296
votes

What's the easiest way to shuffle an array with python?

11
+1 for migrating the most useful bits of the python documentation to the always superior SO Q&A format.charleslparker
is there an option that doesn't mutate the original array but return a new shuffled array?Charlie Parker
you can get a new array (unmodified) with new_array = random.sample( array, len(array) ).Charlie Parker

11 Answers

531
votes
import random
random.shuffle(array)
111
votes
import random
random.shuffle(array)
44
votes

Alternative way to do this using sklearn

from sklearn.utils import shuffle
X=[1,2,3]
y = ['one', 'two', 'three']
X, y = shuffle(X, y, random_state=0)
print(X)
print(y)

Output:

[2, 1, 3]
['two', 'one', 'three']

Advantage: You can random multiple arrays simultaneously without disrupting the mapping. And 'random_state' can control the shuffling for reproducible behavior.

21
votes

The other answers are the easiest, however it's a bit annoying that the random.shuffle method doesn't actually return anything - it just sorts the given list. If you want to chain calls or just be able to declare a shuffled array in one line you can do:

    import random
    def my_shuffle(array):
        random.shuffle(array)
        return array

Then you can do lines like:

    for suit in my_shuffle(['hearts', 'spades', 'clubs', 'diamonds']):
17
votes

Just in case you want a new array you can use sample:

import random
new_array = random.sample( array, len(array) )
12
votes

When dealing with regular Python lists, random.shuffle() will do the job just as the previous answers show.

But when it come to ndarray(numpy.array), random.shuffle seems to break the original ndarray. Here is an example:

import random
import numpy as np
import numpy.random

a = np.array([1,2,3,4,5,6])
a.shape = (3,2)
print a
random.shuffle(a) # a will definitely be destroyed
print a

Just use: np.random.shuffle(a)

Like random.shuffle, np.random.shuffle shuffles the array in-place.

5
votes

You can sort your array with random key

sorted(array, key = lambda x: random.random())

key only be read once so comparing item during sort still efficient.

but look like random.shuffle(array) will be faster since it written in C

3
votes

In addition to the previous replies, I would like to introduce another function.

numpy.random.shuffle as well as random.shuffle perform in-place shuffling. However, if you want to return a shuffled array numpy.random.permutation is the function to use.

1
votes

I don't know I used random.shuffle() but it return 'None' to me, so I wrote this, might helpful to someone

def shuffle(arr):
    for n in range(len(arr) - 1):
        rnd = random.randint(0, (len(arr) - 1))
        val1 = arr[rnd]
        val2 = arr[rnd - 1]

        arr[rnd - 1] = val1
        arr[rnd] = val2

    return arr
0
votes
# arr = numpy array to shuffle

def shuffle(arr):
    a = numpy.arange(len(arr))
    b = numpy.empty(1)
    for i in range(len(arr)):
        sel = numpy.random.random_integers(0, high=len(a)-1, size=1)
        b = numpy.append(b, a[sel])
        a = numpy.delete(a, sel)
    b = b[1:].astype(int)
    return arr[b]
0
votes

Be aware that random.shuffle() should not be used on multi-dimensional arrays as it causes repetitions.

Imagine you want to shuffle an array along its first dimension, we can create the following test example,

import numpy as np
x = np.zeros((10, 2, 3))

for i in range(10):
   x[i, ...] = i*np.ones((2,3))

so that along the first axis, the i-th element corresponds to a 2x3 matrix where all the elements are equal to i.

If we use the correct shuffle function for multi-dimensional arrays, i.e. np.random.shuffle(x), the array will be shuffled along the first axis as desired. However, using random.shuffle(x) will cause repetitions. You can check this by running len(np.unique(x)) after shuffling which gives you 10 (as expected) with np.random.shuffle() but only around 5 when using random.shuffle().