有没有办法将数组翻译成字典答案

【问题标题】：Is there a way to Translate an array into a dictionary有没有办法将数组翻译成字典
【发布时间】：2019-07-03 15:27:21
【问题描述】：

我遇到的问题是我不确定如何将我的double 数组转换为Dictionary。我有一些很重的 LINQ Where 语句，所以我想使用我的 Dictionary 的键来查找特定值。我有两个带有 getter 和 setter 的类，然后是我的计算器。

我在尝试查找或Dictionary 时摆弄了一下，但没有任何运气。

public class Product
{
    public int EarliestOriginYear { get; set; }
    public int NoOfDevelopmentYears { get; set; }
    public string ProductName { get; set; }
    public IEnumerable<ProductIncrementalValue> ProductIncrementalValues { get; set; }
}

public class ProductIncrementalValue
{
    public string ProductName { get; set; }
    public int OriginYear { get; set; }
    public int DevelopmentYear { get; set; }
    public double IncrementalValue { get; set; }
}

public IList<double> Calculate(Product product)
{
     IList<double> cumulativeDataTriangle = new List<double>();
     if (!product.ProductIncrementalValues.Any())
        return cumulativeDataTriangle;

     for (int i = 0; i < product.NoOfDevelopmentYears; i++)
     {
        // This is what I want to change (where statements)
        var incrementalValues = product.ProductIncrementalValues
            .Where(v => v.OriginYear == product.EarliestOriginYear + i)
            .ToList(); 

        double previous = 0;

        for (int j = 0; j < product.NoOfDevelopmentYears - i; j++)
        {
            // This is what I want to change
            double incrementalValue = incrementalValues.Where(val =>
                val.DevelopmentYear == val.OriginYear + j)
                .Select(x => x.IncrementalValue)
                .FirstOrDefault();

            var tmp = incrementalValue + previous;
            cumulativeDataTriangle.Add(tmp);
            previous = tmp;
        }
     }

     return cumulativeDataTriangle;
}

【问题讨论】：

double 不是一个很好的候选者，因为舍入错误：字典可以有，比如 0.9999999999999 键，当你要求它 1.00000000000001 时跨度>

标签： c# arrays dictionary

【解决方案1】：

您可以在循环之前按OriginYear 和DevelopmentYear 对产品进行分组。这样的事情可能会有所帮助：

public IList<double> Calculate(Product product)
{
    IList<double> cumulativeDataTriangle = new List<double>();
    if (!product.ProductIncrementalValues.Any())
        return cumulativeDataTriangle;

    var lookup = product.ProductIncrementalValues.ToLookup(v => (v.OriginYear, v.DevelopmentYear), v => v.IncrementalValue); 
    for (int i = 0; i < product.NoOfDevelopmentYears; i++)
    {
        var originYear = product.EarliestOriginYear + i;
        double previous = 0;
        for (int j = 0; j < product.NoOfDevelopmentYears - i; j++)
        {
            var developmentYear = originYear + j;
            var incrementalValues = lookup[(originYear, developmentYear)];
            double incrementalValue = incrementalValues.FirstOrDefault();

            var tmp = incrementalValue + previous;
            cumulativeDataTriangle.Add(tmp);
            previous = tmp;
        }
    }

    return cumulativeDataTriangle;
}

【讨论】：

【解决方案2】：

感谢上一个答案，我设法基于OriginYear 和DevelopmentYear 实现了Dictionary，它返回了IncrementalValue。

public IList<double> Calculate(Product product)
        {
            IList<double> cumulativeDataTriangle = new List<double>();
            if (!product.ProductIncrementalValues.Any())
                return cumulativeDataTriangle;

            var lookup = product.ProductIncrementalValues.
                ToDictionary(v => (v.OriginYear, v.DevelopmentYear), v => v.IncrementalValue);
            for (int i = 0; i < product.NoOfDevelopmentYears; i++)
            {
                var originYear = product.EarliestOriginYear + i;
                double previous = 0;
                for (int j = 0; j < product.NoOfDevelopmentYears - i; j++)
                {
                    var developmentYear = originYear + j;
                    double incrementalValue;
                    lookup.TryGetValue((originYear, developmentYear), out incrementalValue);

                    var tmp = incrementalValue + previous;
                    cumulativeDataTriangle.Add(tmp);
                    previous = tmp;
                }
            }

            return cumulativeDataTriangle;
        }

【讨论】：