Question

我有两个包含两个字段的列表。我事先并不知道哪一个更大。

我想实现这个目标：

        newList              oldList
Fruit       Qty     Diff    Fruit       Qty
--------------------------------------------
apple       3       +1      apple       2
                    -2      pear        2
peach       4       +3      peach       1
melon       5        0      melon       5
coconut     2       +2      
mango       4        0      mango       4
banana      2       -1      banana      3
lychi       1       +1      
                    -3      pineapple   3

左边的一个你可以看到oldList右边的newList。如果我将oldList与newList进行比较，我希望看到两个列表中每个可能的行和数量的差异。

我写了这个，这会给我两个列表中没有包含的水果（xor）

var difference = newList.Select(x => x.Fruit)
                        .Except(oldList.Select(x => x.Fruit))
                        .ToList()
                        .Union(oldList.Select(x => x.Fruit)
                        .Except(newList.Select(x => x.Fruit))
                        .ToList());

但我也失去了如何将它与数量相结合。

Answer 1

我没有看到LINQ表现良好且可读的方法。我更喜欢混合使用LINQ和循环：

var oldGrouped = oldList.ToDictionary(x => x.Fruit, x => x.Qty);
var newGrouped = newList.ToDictionary(x => x.Fruit, x => x.Qty);

var result = new List<FruitSummary>();

foreach(var oldItem in oldGrouped)
{
    int newQuantity;
    newGrouped.TryGetValue(oldItem.Key, out newQuantity)
    var summary = new FruitSummary { OldFruit = oldItem.Key, OldQty = oldItem.Value };
    if(newQuantity != 0)
    {
        summary.NewFruit = oldItem.Key;
        summary.NewQty = newQuantity;
    }
    summary.Diff = oldItem.Value - newQuantity;
    newGrouped.Remove(oldItem.Key);
    result.Add(summary);
}

foreach(var newItem in newGrouped)
{
    result.Add(new FruitSummary { Diff = -newItem.Value,
                                  NewFruit = newItem.Key,
                                  NewQty = newItem.Value });
}

班级FruitSummary如下所示：

public class FruitSummary
{
    public string OldFruit { get; set; }
    public string NewFruit { get; set; }
    public int OldQty { get; set; }
    public int NewQty { get; set; }
    public int Diff { get; set; }
}

Answer 2

我假设您有Fruit这样的课程：

public class Fruit
{
    public string Name {get; set;}
    public int Quantity {get; set;}
}

你有一个旧的和新的列表，如这些

List<Fruit> oldList = ...
List<Fruit> newList = ...

然后这个LINQ monstrum完成了这项工作，虽然它可能不是最高效的解决方案（但有多少种水果？）：

var result = 
    oldList.Join(newList, 
        oldFruit => oldFruit.Name,
        newFruit => newFruit.Name,
        (oldFruit, newFruit) => new {
            Name = oldFruit.Name,
            Diff = oldFruit.Quantity - newFruit.Quantity}).
    Concat(oldList.
        Where(oldFruit => newList.All(newFruit => newFruit.Name != oldFruit.Name)).
        Select(oldFruit => new {
            Name = oldFruit.Name,
            Diff = -oldFruit.Quantity})).
    Concat(newList.Where(newFruit => oldList.All(oldFruit => oldFruit.Name != newFruit.Name)).
        Select(newFruit => new {
            Name = newFruit.Name,
            Diff = newFruit.Quantity}));

您可以输出如下结果：

Console.WriteLine(string.Join(Environment.NewLine, result.Select(r => $"{r.Name} {r.Diff}")));

请注意，我想出了这个，因为我喜欢linq＆＃34;一个衬里＆＃34;，但使用foreach似乎更可读，甚至可能比这个查询更快。

Answer 3

这样的东西？

var difference = newList
    .Concat(oldList.Select(x => new { x.Fruit, Qty = -x.Qty }))
    .GroupBy(x => x.Fruit)
    .Select(g => new {Fruit = g.Key, Qty = g.Sum(x => x.Qty) });

当我运行foreach(var d in difference) Console.WriteLine(d);时，我得到：

{ Fruit = apple, Qty = 1 }
{ Fruit = peach, Qty = 3 }
{ Fruit = melon, Qty = 0 }
{ Fruit = coconut, Qty = 2 }
{ Fruit = mango, Qty = 0 }
{ Fruit = banana, Qty = -1 }
{ Fruit = lychi, Qty = 1 }
{ Fruit = pear, Qty = -2 }
{ Fruit = pineapple, Qty = -3 }

使用linq

3 个答案: