操作一次，總結費

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)));

加權平均有沒有一種方法我可以schoolFeesResult運行一次，以計算出每個不同類型的費用（Tuition，Travelling，Residence）的加權平均。基本上我不想讓(x.Amount.Value/totalFees)在我的代碼中出現3次？操作一次，總結費

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2013-01-22 General_9

又一個腦破解

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，我喜歡它。注意：這也將枚舉'schoolFeesResult'三次。 –

@DanielHilgarth謝謝！是的，它枚舉了三次，但它不錯:)順便說一句，沒有什麼區別 - 枚舉三次（原始代碼），但我不計算每個'重量*費用':) –

它會計算每個重量（' x.Amount.Value/totalFees'）三次，就像原始代碼一樣。就像原始代碼一樣，它只會計算每個單獨的「重量*費用」。所以，就計算而言，它是一樣的。只是更多的風格:) –

我用這個實現的WeightedAverage作爲一個擴展的方法來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; 

}

有幾個重載處理single，single?當然double。也許你可以調整它來適應你想要達到的目標。

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2013-01-22 12:35:01 Jamiec

我想你可以把它放在另一個查詢，恕我直言，這也更redable然後：

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)));

當然，我不能對它進行測試。

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2013-01-22 12:37:52

你可以使用這樣的事情：

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));

你甚至能走得更遠：

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);

但我懷疑，這第二個版本是一個好主意。它使代碼難以理解。我純粹出於學術原因添加了它，以展示什麼是可能的。

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2013-01-22 12:44:01

我想你需要'ToList'或每個費將被計算三次 –

@lazyberezovsky：這是正確的。如果這是一個問題，你可以添加'ToList'沒有問題。我沒有添加它，因爲我認爲這裏沒有真正的區別。它會使代碼看起來不那麼「好」:-)然而，我添加了一條評論，因爲它確實是一個重要的觀點。 –

Ha-ha :) +1代碼。如果代碼看起來不錯，誰會關心性能？ –

操作一次，總結費

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