获取交叉numpy的二维数组的行索引 [英] Get indices of intersecting rows of Numpy 2d Array

问题描述

我希望得到一个主numpy的二维数组A的相交行的索引，另有一架B。

I want to get the indices of the intersecting rows of a main numpy 2d array A, with another one B.

A=array([[1, 2],
         [3, 4],
         [5, 6],
         [7, 8],
         [9, 10]])

B=array([[1, 4],
         [1, 2],
         [5, 6],
         [6, 3]])

result=[0,2]

如果本应返回[0,2]基于阵列A的指数。

Where this should return [0,2] based on the indices of array A.

如何才能实现这一二维数组有效地完成？

How can this be done efficiently for 2d arrays?

感谢您！

修改

我已经试过功能：

k[np.in1d(k.view(dtype='i,i').reshape(k.shape[0]),k2.view(dtype='i,i').
reshape(k2.shape[0]))]

为in1d二维数组？但我得到一个重塑错误。我的数据类型是浮点数（有两位小数）。
此外，我也试图与套，但表现也是相当缓慢的。

from numpy in1d for 2D arrays? but I get a reshape error. My datatype is floats (with two decimals). Moreover, I also tried with sets but the performance is quite slow.

推荐答案

通过最小的变化，你可以得到你的工作方式：

With minimal changes, you can get your approach to work:

In [15]: A
Out[15]: 
array([[ 1,  2],
       [ 3,  4],
       [ 5,  6],
       [ 7,  8],
       [ 9, 10]])

In [16]: B
Out[16]: 
array([[1, 4],
       [1, 2],
       [5, 6],
       [6, 3]])

In [17]: np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1))
Out[17]: array([ True, False,  True, False, False], dtype=bool)

In [18]: np.nonzero(np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1)))
Out[18]: (array([0, 2], dtype=int64),)

In [19]: np.nonzero(np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1)))[0]
Out[19]: array([0, 2], dtype=int64)

如果您的数组不是花车，和都是连续的，那么下面会更快：

If your arrays are not floats, and are both contiguous, then the following will be faster:

In [21]: dt = np.dtype((np.void, A.dtype.itemsize * A.shape[1]))

In [22]: np.nonzero(np.in1d(A.view(dt).reshape(-1), B.view(dt).reshape(-1)))[0]
Out[22]: array([0, 2], dtype=int64)

和快速时间：

In [24]: %timeit np.nonzero(np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1)))[0]
10000 loops, best of 3: 75 µs per loop

In [25]: %timeit np.nonzero(np.in1d(A.view(dt).reshape(-1), B.view(dt).reshape(-1)))[0]
10000 loops, best of 3: 29.8 µs per loop

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