我有一个大小为 N x N 的矩阵定义为 M = np.zeros((N,N)) 和两个坐标 [x0,y0 ] 和 [x1,y1]。现在我想把这两点联系起来。
N=5 的例子是:
让我们在矩阵中将坐标设置为 2
[x0,y0] = [0,3]
[x1,y1] = [4,2]
那么矩阵应该是这样的:
M =
0 0 0 2 0
0 0 0 1 0
0 0 1 0 0
0 0 1 0 0
0 0 2 0 0
这样做的简单方法是什么？
请您参考如下方法:
这是一个使用 NumPy 的简单实现，只需计算直线方程:
import numpy as np
def draw_line(mat, x0, y0, x1, y1, inplace=False):
if not (0 <= x0 < mat.shape[0] and 0 <= x1 < mat.shape[0] and
0 <= y0 < mat.shape[1] and 0 <= y1 < mat.shape[1]):
raise ValueError('Invalid coordinates.')
if not inplace:
mat = mat.copy()
if (x0, y0) == (x1, y1):
mat[x0, y0] = 2
return mat if not inplace else None
# Swap axes if Y slope is smaller than X slope
transpose = abs(x1 - x0) < abs(y1 - y0)
if transpose:
mat = mat.T
x0, y0, x1, y1 = y0, x0, y1, x1
# Swap line direction to go left-to-right if necessary
if x0 > x1:
x0, y0, x1, y1 = x1, y1, x0, y0
# Write line ends
mat[x0, y0] = 2
mat[x1, y1] = 2
# Compute intermediate coordinates using line equation
x = np.arange(x0 + 1, x1)
y = np.round(((y1 - y0) / (x1 - x0)) * (x - x0) + y0).astype(x.dtype)
# Write intermediate coordinates
mat[x, y] = 1
if not inplace:
return mat if not transpose else mat.T
一些测试:
print(draw_line(np.zeros((5, 5)), 0, 3, 4, 2))
#[[0. 0. 0. 2. 0.]
# [0. 0. 0. 1. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 2. 0. 0.]]
print(draw_line(np.zeros((5, 5)), 4, 2, 0, 3))
#[[0. 0. 0. 2. 0.]
# [0. 0. 0. 1. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 1. 0. 0.]
# [0. 0. 2. 0. 0.]]
print(draw_line(np.zeros((5, 5)), 1, 0, 3, 4))
#[[0. 0. 0. 0. 0.]
# [2. 0. 0. 0. 0.]
# [0. 1. 1. 1. 0.]
# [0. 0. 0. 0. 2.]
# [0. 0. 0. 0. 0.]]
