Python: intersection indices numpy array

Question

How can I get the indices of intersection points between two numpy arrays? I can get intersecting values with intersect1d:

import numpy as np  a = np.array(xrange(11)) b = np.array([2, 7, 10]) inter = np.intersect1d(a, b) # inter == array([ 2,  7, 10]) 

But how can I get the indices into a of the values in inter?

Answer 1

You could use the boolean array produced by in1d to index an arange. Reversing a so that the indices are different from the values:

>>> a[::-1] array([10,  9,  8,  7,  6,  5,  4,  3,  2,  1,  0]) >>> a = a[::-1] 

intersect1d still returns the same values...

>>> numpy.intersect1d(a, b) array([ 2,  7, 10]) 

But in1d returns a boolean array:

>>> numpy.in1d(a, b) array([ True, False, False,  True, False, False, False, False,  True,        False, False], dtype=bool) 

Which can be used to index a range:

>>> numpy.arange(a.shape[0])[numpy.in1d(a, b)] array([0, 3, 8]) >>> indices = numpy.arange(a.shape[0])[numpy.in1d(a, b)] >>> a[indices] array([10,  7,  2]) 

To simplify the above, though, you could use nonzero -- this is probably the most correct approach, because it returns a tuple of uniform lists of X, Y... coordinates:

>>> numpy.nonzero(numpy.in1d(a, b)) (array([0, 3, 8]),) 

Or, equivalently:

>>> numpy.in1d(a, b).nonzero() (array([0, 3, 8]),) 

The result can be used as an index to arrays of the same shape as a with no problems.

>>> a[numpy.nonzero(numpy.in1d(a, b))] array([10,  7,  2]) 

But note that under many circumstances, it makes sense just to use the boolean array itself, rather than converting it into a set of non-boolean indices.

Finally, you can also pass the boolean array to argwhere, which produces a slightly differently-shaped result that's not as suitable for indexing, but might be useful for other purposes.

>>> numpy.argwhere(numpy.in1d(a, b)) array([[0],        [3],        [8]])