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在Python中使用Numpy填充_Numpy_Padding - Fatal编程技术网

在Python中使用Numpy填充

在Python中使用Numpy填充,numpy,padding,Numpy,Padding,取一个三维数组。如下面的示例所示,从结束位置填充实例。也就是说,我需要填充沿数组边缘值镜像的话语反射 数组a: array ([array([[3, 1, 4, 1], [5, 9, 2, 6], [5, 3, 5, 8]]), array([[9, 7, 9, 3], [2, 3, 8, 4]]), array([[6, 2, 6, 4], [3, 3, 8, 3], [2, 7, 9, 5], [0, 2, 8, 8]])], dtype=ob

取一个三维数组。如下面的示例所示,从结束位置填充实例。也就是说,我需要填充沿数组边缘值镜像的话语反射

数组a:

array ([array([[3, 1, 4, 1],
   [5, 9, 2, 6],
   [5, 3, 5, 8]]),
   array([[9, 7, 9, 3],
   [2, 3, 8, 4]]),
   array([[6, 2, 6, 4],
   [3, 3, 8, 3],
   [2, 7, 9, 5],
   [0, 2, 8, 8]])], dtype=object)

dim1 = a.shape[0]    # n
dim2 = max([i.shape[0] for i in a]) # m
dim3 = a[0].shape[1] # k
填充后最终矩阵的尺寸:

结果=np.零((dim1、dim2、dim3))

输出:

      [[[3. 1. 4. 1.]
        [5. 9. 2. 6.]
        [5. 3. 5. 8.]
       [5. 3. 5. 8.]]
      [[9. 7. 9. 3.]
       [2. 3. 8. 4.]
       [2. 3. 8. 4.]
       [9. 7. 9. 3.]]

      [[6. 2. 6. 4.]
       [3. 3. 8. 3.]
       [2. 7. 9. 5.]
       [0. 2. 8. 8.]]]

如何使用numpy.pad获得输出?

诀窍是使用
pad\u width
参数,在每个维度上填充特定数量的行。在这种情况下,
pad\u width
应具有形状(2,2)

for b in a:
    extra_lines = dim2 - b.shape[0]
    c = np.pad(b, pad_width=[[0, extra_lines], [0, 0]], mode='symmetric')
    print(c, '\n')

[[3 1 4 1]
 [5 9 2 6]
 [5 3 5 8]
 [5 3 5 8]] 

[[9 7 9 3]
 [2 3 8 4]
 [2 3 8 4]
 [9 7 9 3]] 

[[6 2 6 4]
 [3 3 8 3]
 [2 7 9 5]
 [0 2 8 8]]