python - 网格化一个numpy数组

Question

xx和yy对应9个矩形的顶点。如何获取每个矩形的 4 个坐标作为元组？

>>> import numpy as np
>>> xx, yy = np.mgrid[0:4, 10:14]
>>> xx
array([[0, 0, 0, 0],
       [1, 1, 1, 1],
       [2, 2, 2, 2],
       [3, 3, 3, 3]])
>>> yy
array([[10, 11, 12, 13],
       [10, 11, 12, 13],
       [10, 11, 12, 13],
       [10, 11, 12, 13]])
>>> xx, yy = xx.ravel(), yy.ravel()
>>> xx
array([0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])
>>> yy
array([10, 11, 12, 13, 10, 11, 12, 13, 10, 11, 12, 13, 10, 11, 12, 13])
>>> lists = list(zip(xx, yy))
>>> lists
[(0, 10), (0, 11), (0, 12), (0, 13), (1, 10), (1, 11), (1, 12), (1, 13), (2, 10), (2, 11), (2, 12), (2, 13), (3, 10), (3, 11), (3, 12), (3, 13)]

第一个矩形的坐标为:

[(0, 10), (0, 11), (1, 11), (1, 10)].

同样，我想以“list of lists”的格式获取其他8个矩形的坐标。

矩形的坐标顺序并不重要。

首选 NumPy/SciPy 方法。

Answer 1

一种方式，不使用 mgrid:

x = np.arange(4)
y = np.arange(10, 14)
d = [[0,0],[0,1],[1,1],[1,0]]  # define order to go around rectangle here

for ix in range(len(x) - 1):
    for iy in range(len(y) - 1):
        print [(x[ix + dx], y[iy + dy]) for dx, dy in d]

结果:

[(0, 10), (0, 11), (1, 11), (1, 10)]
[(0, 11), (0, 12), (1, 12), (1, 11)]
[(0, 12), (0, 13), (1, 13), (1, 12)]
[(1, 10), (1, 11), (2, 11), (2, 10)]
[(1, 11), (1, 12), (2, 12), (2, 11)]
[(1, 12), (1, 13), (2, 13), (2, 12)]
[(2, 10), (2, 11), (3, 11), (3, 10)]
[(2, 11), (2, 12), (3, 12), (3, 11)]
[(2, 12), (2, 13), (3, 13), (3, 12)]

如果你坚持只使用 numpy，你可以这样做:

xx, yy = np.mgrid[0:4, 10:14]
top = xx[:-1,1:]
bot = xx[1:,1:]
left = yy[1:,:-1]
right = yy[1:,1:]
np.column_stack(v.ravel() 
    for v in (top, left, top, right, bot, right, bot, left))

结果是一个单一的矩阵，但是列和你想要的元组完全一样:

array([[ 0, 10,  0, 11,  1, 11,  1, 10],
       [ 0, 11,  0, 12,  1, 12,  1, 11],
       [ 0, 12,  0, 13,  1, 13,  1, 12],
       [ 1, 10,  1, 11,  2, 11,  2, 10],
       [ 1, 11,  1, 12,  2, 12,  2, 11],
       [ 1, 12,  1, 13,  2, 13,  2, 12],
       [ 2, 10,  2, 11,  3, 11,  3, 10],
       [ 2, 11,  2, 12,  3, 12,  3, 11],
       [ 2, 12,  2, 13,  3, 13,  3, 12]])