Question

如果我有一个[[4,3,3]）的3D数组，像这样：

[[0,1,2]   [[9,10,11 ]    [[18,19,20]   [[27,28,29] 
 [3,4,5]    [12,13,14]     [21,22,23]    [30,31,32]
 [6,7,8]] , [15,16,17]] ,  [24,25,26]] , [33,34,35]]

我将如何将其转换为（[6,6]）的2D数组，以使数组的第一半在160x160的上半部分在第二半在底端：

[[0,1,2,9,10,11]
[3,4,5,12,13,14]
[6,7,8,15,16,17]
[18,19,20,27,28,29] 
[21,22,23,30,31,32]
[24,25,26,33,34,35]]

我的数组创建：

qDCTReversed = np.zeros((400,8,8), dtype=np.int)

我需要一个（160,160）数组。

Answer 1

不使用for循环的非常快的单行解决方案是：

# initialization
qDCTReversed = np.arange(4*3*3).reshape((4,3,3)) 

# calculation
qDCTReversed = qDCTReversed.reshape((2,2,3,3)).transpose((0,2,1,3)).reshape((6,6))

或用于(400,8,8)数组：

qDCTReversed.reshape((20,20,8,8)).transpose((0,2,1,3)).reshape((160,160))

速度比较：

Mstaino's answer：0.393毫秒

yatu's answer：0.138毫秒

此答案：0.016毫秒

Answer 2

您可以这样循环遍历列表：

a = [[[ 0, 1, 2], [ 9,10,11]],
     [[ 3, 4, 5], [12,13,14]],
     [[ 6, 7, 8], [15,16,17]],
     [[18,19,20], [27,28,29]],
     [[21,22,23], [30,31,32]],
     [[24,25,26], [33,34,35]]]

b = [[i for j in k for i in j ] for k in a]
print(b)

输出：

[ 0,  1,  2,  9, 10, 11]
[ 3,  4,  5, 12, 13, 14]
[ 6,  7,  8, 15, 16, 17]
[18, 19, 20, 27, 28, 29]
[21, 22, 23, 30, 31, 32]
[24, 25, 26, 33, 34, 35]

Answer 3

您要求的重塑可以通过以下方式完成：

x = np.arange(36).reshape((4,3,3))
np.vstack(np.hstack(x[2*i:2+2*i]) for i in range(x.shape[0]//2))
>>array([[ 0,  1,  2,  9, 10, 11],
   [ 3,  4,  5, 12, 13, 14],
   [ 6,  7,  8, 15, 16, 17],
   [18, 19, 20, 27, 28, 29],
   [21, 22, 23, 30, 31, 32],
   [24, 25, 26, 33, 34, 35]])

将3D数组重塑为2D数组Python

3 个答案: