计算来自（矩阵 - 常数）的矩阵的快速方法

Question

I would like to find a faster way to compute the matrix v . Note that if height = width then the matrix v is symmetric, v = vT . The width and the height are arbitrary positive integers.

v = np.zeros((height, width)) # rows columns

# region center
cr = np.round(height / 2)
cc = np.round(width / 2)

# there has to be a quicker way to do this
for w in range(width):
    v[:, w] = [np.sqrt((w - cc) ** 2 + (h - cr) ** 2) for h in range(height)]

Answer 1

You can simplify this by using numpy's broadcasting :

out = np.sqrt((np.arange(width)-cc)**2 + (np.arange(height)[:,None]-cr)**2) 

---

width=5
height=8
v = np.zeros((height, width)) # rows columns
# region center
cr = np.round(height / 2)
cc = np.round(width / 2)
# there has to be a quicker way to do this
for w in range(width):
    v[:, w] = [np.sqrt((w - cc) ** 2 + (h - cr) ** 2) for h in range(height)]

np.allclose(v, out)
# True

Answer 2

You can use scipy.spatial.distance.cdist , with numpy.indices :

from scipy.spatial.distance import cdist
import numpy as np

shape = (height, width)
v = cdist(np.indices(shape).reshape(2, -1).T, [[cr, cc]]).reshape(shape)

This yields identical results to @yatu's broadcasting solution in all cases, but strangely is ~3x slower for a large variety of input sizes. That being said, it is likely a bit more space efficient, since it allocates only one array of the output size rather than two (second one being the sum fed into the square root). This should not be a consideration for any normal input you encounter, so use the broadcasting solution. This is just here for completeness.

Answer 3

You can do this without any loop. Generate a matrix of row indexes and a matrix of column indexes and apply the formula between the two matrices:

rows,cols = np.indices((height,width))
v         = np.sqrt((rows-height//2)**2+(cols-width//2)**2)