什麼是正確的方式來創建轉型的numpy的陣列矩陣

Question

任意的旋轉角度的列表（可以說繞X軸）：

import numpy as np 
x_axis_rotations = np.radians([0,10,32,44,165]) 

我可以創建這樣做符合這些角矩陣陣列所以：

matrices = [] 
for angle in x_axis_rotations: 
    matrices.append(np.asarray([[1 , 0 , 0],[0, np.cos(angle), -np.sin(angle)], [0, np.sin(angle), np.cos(angle)]])) 
matrices = np.array(matrices) 

這會工作，但它並沒有利用numpy的的優勢在應對大陣......所以，如果我的角度陣列以百萬計，做這種方式會不會很快速。

是否有更好（更快）的方法來從輸入數組中創建變換矩陣的數組？

Answer 1

這裏有一個直接而簡單的方法：

c = np.cos(x_axis_rotations) 
s = np.sin(x_axis_rotations) 
matrices = np.zeros((len(x_axis_rotations), 3, 3)) 
matrices[:, 0, 0] = 1 
matrices[:, 1, 1] = c 
matrices[:, 1, 2] = -s 
matrices[:, 2, 1] = s 
matrices[:, 2, 2] = c 

---

時間，好奇：

In [30]: angles = 2 * np.pi * np.random.rand(1000) 

In [31]: timeit OP(angles) 
100 loops, best of 3: 5.46 ms per loop 

In [32]: timeit askewchan(angles) 
10000 loops, best of 3: 39.6 µs per loop 

In [33]: timeit divakar(angles) 
10000 loops, best of 3: 93.8 µs per loop 

In [34]: timeit divakar_oneline(angles) 
10000 loops, best of 3: 56.1 µs per loop 

In [35]: timeit divakar_combine(angles) 
10000 loops, best of 3: 43.9 µs per loop 

所有比你的循環更快，所以使用哪個你最喜歡:)

Answer 2

您可以使用linear indexing助陣，像這樣 -

# Get cosine and sine values in one-go 
cosv = np.cos(x_axis_rotations) 
sinv = np.sin(x_axis_rotations) 

# Get size parameter 
N = x_axis_rotations.size 

# Initialize output array 
out = np.zeros((N,3,3)) 

# Set the first element in each 3D slice as 1 
out[:,0,0] = 1 

# Calculate the first of positions where cosine valued elements are to be put 
idx1 = 4 + 9*np.arange(N)[:,None] 

# One by one put those 4 values in 2x2 blocks across all 3D slices 
out.ravel()[idx1] = cosv 
out.ravel()[idx1+1] = -sinv 

out.ravel()[idx1+3] = sinv 
out.ravel()[idx1+4] = cosv 

或者，你可以設置中的元素一去，你有zeros初始化輸出數組後，並在每個切片組的第一個元素爲1，像這樣 -

out.reshape(N,-1)[:,[4,5,7,8]] = np.column_stack((cosv,-sinv,sinv,cosv)) 

上述兩種方法之間，兩個中景方法可能演變，再初始化後W擺正第i個零和設置在每個3D片的第一個元素爲1，像這樣 -

out.reshape(N,-1)[:,[4,8]] = cosv[:,None] 
out.reshape(N,-1)[:,[5,7]] = np.column_stack((-sinv[:,None],sinv[:,None])) 

最後一個是 -

out.reshape(N,-1)[:,[4,8]] = cosv[:,None] 
out.reshape(N,-1)[:,5] = -sinv 
out.reshape(N,-1)[:,7] = sinv