work out the Lambda as I plan to use it as a compiled Lambda.

Can someone advise me on how best to do this?

Upvotes: 0

Views: 156

Answers (2)

Alex
Alex

Reputation: 13234

I think this should do the trick.

var processedArray = 
    (from key in inputArray
     where dic.ContainsKey(key)
     select dic[key]).ToArray();

Or using the fluent/dot syntax:

var processedArray = inputArray
    .Where(key => dic.ContainsKey(key))
    .Select(key => dic[key])
    .ToArray(); 

If you want to still have an element in your processedArray output in the case where the dictionary misses a key (e.g. for the "input" string in your example), you could do it like this for example (using null for a missing entry in the dictionary):

var processedArray = inputArray
    .Select(key => dic.ContainsKey(key) ? dic[key] : null)
    .ToArray(); 

Upvotes: 3

user1726343
user1726343

Reputation:

You should be able to select these just by accessing the dictionary indices:

inputArray.Select(k => dic[k]);

Upvotes: 2

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