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)]