问题

我有一个(N,K)维的numpy.ndarray "master"，我想要找到几个K维数组的出现次数，假设我有M个，它们存储在一个(M,K)维数组 "search" 中。

假设例如 N=9, K=2, M=3 并且

```python
master = array([[0, 0], [1, 0], [2, 0], [0, 1], [1, 1], [2, 1], [0, 2], [1, 2], [2, 2]])
search = array([[1, 2], [2, 0], [4, -2]])

我想要的是类似于 array([7, 2]) 或者 array([2, 7])，因为主数组的索引2和7在搜索数组中出现。

我首先尝试使用 np.isin，写成
np.argwhere(np.all(np.isin(master, search), axis=1)).ravel()
但这返回了所有值都属于“search”的索引，但不一定属于同一个元素...

这另一种方法似乎可以工作，但它使用了Python的列表推导式和嵌套循环，所以我认为这是非常低效的：

np.argwhere(np.any(np.array([[np.array_equal(master[i], search[j])
                               for i in range(N)]
                              for j in range(M)]),
                   axis=0)).ravel()

是否有一种方法只使用NumPy的标准函数来做到这一点？我的输入数据非常大，因此理解列表的速度太慢了...

<details>
<summary>英文:</summary>

I have a (N,K)-dimensional numpy.ndarray &quot;master&quot;, in which I want to find the occurences of several K-dimensional arrays, let&#39;s say I have M of them, stored in a (M,K)-dimensional array &quot;search&quot;.

Let&#39;s say for example N=9, K=2, M=3 and

master = array([[0, 0], [1, 0], [2, 0], [0, 1], [1, 1], [2, 1], [0, 2], [1, 2], [2, 2]])
search = array([[1, 2], [2, 0], [4, -2]])

Here what I would like is something like `array([7, 2])` or `array([2, 7])`, because indices 2 and 7 of the master array appear in the search array.

I first tried to use `np.isin` by writing
`np.argwhere(np.all(np.isin(master, search), axis=1)).ravel()`
But this returned the indices where all the values belonged to `search` but not necessarily to the same element...

This other approach seems to work but it uses Python's list comprehension with nested loops so I think this is very sub-optimal :

```python
np.argwhere(np.any(np.array([[np.array_equal(master[i], search[j])
                               for i in range(N)]
                              for j in range(M)]),
                   axis=0)).ravel()

Is there a way to do so by only using Numpy standard functions ? I have pretty big entries so comprehension lists are too slow...

答案1

得分: 1

1. 使用 == 对 search 和 master 中的每个项进行相等性检查。为了逐元素进行操作，我们需要使用 search[:,None] 将 search 转换成一个二维数组。

2. 使用 all 在轴 2 上执行逻辑 AND 操作，检查 search 中的两个项是否都等于 master 中的两个项。

3. 使用 any 在轴 0 上执行逻辑 OR 操作，将结果合并以获取每个 search 项的 True 或 False 值。

4. 最后，使用 np.where 找到索引。

带有命名步骤的代码如下：

import numpy as np

master = np.array([[0, 0], [1, 0], [2, 0], [0, 1], [1, 1], [2, 1], [0, 2], [1, 2], [2, 2]])
search = np.array([[1, 2], [2, 0], [4, -2]])
step1 = master == search[:,None]
step2 = step1.all(2)
step3 = step2.any(0)
step4 = np.where(step3)[0]  # [2, 7]

全部合在一起的代码如下：

import numpy as np

master = np.array([[0, 0], [1, 0], [2, 0], [0, 1], [1, 1], [2, 1], [0, 2], [1, 2], [2, 2]])
search = np.array([[1, 2], [2, 0], [4, -2]])
res = np.where((master == search[:,None]).all(2).any(0))[0]   # [2, 7]

import numpy as np

master = np.array([[0, 0], [1, 0], [2, 0], [0, 1], [1, 1], [2, 1], [0, 2], [1, 2], [2, 2]])
search = np.array([[1, 2], [2, 0], [4, -2]])
step1 = master == search[:,None]
step2 = step1.all(2)
step3 = step2.any(0)
step4 = np.where(step3)[0]  # [2, 7]

import numpy as np

master = np.array([[0, 0], [1, 0], [2, 0], [0, 1], [1, 1], [2, 1], [0, 2], [1, 2], [2, 2]])
search = np.array([[1, 2], [2, 0], [4, -2]])
res = np.where((master == search[:,None]).all(2).any(0))[0]   # [2, 7]