Question

我有两个2D（numpy）数组，我想通过以下方式从中生成3D数组：第一个数组的 n 行中的每行将（第二个）第二个数组乘以产生一个新的矩阵，产生 n 个新数组（形成3D数组）。我认为一个简单的示例将有助于理解：

        $example = collect([
            "tracks" => [
                [
                    "name" => "track 1",
                    "segments" => [
                        [
                            "name" => "track 1 - segment 1",
                            "points" => [
                                ["timestamp" => "2008-09-21T11:00:10Z", "lat"=>1, "lon"=>1],
                                ["timestamp" => "2008-09-21T11:00:20Z", "lat"=>1, "lon"=>1],
                                ["timestamp" => "2008-09-21T11:00:30Z", "lat"=>1, "lon"=>1],
                                ["timestamp" => "2008-09-21T11:00:40Z", "lat"=>1, "lon"=>1],
                                // --- split here ("clone" 'track 1 - segment 1' between segment 1 and 2)
                                ["timestamp" => "2008-09-21T13:00:10Z", "lat"=>1, "lon"=>1],
                                ["timestamp" => "2008-09-21T13:00:20Z", "lat"=>1, "lon"=>1],
                                ["timestamp" => "2008-09-21T13:00:30Z", "lat"=>1, "lon"=>1],
                                // --- and split here
                                ["timestamp" => "2008-09-21T15:00:10Z", "lat"=>1, "lon"=>1],
                                ["timestamp" => "2008-09-21T15:00:20Z", "lat"=>1, "lon"=>1],
                            ]
                        ],
                        [
                            "name" => "track 1 - segment 2",
                            "points" => []
                        ],
                    ],  
                ],
                [
                    "name" => "track 2",
                    "segments" => []
                ]
            ]
        ]);

事实是尺寸是任意的（因此不仅 2x3 和 3x3 ，当然第二个尺寸始终兼容），我正在寻找一种解决方案“用于循环”。例如，我尝试重复数组B然后相乘

A = [[a11 a12 a13]
     [a21 a22 a23]]

B = [[b11 b12 b13]
     [b21 b22 b23]
     [b31 b32 b33]]

# The product "A*B" would result in a matrix C such as

C = [[[a11*b11  a12*b12  a13*b13]
      [a11*b21  a12*b22  a13*b23]
      [a11*b31  a12*b32  a13*b33]]

     [[a21*b11  a22*b12  a23*b13]
      [a21*b21  a22*b22  a23*b23]
      [a21*b31  a22*b32  a23*b33]]]

# Which is equivalent to (in numpy notation)

 C[0] = A[0]*B
 C[1] = A[1]*B

但是新尺寸不兼容。

Answer 1

我认为这就是您想要的：

C = A[:, None] * B

使用numpy进行2D-3D数组乘法

1 个答案: