如果父键是通用的并且日期字段在彼此之间的 n(例如 2)小时内,我需要对集合进行分组。
示例数据:
List<DummyObj> models = new List<DummyObj>()
{
new DummyObj { ParentKey = 1, ChildKey = 1, TheDate = DateTime.Parse("01/01/2020 00:00:00"), Name = "Single item - not grouped" },
new DummyObj { ParentKey = 2, ChildKey = 2, TheDate = DateTime.Parse("01/01/2020 01:00:00"), Name = "Should be grouped with line below" },
new DummyObj { ParentKey = 2, ChildKey = 3, TheDate = DateTime.Parse("01/01/2020 02:00:00"), Name = "Grouped with above" },
new DummyObj { ParentKey = 2, ChildKey = 4, TheDate = DateTime.Parse("01/01/2020 04:00:00"), Name = "Separate item as greater than 2 hours" },
new DummyObj { ParentKey = 2, ChildKey = 5, TheDate = DateTime.Parse("01/01/2020 05:00:00"), Name = "Grouped with above" },
new DummyObj { ParentKey = 3, ChildKey = 6, TheDate = DateTime.Parse("01/01/2020 05:00:00"), Name = "Single item - not grouped" }
};
private class DummyObj
{
public int ParentKey { set; get; }
public int ChildKey { set; get; }
public DateTime TheDate { set; get; }
public string Name { set; get; }
}
结果分组应该是(子键):
{[1]}, {[2,3]}, {[4,5]}, {[6]}
我可以先按父键分组,然后循环比较组内的各个项目,但希望有一个更优雅的解决方案。
一如既往,非常感谢。
public static void Test()
{
var list = GetListFromDb(); //returns List<DummyObj>;
var sortedList = new List<DummyObj>();
foreach(var g in list.GroupBy(x => x.ParentKey))
{
if(g.Count() < 2)
{
sortedList.Add(g.First());
}
else
{
var datesInGroup = g.Select(x => x.TheDate);
var hoursDiff = (datesInGroup.Max() - datesInGroup.Min()).TotalHours;
if(hoursDiff <= 2)
{
string combinedName = string.Join("; ", g.Select(x => x.Name));
g.First().Name = combinedName;
sortedList.Add(g.First());
}
else
{
//now it's the mess
DateTime earliest = g.Select(x => x.TheDate).Min();
var subGroup = new List<DummyObj>();
foreach(var line in g)
{
if((line.TheDate - earliest).TotalHours > 2)
{
//add the current subgroup entry to the sorted group
subGroup.First().Name = string.Join("; ", subGroup.Select(x => x.Name));
sortedList.Add(subGroup.First());
//new group needed and new earliest date to start the group
sortedList = new List<DummyObj>();
sortedList.Add(line);
earliest = line.TheDate;
}
else
{
subGroup.Add(line);
}
}
//add final sub group, i.e. when there's none that are over 2 hours apart or the last sub group
if(subGroup.Count > 1)
{
subGroup.First().Name = string.Join("; ", subGroup.Select(x => x.Name));
sortedList.Add(subGroup.First());
}
else if(subGroup.Count == 1)
{
sortedList.Add(subGroup.First());
}
}
}
}
}
答案 0 :(得分:1)
给你:
List<DummyObj> models = new List<DummyObj>()
{
new DummyObj { ParentKey = 1, ChildKey = 1, TheDate = DateTime.Parse("01/01/2020 00:00:00"), Name = "Single item - not grouped" },
new DummyObj { ParentKey = 2, ChildKey = 2, TheDate = DateTime.Parse("01/01/2020 01:00:00"), Name = "Should be grouped with line below" },
new DummyObj { ParentKey = 2, ChildKey = 3, TheDate = DateTime.Parse("01/01/2020 02:00:00"), Name = "Grouped with above" },
new DummyObj { ParentKey = 2, ChildKey = 4, TheDate = DateTime.Parse("01/01/2020 04:00:00"), Name = "Separate item as greater than 2 hours" },
new DummyObj { ParentKey = 2, ChildKey = 5, TheDate = DateTime.Parse("01/01/2020 05:00:00"), Name = "Grouped with above" },
new DummyObj { ParentKey = 3, ChildKey = 6, TheDate = DateTime.Parse("01/01/2020 05:00:00"), Name = "Single item - not grouped" }
};
List<List<DummyObj>> groups =
models
.GroupBy(x => x.ParentKey)
.Select(xs => xs.OrderBy(x => x.TheDate).ToList())
.SelectMany(xs => xs.Skip(1).Aggregate(new[] { xs.Take(1).ToList() }.ToList(), (a, x) =>
{
if (x.TheDate.Subtract(a.Last().Last().TheDate).TotalHours < 2.0)
{
a.Last().Add(x);
}
else
{
a.Add(new [] { x }.ToList());
}
return a;
}))
.ToList();
string output =
String.Join(", ",
groups.Select(x =>
$"{{[{String.Join(",", x.Select(y => $"{y.ChildKey}"))}]}}"));
这给了我:
{[1]}, {[2,3]}, {[4,5]}, {[6]}