invis invis - 1 month ago 12x
Python Question

Python: intersection indices numpy array

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


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
of the values in


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))