在Python中索引/重塑矩阵以匹配目标矩阵

Question

我有一个NumPy数组，如下所示：

>>> import numpy

>>> foo = numpy.array(
    [[ 1.    ,  0.3491,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       1.    ,  0.1648,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       1.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       1.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       1.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
     [ 0.4269,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.225 ,  0.1637,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.4269,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.2929,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.4078,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
     [ 0.4212,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.1719,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.3856,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.147 ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.2459,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
     [ 0.3581,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.1676,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.2545,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.0619,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.2195,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
     [ 0.3558,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.1658,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.2544,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
       0.2159,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ]])

我想对其重新编制索引和形状，使其成为以下内容：

>>> bar
array(
[[ 1.    ,  1.    ,  1.    ,  1.    ,  1.    ],
 [ 0.4269,  0.225 ,  0.4269,  0.2929,  0.4078],
 [ 0.4212,  0.1719,  0.3856,  0.147 ,  0.2459],
 [ 0.3581,  0.1676,  0.2545,  0.0619,  0.2195],
 [ 0.3558,  0.1658,  0.2544,  0.    ,  0.2159],
 [ 0.3491,  0.1648,  0.    ,  0.    ,  0.    ],
 [ 0.    ,  0.1637,  0.    ,  0.    ,  0.    ]])

有没有一种不使用for循环的有效方法？也许用broadcasting or strides？你知道吗

Answer 1

您可以reshape按Fortran顺序（或使用换位的组合）对数组执行操作，然后slice对数组执行操作以仅提取前7行：

a.reshape(50, -1, order='F')[:7,:]

这里选择的新形状是基于数组中这些形状的位置。当您按Fortran顺序展平数组，然后看到您的模式出现时，这一点就变得很清楚了。你知道吗

Answer 2

首先，创建numpy数组：

import numpy as np
arr = np.asarray(a)

arr

array([[ 1.    ,  0.3491,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  1.    ,  0.1648,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  1.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  1.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  1.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
   [ 0.4269,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.225 ,  0.1637,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.4269,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.2929,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.4078,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
   [ 0.4212,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.1719,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.3856,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.147 ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.2459,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
   [ 0.3581,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.1676,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.2545,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.0619,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.2195,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
   [ 0.3558,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.1658,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.2544,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.2159,  0.    ,
     0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ,  0.    ]])

然后，用：

b = arr[:,0:50:10]

Out[13]: array([[ 1.    ,  1.    ,  1.    ,  1.    ,  1.    ],
   [ 0.4269,  0.225 ,  0.4269,  0.2929,  0.4078],
   [ 0.4212,  0.1719,  0.3856,  0.147 ,  0.2459],
   [ 0.3581,  0.1676,  0.2545,  0.0619,  0.2195],
   [ 0.3558,  0.1658,  0.2544,  0.    ,  0.2159]])

然后，堆叠要保留的其他数据：

c = arr[:,1:50:10]

np.vstack((b,c))
Out[17]: 
array([[ 1.    ,  1.    ,  1.    ,  1.    ,  1.    ],
   [ 0.4269,  0.225 ,  0.4269,  0.2929,  0.4078],
   [ 0.4212,  0.1719,  0.3856,  0.147 ,  0.2459],
   [ 0.3581,  0.1676,  0.2545,  0.0619,  0.2195],
   [ 0.3558,  0.1658,  0.2544,  0.    ,  0.2159],
   [ 0.3491,  0.1648,  0.    ,  0.    ,  0.    ],
   [ 0.    ,  0.1637,  0.    ,  0.    ,  0.    ],
   [ 0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
   [ 0.    ,  0.    ,  0.    ,  0.    ,  0.    ],
   [ 0.    ,  0.    ,  0.    ,  0.    ,  0.    ]])

用更多的切片去掉最后一行。你知道吗