Question

可能重复：
Test whether two IEnumerable<T> have the same values with the same frequencies

我写了

更新 - 更正：

static bool HaveSameItems<T>(this IEnumerable<T> self, IEnumerable<T> other)
{
    return ! 
    ( 
        other.Except(this).Any() ||
        this.Except(other).Any()
    );
}

有没有更短的方式？我知道有SequenceEqual但订单对我来说无关紧要。

Answer 1

即使订单对您无关紧要，也不排除将SequenceEqual作为可行选项。

var lst1 = new [] { 2,2,2,2 };
var lst2 = new [] { 2,3,4,5 };
var lst3 = new [] { 5,4,3,2 };

//your current function which will return true
//when you compare lst1 and lst2, even though
//lst1 is just a subset of lst2 and is not actually equal
//as mentioned by Wim Coenen
(lst1.Count() == lst2.Count() &&
        !lst1.Except(lst2).Any()); //incorrectly returns true

//this also only checks to see if one list is a subset of another
//also mentioned by Wim Coenen
lst1.Intersect(lst2).Any(); //incorrectly returns true

//So even if order doesn't matter, you can make it matter just for
//the equality check like so:
lst1.OrderBy(x => x).SequenceEqual(lst2.OrderBy(x => x)); //correctly returns false
lst3.OrderBy(x => x).SequenceEqual(lst2.OrderBy(x => x)); // correctly returns true

Answer 2

这是一个O(n)解决方案，只能遍历每个序列一次（实际上，它甚至可能不会完全走向第二个序列，它具有提前终止的可能性）：

public static bool HaveSameItems<T>(IEnumerable<T> a, IEnumerable<T> b) {
    var dictionary = a.GroupBy(x => x).ToDictionary(g => g.Key, g => g.Count());
    foreach(var item in b) {
        int value;
        if (!dictionary.TryGetValue(item, out value)) {
            return false;
        }
        if (value == 0) {
            return false;
        }
        dictionary[item] -= 1;
    }
    return dictionary.All(x => x.Value == 0);
}

这个解决方案的一个缺点是它不会很好地与LINQ to SQL，EF，NHiberate等互操作。

如果两个IEnumerable <t>在C＃中具有相同的项目，那么比较最短的方法是什么？</t>

2 个答案: