With new_list = my_list, you don’t actually have two lists. 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 various possibilities:
- You can use the builtin
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()function:new_list = list(old_list) - You can use generic
copy.copy():import copy new_list = copy.copy(old_list)This is a little slower thanlist()because it has to find out the datatype ofold_listfirst. - If the list contains objects and you want to copy them 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.
Example:
import copy
class Foo(object):
def __init__(self, val):
self.val = val
def __repr__(self):
return 'Foo({!r})'.format(self.val)
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('original: %r\nlist.copy(): %r\nslice: %r\nlist(): %r\ncopy: %r\ndeepcopy: %r'
% (a, b, c, d, e, 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)]