如何根据一些属性将对象分割成块?

public class InvestorMailing
{
    public string To { get; set; }

    public IEnumerable<string> Attachments { get; set; }

    public int AttachmentCount { get; set; }

    public long AttachmentSize { get; set; }
}

i have an IList<InvestorMailing> mailingList. if the attachment size is greater than x, then i need to split my object into chunks. is there an easy linq-y way to do this?

我有一个投资邮件列表。如果附件大小大于x，则需要将对象分割成块。有没有一种简单的linq-y方法?

Edited:

编辑:

this is how i'm generating my mailings:

这就是我如何生成我的邮件:

        var groupedMailings = mailingList.GroupBy(g => g.GroupBy);

        var investorMailings = groupedMailings.Select(
            g => new DistinctInvestorMailing
            {
                Id = g.Select(x => x.Id).FirstOrDefault(),
                To = g.Key.Trim(),
                From = g.Select(x => x.From).FirstOrDefault(),
                FromName = g.Select(x => x.FromName).FirstOrDefault(),
                Bcc = g.Select(x => x.Bcc).FirstOrDefault(),
                DeliveryCode = g.Select(x => x.DeliveryCode).FirstOrDefault(),
                Subject = g.Select(x => x.Subject).FirstOrDefault(),
                Body = g.Select(x => x.Body).FirstOrDefault(),
                CommentsOnStatus = g.Select(x => x.CommentsOnStatus).FirstOrDefault(),
                Attachments = g.Select(x => x.AttachmentPath),
                AttachmentCount = g.Select(x => x.AttachmentPath).Count(),
                AttachmentSize = g.Sum(x => x.AttachmentSize),
                MailType = g.Select(x => x.MessageType).FirstOrDefault()
            }
        ).ToList();

3 个解决方案

#1

It should be pretty simple to do it with a standard method. Consider this example:

用标准方法来做应该很简单。考虑一下这个例子:

class Foo
{
    public Foo(int weight) { Weight = weight; }
    public int Weight { get; set; }
}

...

…

IEnumerable<IList<Foo>> GroupFoosByWeight(IList<Foo> foos, int weightLimit)
{
    List<Foo> list = new List<Foo>();
    int sumOfWeight = 0;

    foreach (Foo foo in foos)
    {
        if (sumOfWeight + foo.Weight > weightLimit)
        {
            yield return list;
            sumOfWeight = 0;
            list.Clear();
        }

        list.Add(foo);
        sumOfWeight += foo.Weight;
    }

    if (list.Count > 0)
        yield return list;
}

...

…

List<Foo> foos = new List<Foo>()
{
    new Foo(15), new Foo(32), new Foo(14), new Foo(19), new Foo(27)
};

foreach (IList<Foo> list in GroupFoosByWeight(foos, 35))
{
    Console.WriteLine("{0}\t{1}", list.Count, list.Sum(f => f.Weight));
}

Edit

编辑

I worked on it a bit and produced a LINQ version. It doesn't really save much code in this case, but it's a start.

我做了一点工作，并制作了一个LINQ版本。在这种情况下，它并没有节省很多代码，但这只是一个开始。

int weightLimit = 35;
int fooGroup = 0;
int totalWeight = 0;

Func<Foo, int> groupIncrementer = f =>
{
    if (totalWeight + f.Weight > weightLimit)
    {
        fooGroup++;
        totalWeight = 0;
    }

    totalWeight += f.Weight;

    return fooGroup;
};

var query = from foo in foos
            group foo by new { Group = groupIncrementer(foo) }
                into g
                select g.AsEnumerable();

foreach (IList<Foo> list in query)
{
    Console.WriteLine("{0}\t{1}", list.Count, list.Sum(f => f.Weight));
}

#2

Here's a way to do it using some LINQ to find a chunk that has enough space left to add the attachment:

这里有一种方法，使用LINQ找到一个有足够空间添加附件的块:

var chunks = new List<List<InvestorMailing>>();
int maxAttachmentsSize = 10;

foreach (InvestorMailing mail in mailingList)
{
    var chunkWithSpace = chunks
        .Where(list => list.Sum(x => x.AttachmentSize) +
                       mail.AttachmentSize <= maxAttachmentsSize)
        .FirstOrDefault();

    if (chunkWithSpace != null)
    {
        chunkWithSpace.Add(mail);
    } else {
        chunks.Add(new List<InvestorMailing> { mail });
    }
}

The result is stored in chunks.

结果以块的形式存储。

#3

Yes:

是的:

var highs = mailingList.Where(i => i.AttachmentSize > 10000).ToList();
var lows = mailingList.Where(i => i.AttachmentSize <= 10000).ToList();

How do you need to break them apart aside from this?

你要怎么把它们分开呢?

HTH.

HTH。

#1