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c# - 如何在 ilnumerics 逻辑对称矩阵中查找维度?

转载 作者:太空宇宙 更新时间:2023-11-03 13:34:31 26 4
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我有一个像这样的 ilnumerics 逻辑对称矩阵

0   0   0   0   0   0   1   1   1   1
0 0 0 0 0 1 0 1 1 1
0 0 0 0 0 1 1 0 1 1
0 0 0 0 0 1 1 0 0 1
0 0 0 0 0 1 1 1 1 0
0 1 1 1 1 0 0 0 0 0
1 0 1 1 1 0 0 0 0 0
1 1 0 0 1 0 0 0 0 0
1 1 1 0 1 0 0 0 0 0
1 1 1 1 0 0 0 0 0 0

我想获取值 == 1 的所有维度(行和列)

结果 =(0,6), (0,7), (0,8), (0,9)(1,5), (1,7), (1,8), (1,9)(2,5), (2,6), (2,8), (1,9)(3,5), (3,6), (3,9)(4,5), (4,6), (4,7), (4,8)

在 C# 中使用 ilnumerics 库是否有更快的方法?

编辑:这是我的解决方案

    ILNumerics.ILLogical matrixThreshold;
..... Some C# code
for (int i = 0; i < matrixThreshold.Length; i++)
for (int j = i + 1; j < matrixThreshold.Length; j++)
if (matrixThreshold.GetValue(i, j) == 1) Console.Write("({0},{1}){2}", i, j, Environment.NewLine);

最佳答案

ILLogical L = ILMath.rand(10, 12) > 0.5;

>L
Logical [10,12]
[0]: 0 0 0 1 1 1 0 0 0 0 1 1
[1]: 0 1 1 1 0 1 1 1 1 0 1 0
[2]: 0 1 1 1 1 1 0 1 1 0 1 1
[3]: 0 0 0 0 0 1 0 0 0 0 1 0
[4]: 0 0 0 0 0 0 1 0 1 1 0 0
[5]: 0 0 1 0 0 1 0 0 1 1 1 1
[6]: 1 0 1 0 1 1 0 1 0 1 0 1
[7]: 1 1 1 0 1 1 0 1 1 1 0 1
[8]: 1 0 0 1 1 1 0 0 1 0 0 0
[9]: 1 1 1 1 0 0 0 0 1 1 0 0

成对写出列/行索引:

ILArray<int> C = 1;
ILArray<int> R = ILMath.find(L, 0, C);
// C now holds the column indices, R the row indices
for (int i = 0; i < C.Length; i++) {
System.Diagnostics.Debug.WriteLine("({0},{1})", R.GetValue(i), C.GetValue(i));
}

给出:

(6,0)
(7,0)
(8,0)
(9,0)
(1,1)
(2,1)
(7,1)
(9,1)
(1,2)
(2,2)
(5,2)
(6,2)
(7,2)
(9,2)
(0,3)
(1,3)
(2,3)
(8,3)
(9,3)
(0,4)
(2,4)
(6,4)
(7,4)
(8,4)
(0,5)
(1,5)
(2,5)
(3,5)
(5,5)
(6,5)
(7,5)
(8,5)
(1,6)
(4,6)
(1,7)
(2,7)
(6,7)
(7,7)
(1,8)
(2,8)
(4,8)
(5,8)
(7,8)
(8,8)
(9,8)
(4,9)
(5,9)
(6,9)
(7,9)
(9,9)
(0,10)
(1,10)
(2,10)
(3,10)
(5,10)
(0,11)
(2,11)
(5,11)
(6,11)
(7,11)

关于c# - 如何在 ilnumerics 逻辑对称矩阵中查找维度?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19112215/

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