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.

Gustav Larsson · Accepted Answer · 2013-07-22 20:55:15Z

3

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]

answered Jul 22, 2013 at 20:55

Gustav Larsson

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1 Comment

A.Wan Over a year ago

Could you explain how axis works? I was reading about np.all and didn't understand what they meant by axis

Jaime · Accepted Answer · 2013-07-22 20:59:13Z

1

You can use np.in1d over a view of your original array with all coordinates collapsed into a single variable of dtype np.void:

import numpy as np

A = np.array([[1,2], [1,0], [5,3]])
values = np.array([[1,2], [1,0]])

# Make sure both arrays are contiguous and have common dtype
common_dtype = np.common_type(A, values)
a = np.ascontiguousarray(A, dtype=common_dtype)
vals = np.ascontiguousarray(values, dtype=common_dtype)

a_view = A.view((np.void, A.dtype.itemsize*A.shape[1])).ravel()
values_view = values.view((np.void,
                           values.dtype.itemsize*values.shape[1])).ravel()

Now each item of a_view and values_view is all coordinates for one point packed together, so you can do whatever 1D magic you would use. I don't see how to use np.in1d to find indices though, so I would go the np.searchsorted route:

sort_idx = np.argsort(a_view)
locations = np.searchsorted(a_view, values_view, sorter=sort_idx)
locations = sort_idx[locations]

>>> locations
array([0, 1], dtype=int64)

answered Jul 22, 2013 at 20:59

Jaime

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1 Comment

user1984528 Over a year ago

Thanks, I attempted using a 1d view, but I think my problem was not using np.ascontiguousarray(). As for the np.in1d this would be to generate a mask instead of the indices

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