Question

我有两个可枚举的列表。

List<List_Data> List1 = new List<List_Data>();
List1.Add(new List_Data { Material = "1", Batch = "B1", QTY = 5 });
List1.Add(new List_Data { Material = "1", Batch = "B2", QTY = 5 });
List1.Add(new List_Data { Material = "2", Batch = "B1", QTY = 15 });

List<List_Data> List2 = new List<List_Data>();
List2.Add(new List_Data { Material = "1", Batch = "B1", QTY = 2 });
List2.Add(new List_Data { Material = "3", Batch = "B1", QTY = 5 });
List2.Add(new List_Data { Material = "3", Batch = "B2", QTY = 15 });

我想要的是比较两个列表，并根据物料和批次获得差异QTY（list1.QTY-list2.QTY）。即使其他清单上不存在某项，我也需要根据该物料和批次减去或增加数量。

这是我期望的结果。

Material = "1", Batch = "B1", QTY = 3
Material = "1", Batch = "B2", QTY = 5 
Material = "2", Batch = "B1", QTY = 15
Material = "3", Batch = "B1", QTY = -5
Material = "3", Batch = "B2", QTY = -15

这是我到目前为止所做的，

SendList = (from l1 in List1
            join l2 in List2 on new { l1.Material, l1.Batch } equals new { l2.Material, l2.Batch } into temp
            from l2 in temp.DefaultIfEmpty()
            select new Report_Class
            {
                Material = l1.Material != null ? l1.Material : l2.Material, 
                Batch = l1.Batch != null ? l1.Batch : l2.Batch, 
                Difference = l1 != null && l2 != null ? (l1.QTY - l2.QTY).ToString() : l1 != null ? l1.QTY.ToString() : l2.QTY.ToString(), 

            }).ToList();

问题是它返回list1所有存在的项，但不返回仅存在于列表2中的项。任何帮助将不胜感激。

谢谢。

Answer 1

这是一种方法：

反转List2上的QTY值
将上述结果与List1连接起来。
按Material和Batch对连接列表进行分组，并汇总QTY值

在代码中：

var result = List1.Concat(
             List2.Select(list2Item => new List_Data
             {
                 Material = list2Item.Material,
                 Batch = list2Item.Batch,
                 QTY = list2Item.QTY * -1
             }))
             .GroupBy(item => new { item.Material, item.Batch })
             .Select(grouped => new List_Data
             {
                 Material = grouped.First().Material,
                 Batch = grouped.First().Batch,
                 QTY = grouped.Sum(item => item.QTY)
             })
             .ToList();

即使您具有空QTY，它仍然可以使用。例如，具有以下值：

List<List_Data> List1 = new List<List_Data>();
List1.Add(new List_Data { Material = "1", Batch = "B1", QTY = 5 });
List1.Add(new List_Data { Material = "1", Batch = "B2", QTY = 5 });
List1.Add(new List_Data { Material = "2", Batch = "B1", QTY = 15 });
List1.Add(new List_Data { Material = "3", Batch = "B1", QTY = null });
List1.Add(new List_Data { Material = "3", Batch = "B3", QTY = 4 });

List<List_Data> List2 = new List<List_Data>();
List2.Add(new List_Data { Material = "1", Batch = "B1", QTY = 2 });
List2.Add(new List_Data { Material = "3", Batch = "B1", QTY = 5 });
List2.Add(new List_Data { Material = "3", Batch = "B2", QTY = 15 });
List2.Add(new List_Data { Material = "3", Batch = "B3", QTY = null });

将导致：

Material: "1", Batch: "B1", QTY: 3
Material: "1", Batch: "B2", QTY: 5
Material: "2", Batch: "B1", QTY: 15
Material: "3", Batch: "B1", QTY: -5
Material: "3", Batch: "B3", QTY: 4
Material: "3", Batch: "B2", QTY: -15

Answer 2

如果我们假设，在第二个列表中最多（如果有的话）一个元素具有相同的Material和Bacth值，那么一个幼稚的解决方案可能是以下内容：

// Initially project each element in the list to an element that 
// has also the info in which list this item is contained.
var list1 = List1.Select(x => new {Data = x, List = 1});
var list2 = List2.Select(x => new {Data = x, List = 2});

var result = list1.Concat(list2)
            .GroupBy(x => new {x.Data.Batch, x.Data.Material})
            .Select(gr =>
            {
                var itemsInGroup = gr.Count();
                if (itemsInGroup == 1)
                {
                    var onlyItemInGroup = gr.First();

                    if (onlyItemInGroup.List == 1)
                    {
                        return onlyItemInGroup.Data;
                    }

                    // Item came from the second list. So multiply it's quantity by -1.
                    onlyItemInGroup.Data.QTY *= -1;

                    return onlyItemInGroup.Data;
                }

                // Since for each item in list 1 there is at most one item in the list2
                // and vice versa itemsInGroup now is 2 and it is safe to use First as below
                // to grab the items.

                var itemFromFirstList = gr.First(x => x.List == 1);
                var itemFromSecondList = gr.First(x => x.List == 2);

                return new List_Data
                {
                    Material = gr.Key.Material,
                    Batch = gr.Key.Batch,
                    QTY = itemFromFirstList.Data.QTY - itemFromSecondList.Data.QTY
                };
            }).ToList();

基本上，所有工作都在Select内完成，我们将两个列表串联起来并根据键Material和Batch将结果列表中的项目分组。根据最初的假设，我们提供的选项如下：

该组仅包含一个项目，该项目来自第一个列表。在这种情况下，我们只返回该项目包含的数据。
该组仅包含一个项目，该项目来自第二个列表。在这种情况下，我们必须将值QTY与-1相乘。请记住，您要使用的类型是list1.QTY - list2.QTY，并且第一个列表list1中没有任何关联的元素。因此，您想获得声明的-list2.QTY。
该组包含两个项目，自以来，我们假设一个列表中最多有（如果有）一个关联元素，而另一个列表中则有另一个元素。在这种情况下，我们只需从list2.QTY中减去list1.QTY就可以得到新的数量。

Answer 3

这是另一种解决方案

class MemoryLeak {

    private void startLeaking() throws IOException {
        StringBuilder input = new StringBuilder();
        URLConnection conn = new URL("www.example.com/file.txt").openConnection();

        BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream(), StandardCharsets.UTF_8));        

        while (br.readLine() != null) {
            input.append(br.readLine());
        }
    }

    public static void main(String[] args) throws IOException {
        MemoryLeak ml = new MemoryLeak();
        ml.startLeaking();
    }
}

.net fiddle sample

即使两个列表中都不存在项目，如何获取两个列表中的所有项目？

3 个答案: