Shuffle a list and return a copy

问题

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?

回答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

回答2:

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

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

回答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.

回答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)]

来源：https://stackoverflow.com/questions/30253198/shuffle-a-list-and-return-a-copy