Given the following NumPy array,
> a = array([[1, 2, 3, 4, 5], [1, 2, 3, 4, 5],[1, 2, 3, 4, 5]]) it's simple enough to shuffle a single row,
> shuffle(a[0]) > a array([[4, 2, 1, 3, 5],[1, 2, 3, 4, 5],[1, 2, 3, 4, 5]]) Is it possible to use indexing notation to shuffle each of the rows independently? Or do you have to iterate over the array. I had in mind something like,
> numpy.shuffle(a[:]) > a array([[4, 2, 3, 5, 1],[3, 1, 4, 5, 2],[4, 2, 1, 3, 5]]) # Not the real output though this clearly doesn't work.
2 Answers
Vectorized solution with rand+argsort trick
We could generate unique indices along the specified axis and index into the the input array with advanced-indexing. To generate the unique indices, we would use random float generation + sort trick, thus giving us a vectorized solution. We would also generalize it to cover generic n-dim arrays and along generic axes with np.take_along_axis. The final implementation would look something like this -
def shuffle_along_axis(a, axis): idx = np.random.rand(*a.shape).argsort(axis=axis) return np.take_along_axis(a,idx,axis=axis) Note that this shuffle won't be in-place and returns a shuffled copy.
Sample run -
In [33]: a Out[33]: array([[18, 95, 45, 33], [40, 78, 31, 52], [75, 49, 42, 94]]) In [34]: shuffle_along_axis(a, axis=0) Out[34]: array([[75, 78, 42, 94], [40, 49, 45, 52], [18, 95, 31, 33]]) In [35]: shuffle_along_axis(a, axis=1) Out[35]: array([[45, 18, 33, 95], [31, 78, 52, 40], [42, 75, 94, 49]]) You have to call numpy.random.shuffle() several times because you are shuffling several sequences independently. numpy.random.shuffle() works on any mutable sequence and is not actually a ufunc. The shortest and most efficient code to shuffle all rows of a two-dimensional array a separately probably is
list(map(numpy.random.shuffle, a)) Some people prefer to write this as a list comprehension instead:
[numpy.random.shuffle(x) for x in a] 
