比较LINQ中的列表并返回2个结果对象

Question

I have 2 Lists: List<Car> newCars and List<Car> oldCars .

I need to use the List<Car> newCars as the source of truth and return both Present and Absent cars in the List<Car>oldCars list.

How can I do this?

My approach:

var present = newCars.Where(c => oldCars.All(w => w.Id != c.Id));
var absent = newCars.Where(c => oldCars.All(w => w.Id == c.Id));

I am quite new to LINQ, and I am not sure of the logic that I have used above.

1. Can someone help me out to get this working in a better and optimized way?

2. Can this be done in one single query and return 2 result sets (as tuples)?

I realize that running the same code with equals and not equals as above can be a costly approach.

Answer 1

You can create custom IEqualityComparer :

public class CarEqualityComparer : IEqualityComparer<Car>
{
    public bool Equals(Car x, Car y)
    {
        return x.Id == y.Id;
    }

    public int GetHashCode(Car obj)
    {
        return obj.Id.GetHashCode();
    }
}

And use it like this:

var present = newCars.Intersect(oldCars, new CarEqualityComparer());
var absent = newCars.Except(oldCars, new CarEqualityComparer());

Answer 2

You can utilize an anonymous variable containing the two lists:

var res = new
{
    present = newCars.Where(n => oldCars.Select(o => o.Id).Contains(n.Id)), // present if the old cars list contains the id from the new list
    absent = newCars.Where(n => !oldCars.Select(o => o.Id).Contains(n.Id)) // absent if the old cars list does not contain the id from the new list
};

Answer 3

When you are saying that new cars are the source of truth, that means you want them always in the list, but you may not have an equivalent old car. You are describing a left-join.

This should get you most of what you want.

var results = from n in newCars
    join o in oldCars
    on n.Id == o.Id
    group into cars 
    from oc in cars.DefaultIfEmpty()
    select (NewCar: n, Present: oc != null)

The result is a tuple containing the new car and a boolean Present that is true or false based on whether there is an old car or not.

That gives you a single result set. You wanted two result sets as a tuple. To get this, just extend the above query, grouping by Present . This gives you a result set with exactly two items

var results = (from r in (from n in newCars
        join o in oldCars
        on n.Id equals o.Id
        into cars
        from oc in cars.DefaultIfEmpty()
        select (NewCar: n, Present: oc != null))
    group r by r.Present into g
    orderby g.Key descending 
    select g.Select(x => x.NewCar).ToList()).ToList();

from which you can get a single tuple containing both lists

var finalResult = (Present: results[0], Absent: results[1]);

Answer 4

Using an extension method to treat adding to a list (a tiny bit) more functionally, you can use LINQ Aggregate to do this in one pass:

var idsFromOldCars = oldCars.Select(c => c.Id).ToHashSet();
var(present, absent) = newCars.Aggregate((p:new List<Car>(),a:new List<Car>()), (pa,nc) => idsFromOldCars.Contains(nc.Id) ? (pa.p.AfterAdd(nc),pa.a) : (pa.p,pa.a.AfterAdd(nc)));

Where AfterAdd is defined in a static class as:

public static class ListExt {
    public static List<T> AfterAdd<T>(this List<T> head, params T[] tail) {
        head.AddRange(tail);
        return head;
    }
}

But it is much clearer to just use a foreach loop:

    var idsFromOldCars = oldCars.Select(c => c.Id).ToHashSet();
    var present = new List<Car>();
    var absent = new List<Car>();
    foreach (var nc in newCars) {
        if (idsFromOldCars.Contains(nc.Id))
            present.Add(nc);
        else
            absent.Add(nc);
    }

In both cases, creating the HashSet ensures you don't have to scan through the oldCars list to find each newCar . If you knew the lists were ordered by Id , you could do a more complicated one-pass solution traveling down the lists in parallel, but that hardly seems worth the complexity even then.

NOTE: A Join effectively does something similar, converting the second List into a Lookup (like a Dictionary ) that can be used to find the matches with the first List .