Python – Better way to shuffle two numpy arrays in unison


I have two numpy arrays of different shapes, but with the same length (leading dimension). I want to shuffle each of them, such that corresponding elements continue to correspond — i.e. shuffle them in unison with respect to their leading indices.

This code works, and illustrates my goals:

def shuffle_in_unison(a, b):
    assert len(a) == len(b)
    shuffled_a = numpy.empty(a.shape, dtype=a.dtype)
    shuffled_b = numpy.empty(b.shape, dtype=b.dtype)
    permutation = numpy.random.permutation(len(a))
    for old_index, new_index in enumerate(permutation):
        shuffled_a[new_index] = a[old_index]
        shuffled_b[new_index] = b[old_index]
    return shuffled_a, shuffled_b

For example:

>>> a = numpy.asarray([[1, 1], [2, 2], [3, 3]])
>>> b = numpy.asarray([1, 2, 3])
>>> shuffle_in_unison(a, b)
(array([[2, 2],
       [1, 1],
       [3, 3]]), array([2, 1, 3]))

However, this feels clunky, inefficient, and slow, and it requires making a copy of the arrays — I'd rather shuffle them in-place, since they'll be quite large.

Is there a better way to go about this? Faster execution and lower memory usage are my primary goals, but elegant code would be nice, too.

One other thought I had was this:

def shuffle_in_unison_scary(a, b):
    rng_state = numpy.random.get_state()

This works…but it's a little scary, as I see little guarantee it'll continue to work — it doesn't look like the sort of thing that's guaranteed to survive across numpy version, for example.

Best Solution

Your can use NumPy's array indexing:

def unison_shuffled_copies(a, b):
    assert len(a) == len(b)
    p = numpy.random.permutation(len(a))
    return a[p], b[p]

This will result in creation of separate unison-shuffled arrays.