我有一个元组,我正在使用LINQ
进行迭代List<Tuple<string[], double[]>> tt = new List<Tuple<string[], double[]>>();
var t1 = new Tuple<string[], double[]>(
new string[] { "java", "c#", "nn" },
new double[] { 2, 3, 0 });
var t2 = new Tuple<string[], double[]>(
new string[] { "java", "c#", "nn" },
new double[] { 0, 3, 1 });
var t3 = new Tuple<string[], double[]>(
new string[] { "java", "c#", "nn" },
new double[] { 2, 1, 0 });
tt.Add(t1);
tt.Add(t2);
tt.Add(t3);
var XX = (from tuples in tt
let rows = tuples.Item2.ToArray()
let result = rows.Select(x => x/rows.Count(i => i != 0)).ToArray()
select new Tuple<string[], double[]>(tuples.Item1, result)
).ToList();
如果我想使用.Asparallel
,我可以采用以下方法之一:
我可以在循环列表时添加它:
var XX = (from tuples in tt
.AsParallel() //Parallel added here
let rows = tuples.Item2.ToArray()
let result = rows.Select(x => x/rows.Count(i => i != 0)).ToArray()
select new Tuple<string[], double[]>(tuples.Item1, result)
).ToList();
我可以在使用.ToArray()
或.ToList
var XX = (from tuples in tt
let rows = tuples.Item2.ToArray()
let result = rows.Select(x => x / rows.Count(i => i != 0)).AsParallel().ToArray() //Parallel added right here
select new Tuple<string[], double[]>(tuples.Item1, result)
).ToList();
使用.Asparallel
的最有效方法是什么?所有这些方法都没有错误,但我不确定它们是否会影响性能。
并且这样添加.Asparallel
是一个坏主意吗?:
var XX = (from tuples in tt
.AsParallel() //added here
let rows = tuples.Item2.AsParallel().ToArray() //here
let result = rows.Select(x => x / rows.Count(i => i != 0)).AsParallel().ToArray()//here
select new Tuple<string[], double[]>(tuples.Item1, result)
).AsParallel().ToList(); // and here
答案 0 :(得分:1)
在from
子句中启动LINQ查询时,应添加let
。这将使两个ToList
语句并行执行,之后在最后调用AsParallel
时将它们组合在一起。
您还可以在查询中调用Select
之前添加其他var XX = (from tuples in tt.AsParallel()
let rows = tuples.Item2.ToArray()
let result = rows.AsParallel().Select(x => x/rows.Count(i => i != 0)).ToArray()
select new Tuple<string[], double[]>(tuples.Item1, result)
).ToList();
。
AsParallel
但是,您应该记住YKLOI -6 01 123456789012345678901234
YKLOI -5 25 123456789012345678901234
YKLOI -4 05 123456789012345678901234
YKLOI -3 75 123456789012345678901234
YKLOI -2 83 123456789012345678901234
YKLOI -1 05 123456789012345678901234
YKLOI 0 34 123456789012345678901234
YKLOI 1 28 123456789012345678901234
YKLJW -6 87 569845874254658425485
YKLJW -5 87 569845874254658425485
...
有一些开销,所以在某些情况下它可能比常规的非并行处理慢。取决于元素的数量和计算的成本。