添加不同大小/形状的置换NumPy矩阵

Question

简而言之：我有两个矩阵（或数组）：

import numpy

block_1 = numpy.matrix([[ 0, 0, 0, 0, 0],
                        [ 0, 0, 0, 0, 0],
                        [ 0, 0, 0, 0, 0],
                        [ 0, 0, 0, 0, 0]])

block_2 = numpy.matrix([[ 1, 1, 1],
                        [ 1, 1, 1],
                        [ 1, 1, 1],
                        [ 1, 1, 1]])

我有block_2在block_1元素坐标系中的位移。

pos = (1,1)

我希望能够（快速）添加它们，以获得：

[[0 0 0 0 0]
 [0 1 1 1 0]
 [0 1 1 1 0]
 [0 1 1 1 0]]

长话短说：我想用一种快速的方法把两个不同形状的矩阵加在一起，其中一个矩阵可以移位。得到的矩阵必须具有第一个矩阵的形状，并且两个矩阵之间的重叠元素求和。如果没有重叠，则只返回第一个未变异的矩阵。

我有一个很好的函数，但它有点难看，从元素上来说：

def add_blocks(block_1, block_2, pos):
    for i in xrange(0, block_2.shape[0]):
        for j in xrange(0, block_2.shape[1]):
            if (i + pos[1] >= 0) and (i + pos[1] < block_1.shape[0])
               and (j + pos[0] >= 0) and (j + pos[0] < block_1.shape[1]):
                block_1[pos[1] + i, pos[0] + j] += block_2[i,j]
    return block_1

广播或切片也许能做到这一点吗？

我觉得也许我遗漏了一些显而易见的东西。

Answer 1

一个类似于MATLAB的简单解决方案是：

import numpy as np

block1 = np.zeros((5,4))
block2 = np.ones((3,2))

block1[1:4,2:4] += block2  # use array slicing

print(block1)

[[0. 0. 0. 0.]
 [0. 0. 1. 1.]
 [0. 0. 1. 1.]
 [0. 0. 1. 1.]
 [0. 0. 0. 0.]]

---

所以把它打包成一个可重用的函数：

import numpy as np

def addAtPos(mat1, mat2, xypos):
    """
    Add two matrices of different sizes in place, offset by xy coordinates
    Usage:
      - mat1: base matrix
      - mat2: add this matrix to mat1
      - xypos: tuple (x,y) containing coordinates
    """
    x, y = xypos
    ysize, xsize = mat2.shape
    xmax, ymax = (x + xsize), (y + ysize)
    mat1[y:ymax, x:xmax] += mat2
    return mat1

block1 = np.zeros((5,4))
block2 = np.ones((3,2))
pos = (2,1)
print(addAtPos(block1, block2, pos))

[[0. 0. 0. 0.]
 [0. 0. 1. 1.]
 [0. 0. 1. 1.]
 [0. 0. 1. 1.]
 [0. 0. 0. 0.]]

Answer 2

只需找到重叠区域，然后使用切片添加数组。

b1 = np.zeros((4,5))
b2 = np.ones((4,3))
pos_v, pos_h = 2, 3  # offset
v_range1 = slice(max(0, pos_v), max(min(pos_v + b2.shape[0], b1.shape[0]), 0))
h_range1 = slice(max(0, pos_h), max(min(pos_h + b2.shape[1], b1.shape[1]), 0))

v_range2 = slice(max(0, -pos_v), min(-pos_v + b1.shape[0], b2.shape[0]))
h_range2 = slice(max(0, -pos_h), min(-pos_h + b1.shape[1], b2.shape[1]))

b1[v_range1, h_range1] += b2[v_range2, h_range2]

它们已添加到位，但您也可以创建一个新数组。不过，我可能漏掉了一些角落里的案子，但看起来效果不错。

Answer 3

这很好，下面是如何通过在jorgeca的代码中添加几行来扩展对3D矩阵的添加：

import numpy as np

#two 3d arrays, of different size.
b1 = np.zeros((5,5,5), dtype=np.int) # a 5x5x5 matrix of zeroes
b2 = np.ones((3,3,3), dtype=np.int)  # a 3x3x3 matrix of ones

pos_v, pos_h, pos_z = 2, 2, 2  # a 3d offset -> to plonk b2 in the corner of b1

v_range1 = slice(max(0, pos_v), max(min(pos_v + b2.shape[0], b1.shape[0]), 0))
h_range1 = slice(max(0, pos_h), max(min(pos_h + b2.shape[1], b1.shape[1]), 0))
z_range1 = slice(max(0, pos_z), max(min(pos_z + b2.shape[2], b1.shape[2]), 0))

v_range2 = slice(max(0, -pos_v), min(-pos_v + b1.shape[0], b2.shape[0]))
h_range2 = slice(max(0, -pos_h), min(-pos_h + b1.shape[1], b2.shape[1]))
z_range2 = slice(max(0, -pos_z), min(-pos_z + b1.shape[2], b2.shape[2]))

b1[v_range1, h_range1, z_range1] += b2[v_range2, h_range2, z_range2]

这可能有助于那些想在3d中做同样事情的人（比如我）。