location of array of values in numpy array

Question

Here is a small code to illustrate the problem.

A = array([[1,2], [1,0], [5,3]])
f_of_A = f(A)   # this is precomputed and expensive


values = array([[1,2], [1,0]])


# location of values in A
# if I just had 1d values I could use numpy.in1d here
indices = array([0, 1])


# example of operation type I need (recalculating f_of_A as needed is not an option)
f_of_A[ indices ]

So, basically I think I need some equivalent to in1d for higher dimensions. Does such a thing exist? Or is there some other approach?

Looks like there is also a searchsorted() function, but that seems to work for 1d arrays also. In this example I used 2d points, but any solution would need to work for 3d points also.

Answer 1

Okay, this is what I came up with.

To find the value of one multi-dimensional index, let's say ii = np.array([1,2]), we can do:

n.where((A == ii).all(axis=1))[0]

Let's break this down, we have A == ii, which will give element-wise comparisons with ii for each row of A. We want an entire row to be true, so we add .all(axis=1) to collapse them. To find where these indices happen, we plug this into np.where and get the first value of the tuple.

Now, I don't have a fast way to do this with multiple indices yet (although I have a feeling there is one). However, this will get the job done:

np.hstack([np.where((A == values[i]).all(axis=1))[0] for i in xrange(len(values))])

This basically just calls the above, for each value of values, and concatenates the result.

Update:

Here is for the multi-dimensional case (all in one go, should be fairly fast):

np.where((np.expand_dims(A, -1) == values.T).all(axis=1).any(axis=1))[0]