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()
    numpy.random.shuffle(a)
    numpy.random.set_state(rng_state)
    numpy.random.shuffle(b)

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.

Your “scary” solution does not appear scary to me. Calling shuffle() for two sequences of the same length results in the same number of calls to the random number generator, and these are the only “random” elements in the shuffle algorithm. By resetting the state, you ensure that the calls to the random number generator will give the same results in the second call to shuffle(), so the whole algorithm will generate the same permutation.

If you don’t like this, a different solution would be to store your data in one array instead of two right from the beginning, and create two views into this single array simulating the two arrays you have now. You can use the single array for shuffling and the views for all other purposes.

Example: Let’s assume the arrays a and b look like this:

a = numpy.array([[[  0.,   1.,   2.],
                  [  3.,   4.,   5.]],

                 [[  6.,   7.,   8.],
                  [  9.,  10.,  11.]],

                 [[ 12.,  13.,  14.],
                  [ 15.,  16.,  17.]]])

b = numpy.array([[ 0.,  1.],
                 [ 2.,  3.],
                 [ 4.,  5.]])

We can now construct a single array containing all the data:

c = numpy.c_[a.reshape(len(a), -1), b.reshape(len(b), -1)]
# array([[  0.,   1.,   2.,   3.,   4.,   5.,   0.,   1.],
#        [  6.,   7.,   8.,   9.,  10.,  11.,   2.,   3.],
#        [ 12.,  13.,  14.,  15.,  16.,  17.,   4.,   5.]])

Now we create views simulating the original a and b:

a2 = c[:, :a.size//len(a)].reshape(a.shape)
b2 = c[:, a.size//len(a):].reshape(b.shape)

The data of a2 and b2 is shared with c. To shuffle both arrays simultaneously, use numpy.random.shuffle(c).

In production code, you would of course try to avoid creating the original a and b at all and right away create c, a2 and b2.

This solution could be adapted to the case that a and b have different dtypes.

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.

X = np.array([[1., 0.], [2., 1.], [0., 0.]])
y = np.array([0, 1, 2])
from sklearn.utils import shuffle
X, y = shuffle(X, y, random_state=0)

To learn more, see http://scikit-learn.org/stable/modules/generated/sklearn.utils.shuffle.html

Very simple solution:

randomize = np.arange(len(x))
np.random.shuffle(randomize)
x = x[randomize]
y = y[randomize]

the two arrays x,y are now both randomly shuffled in the same way

James wrote in 2015 an sklearn solution which is helpful. But he added a random state variable, which is not needed. In the below code, the random state from numpy is automatically assumed.

X = np.array([[1., 0.], [2., 1.], [0., 0.]])
y = np.array([0, 1, 2])
from sklearn.utils import shuffle
X, y = shuffle(X, y)

from np.random import permutation
from sklearn.datasets import load_iris
iris = load_iris()
X = iris.data #numpy array
y = iris.target #numpy array

# Data is currently unshuffled; we should shuffle 
# each X[i] with its corresponding y[i]
perm = permutation(len(X))
X = X[perm]
y = y[perm]

Shuffle any number of arrays together, in-place, using only NumPy.

import numpy as np


def shuffle_arrays(arrays, set_seed=-1):
    """Shuffles arrays in-place, in the same order, along axis=0

    Parameters:
    -----------
    arrays : List of NumPy arrays.
    set_seed : Seed value if int >= 0, else seed is random.
    """
    assert all(len(arr) == len(arrays[0]) for arr in arrays)
    seed = np.random.randint(0, 2**(32 - 1) - 1) if set_seed < 0 else set_seed

    for arr in arrays:
        rstate = np.random.RandomState(seed)
        rstate.shuffle(arr)

And can be used like this

a = np.array([1, 2, 3, 4, 5])
b = np.array([10,20,30,40,50])
c = np.array([[1,10,11], [2,20,22], [3,30,33], [4,40,44], [5,50,55]])

shuffle_arrays([a, b, c])

A few things to note:

• The assert ensures that all input arrays have the same length along their first dimension.
• Arrays shuffled in-place by their first dimension – nothing returned.
• Random seed within positive int32 range.
• If a repeatable shuffle is needed, seed value can be set.

After the shuffle, the data can be split using np.split or referenced using slices – depending on the application.

you can make an array like:

s = np.arange(0, len(a), 1)

then shuffle it:

np.random.shuffle(s)

now use this s as argument of your arrays. same shuffled arguments return same shuffled vectors.

x_data = x_data[s]
x_label = x_label[s]

One way in which in-place shuffling can be done for connected lists is using a seed (it could be random) and using numpy.random.shuffle to do the shuffling.

# Set seed to a random number if you want the shuffling to be non-deterministic.
def shuffle(a, b, seed):
   np.random.seed(seed)
   np.random.shuffle(a)
   np.random.seed(seed)
   np.random.shuffle(b)

That’s it. This will shuffle both a and b in the exact same way. This is also done in-place which is always a plus.

EDIT, don’t use np.random.seed() use np.random.RandomState instead

def shuffle(a, b, seed):
   rand_state = np.random.RandomState(seed)
   rand_state.shuffle(a)
   rand_state.seed(seed)
   rand_state.shuffle(b)

When calling it just pass in any seed to feed the random state:

a = [1,2,3,4]
b = [11, 22, 33, 44]
shuffle(a, b, 12345)

Output:

>>> a
[1, 4, 2, 3]
>>> b
[11, 44, 22, 33]

Edit: Fixed code to re-seed the random state

There is a well-known function that can handle this:

from sklearn.model_selection import train_test_split
X, _, Y, _ = train_test_split(X,Y, test_size=0.0)

Just setting test_size to 0 will avoid splitting and give you shuffled data. Though it is usually used to split train and test data, it does shuffle them too.
From documentation

Split arrays or matrices into random train and test subsets

Quick utility that wraps input validation and next(ShuffleSplit().split(X, y)) and application to input data into a single call for splitting (and optionally subsampling) data in a oneliner.

Say we have two arrays: a and b.

a = np.array([[1,2,3],[4,5,6],[7,8,9]])
b = np.array([[9,1,1],[6,6,6],[4,2,0]]) 

We can first obtain row indices by permutating first dimension

indices = np.random.permutation(a.shape[0])
[1 2 0]

Then use advanced indexing. Here we are using the same indices to shuffle both arrays in unison.

a_shuffled = a[indices[:,np.newaxis], np.arange(a.shape[1])]
b_shuffled = b[indices[:,np.newaxis], np.arange(b.shape[1])]

This is equivalent to

np.take(a, indices, axis=0)
[[4 5 6]
 [7 8 9]
 [1 2 3]]

np.take(b, indices, axis=0)
[[6 6 6]
 [4 2 0]
 [9 1 1]]

If you want to avoid copying arrays, then I would suggest that instead of generating a permutation list, you go through every element in the array, and randomly swap it to another position in the array

for old_index in len(a):
    new_index = numpy.random.randint(old_index+1)
    a[old_index], a[new_index] = a[new_index], a[old_index]
    b[old_index], b[new_index] = b[new_index], b[old_index]

This implements the Knuth-Fisher-Yates shuffle algorithm.

This seems like a very simple solution:

import numpy as np
def shuffle_in_unison(a,b):

    assert len(a)==len(b)
    c = np.arange(len(a))
    np.random.shuffle(c)

    return a[c],b[c]

a =  np.asarray([[1, 1], [2, 2], [3, 3]])
b =  np.asarray([11, 22, 33])

shuffle_in_unison(a,b)
Out[94]: 
(array([[3, 3],
        [2, 2],
        [1, 1]]),
 array([33, 22, 11]))

With an example, this is what I’m doing:

combo = []
for i in range(60000):
    combo.append((images[i], labels[i]))

shuffle(combo)

im = []
lab = []
for c in combo:
    im.append(c[0])
    lab.append(c[1])
images = np.asarray(im)
labels = np.asarray(lab)

I extended python’s random.shuffle() to take a second arg:

def shuffle_together(x, y):
    assert len(x) == len(y)

    for i in reversed(xrange(1, len(x))):
        # pick an element in x[:i+1] with which to exchange x[i]
        j = int(random.random() * (i+1))
        x[i], x[j] = x[j], x[i]
        y[i], y[j] = y[j], y[i]

That way I can be sure that the shuffling happens in-place, and the function is not all too long or complicated.

Just use numpy…

First merge the two input arrays 1D array is labels(y) and 2D array is data(x) and shuffle them with NumPy shuffle method. Finally split them and return.

import numpy as np

def shuffle_2d(a, b):
    rows= a.shape[0]
    if b.shape != (rows,1):
        b = b.reshape((rows,1))
    S = np.hstack((b,a))
    np.random.shuffle(S)
    b, a  = S[:,0], S[:,1:]
    return a,b

features, samples = 2, 5
x, y = np.random.random((samples, features)), np.arange(samples)
x, y = shuffle_2d(train, test)