列出与其他列表条件相关的操作

Question

我有两个这样的列表:

public static List<List<List<double>>> listQ = new List<List<List<double>>>();
public static List<List<List<double>>> listVal = new List <List<List<double>>>();

2列表也有相同的大小.我想在listQ关系中对listVal进行一些操作:

• 如果listVal的任何索引具有相同的值,我想在相同的索引中取平均listQ的值.

例如:

listVal[0][0][ 1] = listVal[0][0][2] = 3. 

所以,我想listQ的相同索引是相同的.

(listQ[0][0][ 1] + listQ[0][0][2]) / 2 = (26.5 + 20.4) / 2 = 23.45. 

现在,每个这些索引必须有23.45:

listQ[0][0][ 1] = 23.45

listQ[0][0][ 2] = 23.45

同样地;

listVal[0][2][0] = listVal[0][2][2] = 1.因此listQ[0][2][0] and listQ[0][2][2]必须采取平均值.

我怎样才能做到这一点？

编辑:@Roman Izosimov和@Petko Petkov的解决方案正常运作.哪一个有更高的性能？你怎么看？

Answer 1

lulz 的 LINQ 解决方案:)

List<List<List<double>>> listVal = new List<List<List<double>>>(){
    new List<List<double>>{
        new List<double>(){1,1,3},
        new List<double>(){2,1,2},
        new List<double>(){1,2,3}
    },
    new List<List<double>>{
        new List<double>(){2,1,3},
        new List<double>(){2,4,2},
        new List<double>(){3,1,3}
    },
    new List<List<double>>{
        new List<double>(){4,1,1},
        new List<double>(){4,2,1},
        new List<double>(){4,3,1}
    }
};
List<List<List<double>>> listQ = new List<List<List<double>>>(){
        new List<List<double>>{
        new List<double>(){3,7,4},
        new List<double>(){8,15,23},
        new List<double>(){11,13,17}
    },
    new List<List<double>>{
        new List<double>(){90,3,7},
        new List<double>(){5,7,12},
        new List<double>(){7,14,21}
    },
    new List<List<double>>{
        new List<double>(){32,4,1},
        new List<double>(){55,12,8},
        new List<double>(){3,5,8}
    }
};


//Linq awesomeness
var qry = listVal.SelectMany((l1, i0) =>
                l1.SelectMany((l2, i1) =>
                    l2.Select((ele, i2) =>
                        new { i0, i1, i2, gVal = ele, qVal = listQ[i0][i1][i2] })))
                .GroupBy(x => new { x.i0, x.i1, x.gVal }) // if you want to average across the innermost lists only 
                //.GroupBy(x => x.gVal)                   //if you want to average acreoss the whole data
                .SelectMany(x => x.Select(e => new { e.i0, e.i1, e.i2, avg = x.Average(y => y.qVal) }));
foreach (var e in qry)
{
    listQ[e.i0][e.i1][e.i2] = e.avg;
}