Question

我有以下矩阵：

stdin

它的尺寸为：x = \ np.array([[[[0.99256822, 0.63019905], [0.77484078, 0.27471319]], [[0.94722451, 0.95948516], [0.81838252, 0.48979609]], [[0.81673764, 0.9388614], [0.57575844, 0.82265243]]], [[[0.95485566, 0.94870753], [0.92680463, 0.90044481]], [[0.90128127, 0.98683992], [0.9115591, 0.85900321]], [[0.949711, 0.85709163], [0.70392261, 0.91043368]]]])。我想要做的是将它与以下矩阵相乘：

2,3,2,2

尺码为y = \ np.array([[[[ 0., 0., 0.63019905, 0. ], [ 0., 0.99256822, 0., 0. ], [ 0.77484078, 0., 0., 0.27471319], [ 0., 0., 0., 0. ]], [[ 0., 0., 0., 0. ], [ 0.94722451, 0., 0., 0.95948516], [ 0.81838252, 0., 0., 0. ], [ 0., 0., 0.48979609, 0. ]], [[ 0., 0., 0., 0. ], [ 0., 0.81673764, 0., 0.9388614 ], [ 0., 0., 0., 0.82265243], [ 0.57575844, 0., 0., 0. ]]], [[[ 0., 0.95485566, 0., 0. ], [ 0., 0., 0., 0.94870753], [ 0., 0.92680463, 0., 0. ], [ 0., 0., 0., 0.90044481]], [[ 0., 0.90128127, 0., 0. ], [ 0., 0., 0., 0.98683992], [ 0., 0.9115591, 0., 0. ], [ 0., 0., 0., 0.85900321]], [[ 0., 0., 0., 0.85709163], [ 0., 0.949711, 0., 0. ], [ 0., 0.70392261, 0., 0.91043368], [ 0., 0., 0., 0. ]]]])。所以我需要做的是以这样的方式填充第一个矩阵，即我们将每个条目复制4次，以便可以进行乘法（结果中的3个将详细说明为0，最终结果将是乘法我想要）。因此，我需要将第一个矩阵转换为如下所示：

2,3,4,4

依旧......

更新：

    [[[[ 0.99256822          0.99256822          0.63019905  0.63019905        ]
       [ 0.99256822          0.99256822          0.63019905  0.63019905        ]
        [ 0.77484078         0.77484078          0.27471319  0.27471319]
        [ 0.77484078         0.77484078          0.27471319  0.27471319        ]]

这是我在使用它的场景。

Answer 1

您正在寻找的功能是np.repeat。我在最后两个维度上重复了x矩阵两次，如下所示：

>>> x = np.repeat(x, 2, axis=(2))
>>> x = np.repeat(x, 2, axis=(3))
>>> x
  [[[[ 0.99256822  0.99256822  0.63019905  0.63019905]
   [ 0.99256822  0.99256822  0.63019905  0.63019905]
   [ 0.77484078  0.77484078  0.27471319  0.27471319]
   [ 0.77484078  0.77484078  0.27471319  0.27471319]]

  [[ 0.94722451  0.94722451  0.95948516  0.95948516]
   [ 0.94722451  0.94722451  0.95948516  0.95948516]
   [ 0.81838252  0.81838252  0.48979609  0.48979609]
   [ 0.81838252  0.81838252  0.48979609  0.48979609]]
   ...
>>> x.shape
(2, 3, 4, 4)
>>> y.shape
(2, 3, 4, 4)

既然两个矩阵具有相似的形状，它们可以成倍增加：

>>> x * y
[[[[ 0.          0.          0.39715084  0.        ]
   [ 0.          0.98519167  0.          0.        ]
   [ 0.60037823  0.          0.          0.07546734]
   [ 0.          0.          0.          0.        ]]

  [[ 0.          0.          0.          0.        ]
   [ 0.89723427  0.          0.          0.92061177]
   [ 0.66974995  0.          0.          0.        ]
   [ 0.          0.          0.23990021  0.        ]]

  [[ 0.          0.          0.          0.        ]
   [ 0.          0.66706037  0.          0.88146073]
   [ 0.          0.          0.          0.67675702]
   [ 0.33149778  0.          0.          0.        ]]]


 [[[ 0.          0.91174933  0.          0.        ]
   [ 0.          0.          0.          0.90004598]
   [ 0.          0.85896682  0.          0.        ]
   [ 0.          0.          0.          0.81080086]]

  [[ 0.          0.81230793  0.          0.        ]
   [ 0.          0.          0.          0.97385303]
   [ 0.          0.83093999  0.          0.        ]
   [ 0.          0.          0.          0.73788651]]

  [[ 0.          0.          0.          0.73460606]
   [ 0.          0.90195098  0.          0.        ]
   [ 0.          0.49550704  0.          0.82888949]
   [ 0.          0.          0.          0.        ]]]]

Answer 2

我假设a和b分别为两个数组。

方法＃1

要获得重复版本，我们可以使用a将6D扩展为np.broadcast_to，然后重新转换为4D -

a6D = a[:,:,:,None,:,None]
m,n,p,q = a.shape
r,s = b.shape[-2:]
a_repeated = np.broadcast_to(a6D, (m,n,p,r//p,q,s//q)).reshape(b.shape)

然后，使用a_repeated进行元素乘法与b。

方法＃2（记忆效率高）

您可以通过添加新轴将a扩展到6D，从而避免任何实际的重复或平铺以提高内存效率，使用6D重新整形{{1}执行逐元素乘法最后重塑回b输出。因此，对于4D数组4D和a，我们会有 -

数组填充Numpy

2 个答案:

方法＃1

方法＃2（记忆效率高）