Question

我在numpy中有两组3D点，我想创建一个矩阵和点的矢量表示，如下所示：

| X1 Y1 Z1 0  0  0  0  0  0  1 0 0|     | X1 |
| 0  0  0  X1 Y1 Z1 0  0  0  0 1 0|     | Y1 |
| 0  0  0  0  0  0  X1 Y1 Z1 0 0 1|     | Z1 |
| X2 Y2 Z2 0  0  0  0  0  0  1 0 0|     | X2 |
| 0  0  0  X2 Y2 Z2 0  0  0  0 1 0|     | Y2 |
| 0  0  0  0  0  0  X2 Y2 Z2 0 0 1|     | Z2 |

用法类似于：

import numpy as np
pts = np.random.rand(10, 3)

因此，矩阵现在的形状为(30, 12). 30行（每点3个）和12列。在这种情况下，矩阵长度为30个元素。有没有办法在python中实现这一点而不编写明确的for循环？

Answer 1

Kronecker product（np.kron）对于构建这样的块矩阵非常有用：

import numpy as np

pts = np.arange(1, 31).reshape(10, 3)

n, d = pts.shape
I = np.eye(d, dtype=pts.dtype)

# the first d**2 columns of xyz values
xyzcols = np.kron(I, pts[:, None]).reshape(-1, d * d)

# the final d columns of ones
eyecols = np.tile(I, n).T

# concatenate
out = np.hstack([xyzcols, eyecols])

print(repr(out[:6]))
# array([[1, 2, 3, 0, 0, 0, 0, 0, 0, 1, 0, 0],
#        [0, 0, 0, 1, 2, 3, 0, 0, 0, 0, 1, 0],
#        [0, 0, 0, 0, 0, 0, 1, 2, 3, 0, 0, 1],
#        [4, 5, 6, 0, 0, 0, 0, 0, 0, 1, 0, 0],
#        [0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 1, 0],
#        [0, 0, 0, 0, 0, 0, 4, 5, 6, 0, 0, 1]])

Answer 2

一种滥用linear indexing和广播索引的矢量化方法 -

m,n = pts.shape

idx1 = np.arange(n)[:,None] + np.arange(n)*(n*(n+2))
idx2  = idx1 + np.arange(m)[:,None,None]*(n*n*(n+1))
idx3 = (n*n + np.arange(n)*(n*(n+1)+1)) + np.arange(m)[:,None]*(n*n*(n+1))

out = np.zeros((m*n,n*(n+1)),dtype=pts.dtype)
out.ravel()[idx2] = pts[:,:,None]
out.ravel()[idx3] = 1

示例运行 -

In [550]: pts
Out[550]: 
array([[47, 34],
       [36, 25],
       [29, 38],
       [35, 20],
       [37, 48]])

In [551]: m,n = pts.shape
     ...: 
     ...: idx1 = np.arange(n)[:,None] + np.arange(n)*(n*(n+2))
     ...: idx2  = idx1 + np.arange(m)[:,None,None]*(n*n*(n+1))
     ...: idx3=(n*n + np.arange(n)*(n*(n+1)+1)) + np.arange(m)[:,None]*(n*n*(n+1))
     ...: 
     ...: out = np.zeros((m*n,n*(n+1)),dtype=pts.dtype)
     ...: out.ravel()[idx2] = pts[:,:,None]
     ...: out.ravel()[idx3] = 1
     ...: 

In [552]: out
Out[552]: 
array([[47, 34,  0,  0,  1,  0],
       [ 0,  0, 47, 34,  0,  1],
       [36, 25,  0,  0,  1,  0],
       [ 0,  0, 36, 25,  0,  1],
       [29, 38,  0,  0,  1,  0],
       [ 0,  0, 29, 38,  0,  1],
       [35, 20,  0,  0,  1,  0],
       [ 0,  0, 35, 20,  0,  1],
       [37, 48,  0,  0,  1,  0],
       [ 0,  0, 37, 48,  0,  1]])

在numpy中创建这个块矩阵

2 个答案: