操作一次，总结费

return new SchoolFees(
     new Percentage(schoolFeesResult.Sum(x => (x.Amount.Value/totalFees) * x.TuitionFee.Value)), 
     new Percentage(schoolFeesResult.Sum(x => (x.Amount.Value/totalFees) * x.TravellingFee.Value)), 
     new Percentage(schoolFeesResult.Sum(x => (x.Amount.Value/totalFees) * x.ResidentialFee.Value)));

加权平均有没有一种方法我可以schoolFeesResult运行一次，以计算出每个不同类型的费用（Tuition，Travelling，Residence）的加权平均。基本上我不想让(x.Amount.Value/totalFees)在我的代码中出现3次？操作一次，总结费

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2013-01-22 General_9

又一个脑破解

Func<Func<Fee, decimal>, decimal> totalFee = feeSelector => 
    schoolFeesResult.Sum(x => x.Amount.Value/totalFees * feeSelector(x)); 

return new SchoolFees(
    new Percentage(totalFee(f => f.TuitionFee.Value)), 
    new Percentage(totalFee(f => f.TravellingFee.Value)), 
    new Percentage(totalFee(f => f.ResidentialFee.Value)) 
);

或者更短：

Func<Func<Fee, decimal>, Percentage> percentageOf = feeSelector => 
    new Percentage(schoolFeesResult.Sum(x => 
     x.Amount.Value/totalFees * feeSelector(x))); 

return new SchoolFees(
    percentageOf(f => f.TuitionFee.Value), 
    percentageOf(f => f.TravellingFee.Value), 
    percentageOf(f => f.ResidentialFee.Value) 
);

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2013-01-22 13:16:17

+1：很好，我喜欢它。注意：这也将枚举'schoolFeesResult'三次。 –

@DanielHilgarth谢谢！是的，它枚举了三次，但它不错:)顺便说一句，没有什么区别 - 枚举三次（原始代码），但我不计算每个'重量*费用':) –

它会计算每个重量（' x.Amount.Value/totalFees'）三次，就像原始代码一样。就像原始代码一样，它只会计算每个单独的“重量*费用”。所以，就计算而言，它是一样的。只是更多的风格:) –

我用这个实现的WeightedAverage作为一个扩展的方法来IEnumerable<T>：

public static double? WeightedAverage<TSource>(this IEnumerable<TSource> source 
                 , Func<TSource, float> weightField 
                 , Func<TSource, double> propertyToWeight) 
{ 
    var total = source.Sum(weightField); 

    var sum = source.Select(item => weightField(item) * propertyToWeight(item)).Sum(); 
    return sum/total; 

}

有几个重载处理single，single?当然double。也许你可以调整它来适应你想要达到的目标。

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2013-01-22 12:35:01 Jamiec

我想你可以把它放在另一个查询，恕我直言，这也更redable然后：

var percentages = schoolFeesResult 
    .Select(x => new { SFR = x, AmoundDivFees = (x.Amount.Value/totalFees)}) 
    .Select(x => new { 
     TuitionFee = x.AmoundDivFees * x.SFR.TuitionFee.Value, 
     TravellingFee = x.AmoundDivFees * x.SFR.TravellingFee.Value, 
     ResidentialFee = x.AmoundDivFees * x.SFR.ResidentialFee.Value 
    }); 
return new SchoolFees(
    new Percentage(percentages.Sum(x => x.TuitionFee)), 
    new Percentage(percentages.Sum(x => x.TravellingFee)), 
    new Percentage(percentages.Sum(x => x.ResidentialFee)));

当然，我不能对它进行测试。

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2013-01-22 12:37:52

你可以使用这样的事情：

var fees = from fee in schoolFeesResult 
      let weight = fee.Amount.Value/totalFees 
      select new 
      { 
       TuitionFee = weight * fee.TuitionFee.Value, 
       TravellingFee = weight * fee.TravellingFee.Value, 
       ResidentialFee = weight * fee.ResidentialFee.Value 
      }; 

// if the calculation of the fees is a performance bottleneck, 
// uncomment the next line: 
// fees = fees.ToList(); 

return new SchoolFees(
    new Percentage(fees.Sum(x => x.TuitionFee), 
    new Percentage(fees.Sum(x => x.TravellingFee), 
    new Percentage(fees.Sum(x => x.ResidentialFee));

你甚至能走得更远：

var fees = (from fee in schoolFeesResult 
      let weight = fee.Amount.Value/totalFees 
      group fee by 1 into g 
      select new 
      { 
       TuitionFee = g.Sum(x => weight * x.TuitionFee.Value), 
       TravellingFee = g.Sum(x => weight * x.TravellingFee.Value), 
       ResidentialFee = g.Sum(x => weight * x.ResidentialFee.Value) 
      }).Single(); 

return new SchoolFees(
    new Percentage(fees.TuitionFee, 
    new Percentage(fees.TravellingFee, 
    new Percentage(fees.ResidentialFee);

但我怀疑，这第二个版本是一个好主意。它使代码难以理解。我纯粹出于学术原因添加了它，以展示什么是可能的。

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2013-01-22 12:44:01

我想你需要'ToList'或每个费将被计算三次 –

@lazyberezovsky：这是正确的。如果这是一个问题，你可以添加'ToList'没有问题。我没有添加它，因为我认为这里没有真正的区别。它会使代码看起来不那么“好”:-)然而，我添加了一条评论，因为它确实是一个重要的观点。 –

Ha-ha :) +1代码。如果代码看起来不错，谁会关心性能？ –

操作一次，总结费

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