Shuffle a list and return a copy

Question

I want to shuffle an array, but all I find was method like random.shuffle(x) , from Best way to randomize a list of strings in Python

Can I do something like

import random
rectangle = [(0,0),(0,1),(1,1),(1,0)]
# I want something like
# disorderd_rectangle = rectangle.shuffle

Now I can only get away with

disorderd_rectangle = rectangle
random.shuffle(disorderd_rectangle)
print(disorderd_rectangle)
print(rectangle)

But it returns

[(1, 1), (1, 0), (0, 1), (0, 0)]
[(1, 1), (1, 0), (0, 1), (0, 0)]

So the original array is also changed. How can I just create another shuffled array without changing the original one?

Answer 1

People here are advising deepcopy, which is surely an overkill. You probably don't mind the objects in your list being same, you just want to shuffle their order. For that, list provides shallow copying directly.

rectangle2 = rectangle.copy()
random.shuffle(rectangle2)

About your misconception: please read http://nedbatchelder.com/text/names.html#no_copies

Answer 2

Usecopy.deepcopy to create a copy of the array, shuffle the copy.

c = copy.deepcopy(rectangle)
random.shuffle(c)

Answer 3

You need to make a copy of the list, by default python only creates pointers to the same object when you write:

disorderd_rectangle = rectangle

But instead use this or the copy method mentioned by Veky.

disorderd_rectangle = rectangle[:]

It will make a copy of the list.

Answer 4

Use a slice to make a shallow copy, then shuffle the copy:

>>> rect = [(0,0),(0,1),(1,1),(1,0)]
>>> sh_rect=rect[:]
>>> random.shuffle(sh_rect)
>>> sh_rect
[(0, 1), (1, 0), (1, 1), (0, 0)]
>>> rect
[(0, 0), (0, 1), (1, 1), (1, 0)]

Answer 5

Use random.sample to shuffle a list without changing the original one.

from random import sample
rect = [(0,0),(0,1),(1,1),(1,0)]
shuffled_rect = sample(rect, len(rect))

The code snippet above will be faster since you don't deep copy your list.