Numpy shuffle multidimensional array by row only, keep column order unchanged

Question

How can I shuffle a multidimensional array by row only in Python (so do not shuffle the columns).

I am looking for the most efficient solution, because my ma

Answer 1

You can also use np.random.permutation to generate random permutation of row indices and then index into the rows of X using np.take with axis=0. Also, np.take facilitates overwriting to the input array X itself with out= option, which would save us memory. Thus, the implementation would look like this -

np.take(X,np.random.permutation(X.shape[0]),axis=0,out=X)

Sample run -

In [23]: X
Out[23]: 
array([[ 0.60511059,  0.75001599],
       [ 0.30968339,  0.09162172],
       [ 0.14673218,  0.09089028],
       [ 0.31663128,  0.10000309],
       [ 0.0957233 ,  0.96210485],
       [ 0.56843186,  0.36654023]])

In [24]: np.take(X,np.random.permutation(X.shape[0]),axis=0,out=X);

In [25]: X
Out[25]: 
array([[ 0.14673218,  0.09089028],
       [ 0.31663128,  0.10000309],
       [ 0.30968339,  0.09162172],
       [ 0.56843186,  0.36654023],
       [ 0.0957233 ,  0.96210485],
       [ 0.60511059,  0.75001599]])

Additional performance boost

Here's a trick to speed up np.random.permutation(X.shape[0]) with np.argsort() -

np.random.rand(X.shape[0]).argsort()

Speedup results -

In [32]: X = np.random.random((6000, 2000))

In [33]: %timeit np.random.permutation(X.shape[0])
1000 loops, best of 3: 510 µs per loop

In [34]: %timeit np.random.rand(X.shape[0]).argsort()
1000 loops, best of 3: 297 µs per loop

Thus, the shuffling solution could be modified to -

np.take(X,np.random.rand(X.shape[0]).argsort(),axis=0,out=X)

---

Runtime tests -

These tests include the two approaches listed in this post and np.shuffle based one in @Kasramvd's solution.

In [40]: X = np.random.random((6000, 2000))

In [41]: %timeit np.random.shuffle(X)
10 loops, best of 3: 25.2 ms per loop

In [42]: %timeit np.take(X,np.random.permutation(X.shape[0]),axis=0,out=X)
10 loops, best of 3: 53.3 ms per loop

In [43]: %timeit np.take(X,np.random.rand(X.shape[0]).argsort(),axis=0,out=X)
10 loops, best of 3: 53.2 ms per loop

So, it seems using these np.take based could be used only if memory is a concern or else np.random.shuffle based solution looks like the way to go.

Answer 2

That's what numpy.random.shuffle() is for :

>>> X = np.random.random((6, 2))
>>> X
array([[ 0.9818058 ,  0.67513579],
       [ 0.82312674,  0.82768118],
       [ 0.29468324,  0.59305925],
       [ 0.25731731,  0.16676408],
       [ 0.27402974,  0.55215778],
       [ 0.44323485,  0.78779887]])

>>> np.random.shuffle(X)
>>> X
array([[ 0.9818058 ,  0.67513579],
       [ 0.44323485,  0.78779887],
       [ 0.82312674,  0.82768118],
       [ 0.29468324,  0.59305925],
       [ 0.25731731,  0.16676408],
       [ 0.27402974,  0.55215778]])

Answer 3

After a bit experiment i found most memory and time efficient way to shuffle data(row wise) of nd-array is, shuffle the index and get the data from shuffled index

rand_num2 = np.random.randint(5, size=(6000, 2000))
perm = np.arange(rand_num2.shape[0])
np.random.shuffle(perm)
rand_num2 = rand_num2[perm]

in more details
Here, I am using memory_profiler to find memory usage and python's builtin "time" module to record time and comparing all previous answers

def main():
    # shuffle data itself
    rand_num = np.random.randint(5, size=(6000, 2000))
    start = time.time()
    np.random.shuffle(rand_num)
    print('Time for direct shuffle: {0}'.format((time.time() - start)))

    # Shuffle index and get data from shuffled index
    rand_num2 = np.random.randint(5, size=(6000, 2000))
    start = time.time()
    perm = np.arange(rand_num2.shape[0])
    np.random.shuffle(perm)
    rand_num2 = rand_num2[perm]
    print('Time for shuffling index: {0}'.format((time.time() - start)))

    # using np.take()
    rand_num3 = np.random.randint(5, size=(6000, 2000))
    start = time.time()
    np.take(rand_num3, np.random.rand(rand_num3.shape[0]).argsort(), axis=0, out=rand_num3)
    print("Time taken by np.take, {0}".format((time.time() - start)))

Result for Time

Time for direct shuffle: 0.03345608711242676   # 33.4msec
Time for shuffling index: 0.019818782806396484 # 19.8msec
Time taken by np.take, 0.06726956367492676     # 67.2msec

Memory profiler Result

Line #    Mem usage    Increment   Line Contents
================================================
    39  117.422 MiB    0.000 MiB   @profile
    40                             def main():
    41                                 # shuffle data itself
    42  208.977 MiB   91.555 MiB       rand_num = np.random.randint(5, size=(6000, 2000))
    43  208.977 MiB    0.000 MiB       start = time.time()
    44  208.977 MiB    0.000 MiB       np.random.shuffle(rand_num)
    45  208.977 MiB    0.000 MiB       print('Time for direct shuffle: {0}'.format((time.time() - start)))
    46                             
    47                                 # Shuffle index and get data from shuffled index
    48  300.531 MiB   91.555 MiB       rand_num2 = np.random.randint(5, size=(6000, 2000))
    49  300.531 MiB    0.000 MiB       start = time.time()
    50  300.535 MiB    0.004 MiB       perm = np.arange(rand_num2.shape[0])
    51  300.539 MiB    0.004 MiB       np.random.shuffle(perm)
    52  300.539 MiB    0.000 MiB       rand_num2 = rand_num2[perm]
    53  300.539 MiB    0.000 MiB       print('Time for shuffling index: {0}'.format((time.time() - start)))
    54                             
    55                                 # using np.take()
    56  392.094 MiB   91.555 MiB       rand_num3 = np.random.randint(5, size=(6000, 2000))
    57  392.094 MiB    0.000 MiB       start = time.time()
    58  392.242 MiB    0.148 MiB       np.take(rand_num3, np.random.rand(rand_num3.shape[0]).argsort(), axis=0, out=rand_num3)
    59  392.242 MiB    0.000 MiB       print("Time taken by np.take, {0}".format((time.time() - start)))

Answer 4

I tried many solutions, and at the end I used this simple one:

from sklearn.utils import shuffle
x = np.array([[1, 2],
              [3, 4],
              [5, 6]])
print(shuffle(x, random_state=0))

output:

[
[5 6]  
[3 4]  
[1 2]
]

if you have 3d array, loop through the 1st axis (axis=0) and apply this function, like:

np.array([shuffle(item) for item in 3D_numpy_array])

Answer 5

I have a question on this (or maybe it is the answer) Lets say we have a numpy array X with shape=(1000,60,11,1) Also suppose that X is an array of images with size 60x11 and channel number =1 (60x11x1).

What if I want to shuffle the order of all these images, and to do that I'll use shuffling on the indexes of X.

def shuffling( X):
 indx=np.arange(len(X))          # create a array with indexes for X data
 np.random.shuffle(indx)
 X=X[indx]
 return X

Will that work? From my knowledge len(X) will return the biggest dimension size.

Answer 6

You can shuffle a two dimensional array A by row using the np.vectorize() function:

shuffle = np.vectorize(np.random.permutation, signature='(n)->(n)')

A_shuffled = shuffle(A)