How do I clone a list so that it doesn't change unexpectedly after assignment?
While using new_list = my_list, any modifications to new_list changes my_list every time. Why is this, and how can I clone or copy the list to prevent it? For example: >>> my_list = [1, 2, 3] >>> new_list = my_list >>> new_list.append(4) >>> my_list [1, 2, 3, 4]
new_list = my_list doesn't actually create a second list. The assignment just copies the reference to the list, not the actual list, so both new_list and my_list refer to the same list after the assignment. To actually copy the list, you have several options: You can use the built-in list.copy() method (available since Python 3.3): new_list = old_list.copy() You can slice it: new_list = old_list[:] Alex Martelli's opinion (at least back in 2007) about this is, that it is a weird syntax and it does not make sense to use it ever. ;) (In his opinion, the next one is more readable). You can use the built-in list() constructor: new_list = list(old_list) You can use generic copy.copy(): import copy new_list = copy.copy(old_list) This is a little slower than list() because it has to find out the datatype of old_list first. If you need to copy the elements of the list as well, use generic copy.deepcopy(): import copy new_list = copy.deepcopy(old_list) Obviously the slowest and most memory-needing method, but sometimes unavoidable. This operates recursively; it will handle any number of levels of nested lists (or other containers). Example: import copy class Foo(object): def __init__(self, val): self.val = val def __repr__(self): return f'Foo({self.val!r})' foo = Foo(1) a = ['foo', foo] b = a.copy() c = a[:] d = list(a) e = copy.copy(a) f = copy.deepcopy(a) # edit orignal list and instance a.append('baz') foo.val = 5 print(f'original: {a}\nlist.copy(): {b}\nslice: {c}\nlist(): {d}\ncopy: {e}\ndeepcopy: {f}') Result: original: ['foo', Foo(5), 'baz'] list.copy(): ['foo', Foo(5)] slice: ['foo', Foo(5)] list(): ['foo', Foo(5)] copy: ['foo', Foo(5)] deepcopy: ['foo', Foo(1)]
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