Shuffling NumPy array along a given axis

问题:

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.

回答1:

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

map(numpy.random.shuffle, a)

回答2:

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]])