如何选择每个类别的最近距离

Question

我正在尝试获得每个类别中最近的位置。任何人都可以帮我这么做吗？

var xml = new XElement("Locations", 
                    locations.OrderBy(n => n.CategoryID)
                        .ThenBy(n => distance(lat, lon, (double)n.Latitude, (double)n.Longitude))
                        .Where(n => (distance(lat, lon, (double)n.Latitude, (double)n.Longitude) <= 5))
                        .Select(location => 
                        new XElement("Location",
                            new XAttribute("CategoryID", location.CategoryID),
                            new XElement("Category", location.Category),
                            new XElement("LocationID", location.LocationID),
                            new XElement("LocationName", location.LocationName),
                            new XElement("Latitude", location.Latitude),
                            new XElement("Longitude", location.Longitude),
                            new XElement("Distance", distance(lat, lon, (double)location.Latitude, (double)location.Longitude)),
                            new XElement("Status", (location.HasManagedHours ? "Managed Hours" : "Open"))
                            )));

Answer 1

我没有测试过，但我建议分组是要走的路，比如：

var xml = new XElement("Locations",                             
  locations
    .GroupBy(n => n.CategoryID)
    .SelectMany(g => g
       .OrderBy(n => distance(lat, lon, (double)n.Latitude, (double)n.Longitude))
       .Take(1))
    .Select(location => 
      new XElement("Location",
          new XAttribute("CategoryID", location.CategoryID),
          new XElement("Category", location.Category),
          new XElement("LocationID", location.LocationID),
          new XElement("LocationName", location.LocationName),
          new XElement("Latitude", location.Latitude),
          new XElement("Longitude", location.Longitude),
          new XElement("Distance", distance(lat, lon, (double)location.Latitude, (double)location.Longitude)),
          new XElement("Status", (location.HasManagedHours ? "Managed Hours" : "Open"))
          )));

有关详细信息，请参阅Projection Operators和Grouping Operators。