Numpy索引，获取宽度为2的条带

Question

我想知道是否有一种方法，它索引/切片一个numpy数组，这样一个可以得到每2个元素的其他波段。换句话说，假设：

test = np.array([[1,2,3,4,5,6,7,8],[9,10,11,12,13,14,15,16]])

我想得到数组：

[[1,  2,  5,  6],
 [9, 10, 13, 14]]

关于如何通过切片/索引实现这一点的想法？你知道吗

Answer 1

给出：

>>> test
array([[ 1,  2,  3,  4,  5,  6,  7,  8],
       [ 9, 10, 11, 12, 13, 14, 15, 16]])

你可以做：

>>> test.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1,  2,  5,  6],
       [ 9, 10, 13, 14]])

它甚至适用于不同形状的初始阵列：

>>> test2
array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16])
>>> test2.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1,  2,  5,  6],
       [ 9, 10, 13, 14]])

>>> test3
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12],
       [13, 14, 15, 16]])
>>> test3.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1,  2,  5,  6],
       [ 9, 10, 13, 14]])

工作原理：

1. Reshape into two columns by however many rows:
>>> test.reshape(-1,2)
array([[ 1,  2],
       [ 3,  4],
       [ 5,  6],
       [ 7,  8],
       [ 9, 10],
       [11, 12],
       [13, 14],
       [15, 16]])

2. Stride the array by stepping every second element
>>> test.reshape(-1,2)[::2]
array([[ 1,  2],
       [ 5,  6],
       [ 9, 10],
       [13, 14]])

3. Set the shape you want of 4 columns, however many rows:
>>> test.reshape(-1,2)[::2].reshape(-1,4)
array([[ 1,  2,  5,  6],
       [ 9, 10, 13, 14]])

Answer 2

通过一些巧妙的重塑并不是那么困难：）

test.reshape((4, 4))[:, :2].reshape((2, 4))