将thetas、rs组合与numpy矢量化

Question

我使用numpy生成θ、半径范围、cos_thetas和sin_thetas。

就获得这些组合而言，我仍然留下了效率非常低的代码。

我曾试图走矢量化的道路，但重塑会让事情变得更糟。

我目前有以下代码：

candidates = []

delta_r = 1
r_min = 50
r_max = 100

num_thetas = 6
# R and Theta ranges
dtheta = int(360 / num_thetas)
## Thetas is bins created from 0 to 360 degree with increment of the dtheta
thetas = np.arange(0, 360, step=dtheta)
## Radius ranges from r_min to r_max 
rs = np.arange(r_min, r_max, step=delta_r)
cos_thetas = np.cos(np.deg2rad(thetas))
sin_thetas = np.sin(np.deg2rad(thetas))     
for r in rs:
for t in range(num_thetas):
candidates.append((r, int(r * cos_thetas[t]), int(r * sin_thetas[t])))

Answer 1

广播是你的朋友：

d = np.zeros((num_thetas*rs.size, 3), order = "F", dtype = int)
d[:,0] = np.repeat(rs, num_thetas)
d[:,1] = np.ravel(rs[:,None] * cos_thetas[:num_thetas]).astype(int)
d[:,2] = np.ravel(rs[:,None] * sin_thetas[:num_thetas]).astype(int)

尽管远非完美，但这种快速解决方案应该至少能带来一个数量级的改善。

Answer 2

像这样的smth怎么样：
这个例子包含较少的变量值，但您可以调整它。输出是一个数组，但您可以很容易地将其强制转换为列表。

import numpy as np
rs = np.linspace(50, 100, 3)
theta = np.linspace(0, 360, 3)
rs_grid, theta = np.meshgrid(rs, theta)
mgrid = np.stack(
[rs_grid,
rs_grid*np.cos(theta),
rs_grid*np.sin(theta)])
mgrid.T.reshape(-1, 3)

输出：

array([[ 50.        ,  50.        ,   0.        ],
[ 50.        , -29.92300345, -40.05763179],
[ 50.        , -14.18455457,  47.94578617],
[ 75.        ,  75.        ,   0.        ],
[ 75.        , -44.88450518, -60.08644768],
[ 75.        , -21.27683186,  71.91867926],
[100.        , 100.        ,   0.        ],
[100.        , -59.84600691, -80.11526357],
[100.        , -28.36910915,  95.89157234]])