Question

我有一个数据集（汽车）：

Brand     DateSold    Amount  Bought/Sold
Toyota    06/07/2015  18.5    Bought
BMW       01/01/2016  25.15   Sold
Mercedes  06/06/2016  20.75   Bought

我想按年份分组并返回金额的总和，即：

Year Amount  
2015 -18.5   
2016 4.4

并将其输出到列表框中。

我可以在没有购买/出售条件的情况下求和：

var AmountsByYear = cars.GroupBy(i => i.Date.Year)
                        .Select(g => new { 
                            Year = g.Key,
                            Total = g.Sum(i => i.Amount)
                        }).ToList();
lstAmountsByYear.DataSource = AmountsByYear;

Answer 1

由于您没有详细说明如何在数据结构中定义“购买/出售”字段，因此建议您使用一个枚举。例如，

public class Car
{
    public string Brand;
    public DateTime Date;
    public double Amount;
    public BusinessType BusinessType;
}

public enum BusinessType
{
    Bought = -1,
    Sold = 1
}

这将使您以最小的变化使用查询，以达到预期的结果。

var AmountsByYear = cars.GroupBy(i => i.Date.Year)
                    .Select(g => new { 
                        Year = g.Key,
                        Total = g.Sum(i => i.Amount*(int)i.BusinessType)
                    }).ToList();

输入数据，

var cars = new List<Car>
{
    new Car{Brand="Toyota", Date = new DateTime(2015,06,07),Amount=18.5,BusinessType=BusinessType.Bought},
    new Car{Brand="BMW", Date = new DateTime(2016,01,01),Amount=25.15,BusinessType=BusinessType.Sold},
    new Car{Brand="Mercedes", Date = new DateTime(2016,06,06),Amount=20.75,BusinessType=BusinessType.Bought},
};

输出

Answer 2

考虑创建Car的扩展方法，该方法将[Amount，bought / sold]的组合转换为Amount的正值或负值：

public static decimal ToProperAmountValue(this Car car)
{
    return car.IsCarBought ? -car.Amount : car.Amount;
}
// TODO: invent proper method name

此外，使用已经为您执行“选择”的正确overload of Enumerable.GroupBy：

var amountsByYear = cars.GroupBy(

    // KeySelector: Group Cars into Groups of same year
    car => car.Date.Year)

    // ResultSelector: take the Key and the Cars with this Key to make a new object
    (year, carsInThisYear => new
    { 
        Year = year,
        Total = carsInThisYear.Sum(car => car.ToProperAmountValue())
    })
    .ToList();

LINQ-分组依据，然后有条件地求和

2 个答案: