查找两个numpy数组之间的索引映射 [英] Find index mapping between two numpy arrays

问题描述

numpy 中是否有一种很好的方法来获取 array1 中每个元素在 array2 中位置的逐元素索引?

Is there a nice way in numpy to get element-wise indexes of where each element in array1 is in array2?

一个例子:

array1 = np.array([1, 3, 4])
array2 = np.arange(-2, 5, 1, dtype=np.int)

np.where(array1[0] == array2)
# (array([3]),)
np.where(array1[1] == array2)
# (array([5]),)
np.where(array1[2] == array2)
# (array([6]),)

我想做

np.where(array1 == array2)
# (array([3 5 6]),)

像这样可能吗?我们保证 array1 中的所有条目都可以在 array2 中找到.

Is something like this possible? We are guaranteed that all entries in array1 can be found in array2.

推荐答案

方法1:使用

Approach #1 : Use np.in1d there to get a mask of places where matches occur and then np.where to get those index positions -

np.where(np.in1d(array2, array1))

方法2::使用 np.searchsorted -

np.searchsorted(array2, array1)

请注意，如果未对 array2 进行排序，则需要使用其他可选参数 sorter .

Please note that if array2 is not sorted, we need to use the additional optional argument sorter with it.

样品运行-

In [14]: array1
Out[14]: array([1, 3, 4])

In [15]: array2
Out[15]: array([-2, -1,  0,  1,  2,  3,  4])

In [16]: np.where(np.in1d(array2, array1))
Out[16]: (array([3, 5, 6]),)

In [17]: np.searchsorted(array2, array1)
Out[17]: array([3, 5, 6])

运行时测试-

In [62]: array1 = np.random.choice(10000,1000,replace=0)

In [63]: array2 = np.sort(np.random.choice(100000,10000,replace=0))

In [64]: %timeit np.where(np.in1d(array2, array1))
1000 loops, best of 3: 483 µs per loop

In [65]: %timeit np.searchsorted(array2, array1)
10000 loops, best of 3: 40 µs per loop

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