问题

Here is the translated code portion:

假设有一个数组

    arr = np.array([[1,2,3], [4,5,6]])

我的目标是在`axis = 0`上重复`N = 2`次元素，所以期望的输出是

    array([[[1., 4.],
        [1., 4.]],

       [[2., 5.],
        [2., 5.]],

       [[3., 6.],
        [3., 6.]]])

我尝试过`np.ones((2,2))*arr.T[:,:,None]`，但输出不同

    array([[[1., 1.],
        [4., 4.]],

       [[2., 2.],
        [5., 5.]],

       [[3., 3.],
        [6., 6.]]])

有没有简单的方法来解决这个问题？我猜这涉及到转置，但我不确定如何实现。

**编辑：** 我找到了答案。它是：

    np.transpose(np.ones((2,2))*arr.T[:,:,None], axes = tuple([0, 2, 1]))

arr = np.array([[1,2,3], [4,5,6]])

array([[[1., 4.],
    [1., 4.]],

   [[2., 5.],
    [2., 5.]],

   [[3., 6.],
    [3., 6.]]])

array([[[1., 1.],
    [4., 4.]],

   [[2., 2.],
    [5., 5.]],

   [[3., 3.],
    [6., 6.]]])

np.transpose(np.ones((2,2))*arr.T[:,:,None], axes = tuple([0, 2, 1]))

答案1

得分: 2

一种方法

```python
(s0, s1) = arr.strides
np.lib.stride_tricks.as_strided(arr, shape=(3, 2, 2), strides=(s1, 0, s0))

请注意，这基本上是免费的：它什么都不做，并返回相同的 arr。只是改变了步幅，使迭代沿着新数组的轴0走向 arr 的轴1；只是对轴1重复；以及沿着 arr 的轴0走向新数组的轴2。

对于所有2D arr 大小和所有重复的泛化

def myrepeat(arr, n):
   (sh0, sh1) = arr.shape
   (st0, st1) = arr.strides
   return np.lib.stride_tricks.as_strided(arr, shape=(sh1, n, sh0), strides=(st1, 0, st0))

# 示例
arr = np.arange(20).reshape(4, 5)
myrepeat(arr, 3)

返回

array([[[ 0,  5, 10, 15],
        [ 0,  5, 10, 15],
        [ 0,  5, 10, 15]],

       [[ 1,  6, 11, 16],
        [ 1,  6, 11, 16],
        [ 1,  6, 11, 16]],

       [[ 2,  7, 12, 17],
        [ 2,  7, 12, 17],
        [ 2,  7, 12, 17]],

       [[ 3,  8, 13, 18],
        [ 3,  8, 13, 18],
        [ 3,  8, 13, 18]],

       [[ 4,  9, 14, 19],
        [ 4,  9, 14, 19],
        [ 4,  9, 14, 19]]])

<details>
<summary>英文:</summary>

One method

```python
(s0,s1)=arr.strides
np.lib.stride_tricks.as_strided(arr, shape=(3, 2, 2), strides=(s1,0,s0))

Note that it is basically free: it does nothing, and return the same arr. Just, a change of strides make iterations goes along arr axis 1 for the new array axis 0; just repeat for axis 1; and along arr axis 0 for the new array axis 2.

A generalization to all sizes of 2D arr, and all repetitions

def myrepeat(arr, n):
   (sh0,sh1)=arr.shape
   (st0,st1)=arr.strides
   return np.lib.stride_tricks.as_strided(arr, shape=(sh1,n,sh0), strides=(st1,0,st0))

# Example
arr=np.arange(20).reshape(4,5)
myrepeat(arr,3)

Returns

array([[[ 0,  5, 10, 15],
        [ 0,  5, 10, 15],
        [ 0,  5, 10, 15]],

       [[ 1,  6, 11, 16],
        [ 1,  6, 11, 16],
        [ 1,  6, 11, 16]],

       [[ 2,  7, 12, 17],
        [ 2,  7, 12, 17],
        [ 2,  7, 12, 17]],

       [[ 3,  8, 13, 18],
        [ 3,  8, 13, 18],
        [ 3,  8, 13, 18]],

       [[ 4,  9, 14, 19],
        [ 4,  9, 14, 19],
        [ 4,  9, 14, 19]]])

</details>



# 答案2
**得分**: 2

[@chrslg][1]的 `stride_tricks` 解决方案可能速度较快，但不容易阅读。相反，我建议使用numpy的 [`np.repeat`][2] 函数，这样可以清晰地表达你的意图。但需要注意的是，你不能直接使用它来获得你想要的结果，首先需要添加另一个维度来分割行。由于这个分割，应该将它应用到 `axis=1` 而不是0。最后的结果需要进行转置以匹配你的需求。

```python
import numpy as np

arr = np.array([[1,2,3], [4,5,6]])
res = np.repeat(arr[:, None, :], 2, axis=1).T
print(res)

输出:

[[[1 4]
  [1 4]]

 [[2 5]
  [2 5]]

 [[3 6]
  [3 6]]]

import numpy as np

arr = np.array([[1,2,3], [4,5,6]])
res = np.repeat(arr[:, None, :], 2, axis=1).T
print(res)

[[[1 4]
  [1 4]]

 [[2 5]
  [2 5]]

 [[3 6]
  [3 6]]]