拥有一组对象。示意性地:
[
{ A = 1, B = 1 }
{ A = 1, B = 2 }
{ A = 2, B = 3 }
{ A = 2, B = 4 }
{ A = 1, B = 5 }
{ A = 3, B = 6 }
]
需要:
[
{ A = 1, Bs = [ 1, 2 ] }
{ A = 2, Bs = [ 3, 4 ] }
{ A = 1, Bs = [ 5 ] }
{ A = 3, Bs = [ 6 ] }
]
LINQ可以吗?
注意:订购非常重要。因此,Bs = [5]
无法与Bs = [1, 2]
答案 0 :(得分:2)
鉴于这些简单的课程:
class C {
public int A;
public int B;
}
class R {
public int A;
public List<int> Bs = new List<int>();
}
你可以这样做:
var cs = new C[] {
new C() { A = 1, B = 1 },
new C() { A = 1, B = 2 },
new C() { A = 2, B = 3 },
new C() { A = 2, B = 4 },
new C() { A = 1, B = 5 },
new C() { A = 3, B = 6 }
};
var rs = cs.
OrderBy(o => o.B).
ThenBy(o => o.A).
Aggregate(new List<R>(), (l, o) => {
if (l.Count > 0 && l.Last().A == o.A) {
l.Last().Bs.Add(o.B);
}
else {
l.Add(new R { A = o.A, Bs = { o.B } });
}
return l;
});
注意:在上面我假设必须对Bs和As进行排序。如果情况并非如此,那么删除排序说明就很简单了:
var rs = cs.
Aggregate(new List<R>(), (l, o) => {
if (l.Count > 0 && l.Last().A == o.A) {
l.Last().Bs.Add(o.B);
}
else {
l.Add(new R { A = o.A, Bs = { o.B } });
}
return l;
});
答案 1 :(得分:2)
所以基本上你想把具有相同A
- 值的东西组合在一起并且是连续的。
您需要将对象列表转换为包含上一个/下一个元素的匿名类型。我使用了两个Selects
来使它更加红色。然后你需要检查两个元素是否是连续的(相邻的索引)。
现在,您拥有GroupBy
所需的全部价值和bool
。
你的对象:
var list = new System.Collections.Generic.List<Foo>(){
new Foo(){ A = 1, B = 1 },
new Foo(){ A = 1, B = 2 },
new Foo(){ A = 2, B = 3 },
new Foo(){ A = 2, B = 4 },
new Foo(){ A = 1, B = 5 },
new Foo(){ A = 3, B = 6 }
};
查询:
var groups = list
.Select((f, i) => new
{
Obj = f,
Next = list.ElementAtOrDefault(i + 1),
Prev = list.ElementAtOrDefault(i - 1)
})
.Select(x => new
{
A = x.Obj.A,
x.Obj,
Consecutive = (x.Next != null && x.Next.A == x.Obj.A)
|| (x.Prev != null && x.Prev.A == x.Obj.A)
})
.GroupBy(x => new { x.Consecutive, x.A });
输出结果:
foreach (var abGroup in groups)
{
int aKey = abGroup.Key.A;
var bList = string.Join(",", abGroup.Select(x => x.Obj.B));
Console.WriteLine("A = {0}, Bs = [ {1} ] ", aKey, bList);
}
以下是有效的演示:http://ideone.com/fXgQ3
答案 2 :(得分:2)
您可以使用The GroupAdjacent Extension Method。
然后,你只需要
var grps = objects.GroupAdjacent(p => new { p.A });
我认为这是实施它的最简单方法。
修改强>
这是我的测试代码。
class Program
{
static void Main(string[] args)
{
var ia = new Dummycls[] {
new Dummycls{ A = 1, B = 1 },
new Dummycls{ A = 1, B = 2 },
new Dummycls{ A = 2, B = 3 },
new Dummycls{ A = 2, B = 4 },
new Dummycls{ A = 1, B = 5 },
new Dummycls{ A = 3, B = 6 },
};
var groups = ia.GroupAdjacent(i => i.A);
foreach (var g in groups)
{
Console.WriteLine("Group {0}", g.Key);
foreach (var i in g)
Console.WriteLine(i.ToString());
Console.WriteLine();
}
Console.ReadKey();
}
}
class Dummycls
{
public int A { get; set; }
public int B { get; set; }
public override string ToString()
{
return string.Format("A={0};B={1}" , A , B);
}
}
结果是
Group 1
A=1;B=1
A=1;B=2
Group 2
A=2;B=3
A=2;B=4
Group 1
A=1;B=5
Group 3
A=3;B=6
答案 3 :(得分:1)
这是执行您想要的方法的结构:
public static IEnumerable<IGrouping<TKey, TElement>> GroupWithKeyBreaks<T, TKey, TElement>(IEnumerable<T> enumerable,
Func<T, TKey> keySelector,
Func<T, TElement> itemSelector)
{
// Error handling goes here
TKey currentKey = default(TKey);
List<TElement> elements = new List<TElement>();
foreach (T element in enumerable)
{
TKey thisKey = keySelector(element);
if (thisKey == null)
{
continue;
}
if (!thisKey.Equals(currentKey) && elements.Count > 0)
{
yield return new SimpleGrouping<TKey, TElement>(currentKey, elements);
elements = new List<TElement>();
}
elements.Add(itemSelector(element));
currentKey = thisKey;
}
// Add the "last" item
if (elements.Count > 0)
{
yield return new SimpleGrouping<TKey, TElement>(currentKey, elements);
}
}
它使用以下助手类:
private class SimpleGrouping<T, U> : IGrouping<T, U>
{
private T key;
private IEnumerable<U> grouping;
T IGrouping<T, U>.Key
{
get { return key; }
}
IEnumerator<U> IEnumerable<U>.GetEnumerator()
{
return grouping.GetEnumerator();
}
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return grouping.GetEnumerator();
}
public SimpleGrouping(T k, IEnumerable<U> g)
{
this.key = k;
this.grouping = g;
}
}
以下是一个示例用法:
foreach (var grouping in data.GroupWithKeyBreaks(x => x.A, x => x.B))
{
Console.WriteLine("Key: " + grouping.Key);
foreach (var element in grouping)
{
Console.Write(element);
}
}
答案 4 :(得分:1)
var result = list.ToKeyValuePairs(x => x.A)
.Select(x => new { A = x.Key, Bs = x.Value.Select(y => y.B) });
foreach (var item in result)
{
Console.WriteLine("A = {0} Bs=[{1}]",item.A, String.Join(",",item.Bs));
}
-
public static class MyExtensions
{
public static IEnumerable<KeyValuePair<S,IEnumerable<T>>> ToKeyValuePairs<T,S>(
this IEnumerable<T> list,
Func<T,S> keySelector)
{
List<T> retList = new List<T>();
S prev = keySelector(list.FirstOrDefault());
foreach (T item in list)
{
if (keySelector(item).Equals(prev))
retList.Add(item);
else
{
yield return new KeyValuePair<S, IEnumerable<T>>(prev, retList);
prev = keySelector(item);
retList = new List<T>();
retList.Add(item);
}
}
if(retList.Count>0)
yield return new KeyValuePair<S, IEnumerable<T>>(prev, retList);
}
}
<强>输出:强>
A = 1 Bs=[1,2]
A = 2 Bs=[3,4]
A = 1 Bs=[5]
A = 3 Bs=[6]
答案 5 :(得分:1)
var groupCounter = 0;
int? prevA = null;
collection
.Select(item => {
var groupId = item.A == prevA ? groupCounter : ++groupCounter;
prevA = item.A;
return new { groupId, item.A, item.B };
})
.GroupBy(item => item.groupId)
.Select(grp => new { A = grp.First().A, Bs = grp.Select(g => g.B) });
答案 6 :(得分:1)
如果您的收藏集在o
,那么:
var trans = o.Aggregate
(
new {
List = new List<Tuple<int, List<int>>>(),
LastSeed = (int?)0
},
(acc, item) =>
{
if (acc.LastSeed == null || item.A != acc.LastSeed)
acc.List.Add(Tuple.Create(item.A, new List<int>()));
acc.List[acc.List.Count - 1].Item2.Add(item.B);
return new { List = acc.List, LastSeed = (int?)item.A};
},
acc => acc.List.Select(
z=>new {A = z.Item1,
B = z.Item2 as IEnumerable<int>
})
);
这会产生IEnumerable<int, IEnumerable<int>>
所需的表格。