我有一个循环,一旦简化,看起来像这样:
Dictionary<Tuple<A,G>,Decimal> results = new Dictionary<Tuple<A,G>,Decimal>();
foreach( A a in collectionA )
foreach( B b in collectionB )
results [Tuple.Create(a, (G)b.groupBy)] += (Decimal) Func(a, b);
有没有办法可以使用Linq查询复制此结果(例如GroupBy
,Sum
和ToDictionary
? (正如this对上一个问题的回答所建议的那样:Concise way to do a plus equals operation on a Dictionary element that might not be initialized)
结果
//Dictionary<EventGroupIDLayerTuple, Decimal> BcEventGroupLayerLosses
使用下面的Yuxiu Li的答案,我能够从链接的问题转换这4个班轮:
BcEventGroupLayerLosses = new Dictionary<EventGroupIDLayerTuple, Decimal>();
foreach( UWBCEvent evt in this.BcEvents.IncludedEvents )
foreach( Layer lyr in this.ProgramLayers )
BcEventGroupLayerLosses.AddOrUpdate(
new EventGroupIDLayerTuple(evt.EventGroupID, lyr),
GetEL(evt.AsIfs, lyr.LimitInMillions, lyr.AttachmentInMillions),
(a, b) => a + b);
进入这一个班轮:
BcEventGroupLayerLosses = this.BcEvents.IncludedEvents
.SelectMany(evt => ProgramLayers, (evt, lyr) => new { evt, lyr })
.GroupBy(g => new EventGroupIDLayerTuple(g.evt.EventGroupID, g.lyr),
g => GetEL(g.evt.AsIfs, g.lyr.LimitInMillions, g.lyr.AttachmentInMillions))
.ToDictionary(g => g.Key, g => g.Sum());
两者都产生了相同的结果。
虽然没有特别可读,但这是一个很好的实验。谢谢大家的帮助!
答案 0 :(得分:4)
Dictionary<Tuple<A, G>, decimal> dictionary =
(from a in collectionA
from b in collectionB
group (decimal)Func(a, b) by Tuple.Create<A, G>(a, b.groupBy))
.ToDictionary(g => g.Key, g => g.Sum());
在声明性语法
中var dictionary = collectionA
.SelectMany(a => collectionB,
(a, b) => new { a, b })
.GroupBy(g => Tuple.Create(g.a, g.b.groupBy),
g => Func(g.a, g.b))
.ToDictionary(g => g.Key, g => g.Sum());
答案 1 :(得分:1)
我怀疑你想要的东西:
// Extract the key/value pair from the nested loop
var result = collectionA.SelectMany(a => collectionB,
(a, b) => new {
Key = Tuple.Create(a, (G)b.groupBy),
Value = (decimal) Func(a, b)
})
// Group by the key, and convert each group's values
// to its sum
.GroupBy(pair => pair.Key,
pair => pair.Value,
(key, values) => new { Key = key,
Value = values.Sum() })
// Make a dictionary from the key/value pairs
.ToDictionary(pair => pair.Key, pair => pair.Value);
这是我的头脑,可能需要更多括号:)我没有时间添加解释,但稍后可以这样做。