如果可能有重复键,如何添加到字典？

Question

需要将来自可能具有重复键的对象的一组键/值对添加到字典中.只应将键的第一个不同实例(以及实例的值)添加到字典中.

下面是一个示例实现,首先出现,以便正常工作.

void Main()
{
    Dictionary<long, DateTime> items = new Dictionary<long, DateTime>();
    items = AllItems.Select(item =>
                    {
                        long value;
                        bool parseSuccess = long.TryParse(item.Key, out value);
                        return new { value = value, parseSuccess, item.Value };
                    })
                    .Where(parsed => parsed.parseSuccess && !items.ContainsKey(parsed.value))
                    .Select(parsed => new { parsed.value, parsed.Value })
                    .Distinct()
                    .ToDictionary(e => e.value, e => e.Value);
    Console.WriteLine(string.Format("Distinct: {0}{1}Non-distinct: {2}",items.Count, Environment.NewLine, AllItems.Count));

}

public List<KeyValuePair<string, DateTime>> AllItems
{
    get
    {
        List<KeyValuePair<string, DateTime>> toReturn = new List<KeyValuePair<string, DateTime>>();
        for (int i = 1000; i < 1100; i++)
        {
            toReturn.Add(new KeyValuePair<string, DateTime>(i.ToString(), DateTime.Now));
            toReturn.Add(new KeyValuePair<string, DateTime>(i.ToString(), DateTime.Now));
        }
        return toReturn;
    }
}

但是,如果修改AllItems以返回更多对,则会发生ArgumentException:"已添加具有相同键的项目."

void Main()
{
    Dictionary<long, DateTime> items = new Dictionary<long, DateTime>();
    var AllItems = PartOne.Union(PartTwo);
    Console.WriteLine("Total items: " + AllItems.Count());
    items = AllItems.Select(item =>
                    {
                        long value;
                        bool parseSuccess = long.TryParse(item.Key, out value);
                        return new { value = value, parseSuccess, item.Value };
                    })
                    .Where(parsed => parsed.parseSuccess && !items.ContainsKey(parsed.value))
                    .Select(parsed => new { parsed.value, parsed.Value })
                    .Distinct()
                    .ToDictionary(e => e.value, e => e.Value);
    Console.WriteLine("Distinct: {0}{1}Non-distinct: {2}",items.Count, Environment.NewLine, AllItems.Count());

}

public IEnumerable<KeyValuePair<string, DateTime>> PartOne
{
    get
    {
        for (int i = 10000000; i < 11000000; i++)
        {
            yield return (new KeyValuePair<string, DateTime>(i.ToString(), DateTime.Now));
        }
    }
}
public IEnumerable<KeyValuePair<string, DateTime>> PartTwo
{
    get
    {
        for (int i = 10000000; i < 11000000; i++)
        {
            yield return (new KeyValuePair<string, DateTime>(i.ToString(), DateTime.Now));
        }
    }
}

完成此任务的最佳方法是什么？请注意,long.TryParse需要在解决方案中使用,因为实际输入可能不包括有效的Int64.

Answer 1

只应将键的第一个不同实例(以及实例的值)添加到字典中.

您可以通过使用该Enumerable.GroupBy方法并获取组中的第一个值来实现此目的:

items = AllItems.Select(item =>
                {
                    long value;
                    bool parseSuccess = long.TryParse(item.Key, out value);
                    return new { Key = value, parseSuccess, item.Value };
                })
                .Where(parsed => parsed.parseSuccess)
                .GroupBy(o => o.Key)
                .ToDictionary(e => e.Key, e => e.First().Value)