Question

我有几个函数可以获取对象列表并从中创建一个图表系列。它看起来是重复的，我觉得我可以用Lambda表达式或模板替换这些看似相似的函数，但是却无法弄清楚如何处理每个对象上的不同属性名称。有人可以帮忙吗？

这就是我所拥有的：

private static Series getBuysSeries(List<Order> buys)
{
    var series = new Series();
    foreach (var buy in buys)
    {
        series.Points.AddXY(buy.Price, buy.Quantity);
    }
    return series;
}

private Series getPriceSeries(List<Trade> trades)
{
    var series = new Series();
    foreach (var trade in trades)
    {
        series.Points.AddXY(trade.DateTimeUTC, trade.UnitPrice);

    }
    return series;
}

我认为它看起来像：

private static Series getBuysSeries(List<Order> buys)
{

    return getSeries<Order>(buys);
}

private Series getPriceSeries(List<Trade> trades)
{
    return getSeries<Trade>(trades);
}

private Series getSeries<T>(List<T> list)
{
    ...do the loop but account for property names...
}

Answer 1

是的，你可以。至少，你可能会。例如 - 将方法名称更改为更常规，并对方法调用使用泛型类型推断：

private static Series GetBuysSeries(List<Order> buys)
{
    return GetSeries(buys, buy => buy.Price, buy => buy.Quantity);
}

private Series GetPriceSeries(List<Trade> trades)
{
    return getSeries<Trade>(trades,
                            trade => trade.DateTimeUtc,
                            trade => trade.UnitPrice);
}

private Series GetSeries<T>(List<T> list,
                            Func<T, double> selector1,
                            Func<T, double> selector2)
{
    var series = new Series();
    foreach (var item in list)
    {
        series.Points.AddXY(selector1(item), selector2(item));
    }
    return series;
}

现在假设AddXY需要两个double值。 trade.DateTimeUTC似乎有点奇怪，但如果不知道如何定义Series.AddXY，很难提供更具体的建议。

如果您确实想在每种情况下调用AddXY的其他重载，那么它会有点棘手。在这种情况下，您可能需要使用动态类型，例如

private Series GetSeries<T>(List<T> list,
                            Func<T, dynamic> selector1,
                            Func<T, dynamic> selector2)
{
    var series = new Series();
    foreach (var item in list)
    {
        // Here the overload is determined at execution time...
        series.Points.AddXY(selector1(item), selector2(item));
    }
    return series;
}

我可以用lambda表达式替换多个类似的方法吗？

1 个答案: