CODEWITHSUNDEEP
linqc#-4.0
Nodoid
Nodoid

Reputation: 1561

Searching a list using an array as the parameter list using LINQ

I currently have some code that looks like this

string[] contains = new string[]{"marge", "homer", "lisa", "bart", "maggie"};
string[] actions = new string[]{"get dye", "mmm beer", "blow saxophone", "have a cow", "bang"};

for (int i = 0; i < someActions.Count; ++i)
{
  if (someActions.Contains(contains[i]))
    callRoutine(actions[i]);
}

(this is a very trivial example - someActions is a List)

I'm wondering if there is a way in LINQ to do the same as loop? I'm thinking of something along the lines of 

int i = position in contains or 0
callRoutine(actions[i]);

The problem is that I don't know how to use an array as the search parameter. Various searches suggest I need to use IEnumerable to do this, but I'm not sure.

Any help would be appreciated here.

Paul

Upvotes: 0

Views: 75

Answers (3)

AD.Net
AD.Net

Reputation: 13399

someActions.Intersect(contains).ForEach(callRoutine);

OR

someActions.Intersect(contains).ForEach(i=>callRoutine(i));

Upvotes: 0

Chris
Chris

Reputation: 707

This wouldn't work nicely with your current data setup.

If you are flexible on your data, you could try this:

var namesToActions = new Dictionary<string, string>()
    {
        { "marge" , "get dye" },
        { "homer", "mmm beer"},
        { "lisa", "blow saxophone"},
        { "bart", "have a cow"},
        { "maggie", "bang"}
    };

someActions.ForEach(a => callRoutine(namesToActions[a]));

Switching to a Dictionary makes it a little easier to perform the type of Linq action you're looking for and provides additional flexibility and quicker lookup times.

Upvotes: 1

Bryan Hong
Bryan Hong

Reputation: 1483

I'm not sure what your purpose is, but if you want to convert your for loop into a linq statement, you can do this:

var i = 0;

someActions.ForEach(x =>
                        {
                            if (someActions.Contains(contains[i]))
                               callRoutine(actions[i]);
                            i++;
                        });

Upvotes: 0

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