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Lambda expressions - set the value of one property in a collection of objects based on the value of another property in the collection

I'm new to lambda expressions and looking to leverage the syntax to set the value of one property in a collection based on another value in a collection

Typically I would do a loop:

class Item
{
    public string Name { get; set; }
    public string Value { get; set; }
}

void Run()
{
    Item item1 = new Item { Name = "name1" };
    Item item2 = new Item { Name = "name2" };
    Item item3 = new Item { Name = "name3" };

    Collection<Item> items = new Collection<Item>() { item1, item2, item3 };

    // This is what I want to simplify.
    for (int i = 0; i < items.Count; i++)
    {
        if (items[i].Name == "name2")
        {
            // Set the value.
            items[i].Value = "value2";
        }
    }
}
like image 699
Michael Rut Avatar asked May 26 '10 19:05

Michael Rut


1 Answers

LINQ is generally more useful for selecting data than for modifying data. However, you could write something like this:

foreach(var item in items.Where(it => it.Name == "name2")) 
  item.Value = "value2";

This first selects items that need to be modified and then modifies all of them using a standard imperative loop. You can replace the foreach loop with ForAll method that's available for lists, but I don't think this gives you any advantage:

items.Where(it => it.Name == "name2").ToList()
     .ForEach(it => it.Value = "value2");

Note that you need to add ToList in the middle, because ForEach is a .NET 2.0 feature that's available only for List<T> type - not for all IEnumerable<T> types (as other LINQ methods). If you like this approach, you can implement ForEach for IEnuerable<T>:

public static void ForEach<T>(this IEnumerable<T> en, Action<T> f) {
  foreach(var a in en) f(a);
}

// Then you can omit the `ToList` conversion in the middle 
items.Where(it => it.Name == "name2")
     .ForEach(it => it.Value = "value2");

Anyway, I'd prefer foreach loop, because that also makes it clear that you're doing some mutation - and it is useful to see this fact easily in the code.

like image 186
Tomas Petricek Avatar answered Oct 12 '22 14:10

Tomas Petricek