不幸的是，没有内置的方法来做到这一点，但它很容易为自己实现。以下是它的核心内容:

public static TSource MinBy<TSource, TKey>(this IEnumerable<TSource> source,
    Func<TSource, TKey> selector)
{
    return source.MinBy(selector, null);
}

public static TSource MinBy<TSource, TKey>(this IEnumerable<TSource> source,
    Func<TSource, TKey> selector, IComparer<TKey> comparer)
{
    if (source == null) throw new ArgumentNullException("source");
    if (selector == null) throw new ArgumentNullException("selector");
    comparer ??= Comparer<TKey>.Default;

    using (var sourceIterator = source.GetEnumerator())
    {
        if (!sourceIterator.MoveNext())
        {
            throw new InvalidOperationException("Sequence contains no elements");
        }
        var min = sourceIterator.Current;
        var minKey = selector(min);
        while (sourceIterator.MoveNext())
        {
            var candidate = sourceIterator.Current;
            var candidateProjected = selector(candidate);
            if (comparer.Compare(candidateProjected, minKey) < 0)
            {
                min = candidate;
                minKey = candidateProjected;
            }
        }
        return min;
    }
}

使用示例:

var firstBorn = People.MinBy(p => p.DateOfBirth ?? DateTime.MaxValue);

请注意，如果序列为空，将抛出异常，如果有多个元素，则返回第一个具有最小值的元素。

或者，你也可以在MinBy.cs中使用MoreLINQ中的实现。(当然，有一个相应的MaxBy。)

通过包管理器控制台安装:

PM>安装包morelinq

我自己也在寻找类似的东西，最好不使用库或对整个列表进行排序。我的解决方案与问题本身相似，只是简化了一点。

var min = People.Min(p => p.DateOfBirth);
var firstBorn = People.FirstOrDefault(p => p.DateOfBirth == min);

通过IEnumerable上的扩展函数返回对象和找到的最小值的方法。它接受一个可以对集合中的对象执行任何操作的Func:

public static (double min, T obj) tMin<T>(this IEnumerable<T> ienum, 
            Func<T, double> aFunc)
        {
            var okNull = default(T);
            if (okNull != null)
                throw new ApplicationException("object passed to Min not nullable");

            (double aMin, T okObj) best = (double.MaxValue, okNull);
            foreach (T obj in ienum)
            {
                double q = aFunc(obj);
                if (q < best.aMin)
                    best = (q, obj);
            }
            return (best);
        }

例子，对象是一个机场，我们想找到最近的机场到给定的(纬度，经度)。机场有纬度(纬度、纬度)功能。

(double okDist, Airport best) greatestPort = airPorts.tMin(x => x.dist(okLat, okLon));

如果要选择具有最小或最大属性值的对象。另一种方法是使用实现IComparable。

public struct Money : IComparable<Money>
{
   public Money(decimal value) : this() { Value = value; }
   public decimal Value { get; private set; }
   public int CompareTo(Money other) { return Value.CompareTo(other.Value); }
}

最大执行将是。

var amounts = new List<Money> { new Money(20), new Money(10) };
Money maxAmount = amounts.Max();

最小实现将是。

var amounts = new List<Money> { new Money(20), new Money(10) };
Money maxAmount = amounts.Min();

通过这种方式，您可以比较任何对象并在返回对象类型的同时获得Max和Min。

希望这能帮助到一些人。

. net 6原生支持MaxBy/MinBy。你可以用简单的

人。MinBy(p => p.出生日期)

注意:我之所以包含这个答案是为了完整性，因为OP没有提到数据源是什么，我们不应该做任何假设。

这个查询给出了正确的答案，但可能会慢一些，因为它可能必须对People中的所有项进行排序，这取决于People是什么数据结构:

var oldest = People.OrderBy(p => p.DateOfBirth ?? DateTime.MaxValue).First();

UPDATE: Actually I shouldn't call this solution "naive", but the user does need to know what he is querying against. This solution's "slowness" depends on the underlying data. If this is a array or List<T>, then LINQ to Objects has no choice but to sort the entire collection first before selecting the first item. In this case it will be slower than the other solution suggested. However, if this is a LINQ to SQL table and DateOfBirth is an indexed column, then SQL Server will use the index instead of sorting all the rows. Other custom IEnumerable<T> implementations could also make use of indexes (see i4o: Indexed LINQ, or the object database db4o) and make this solution faster than Aggregate() or MaxBy()/MinBy() which need to iterate the whole collection once. In fact, LINQ to Objects could have (in theory) made special cases in OrderBy() for sorted collections like SortedList<T>, but it doesn't, as far as I know.