Changing One List Unexpectedly Changes Another, Too

Changing one list unexpectedly changes another, too

Why does v change at all?

vec and v are both references.

When coding vec = v you assign v address to vec.
Therefore changing data in v will also "change" vec.

If you want to have two different arrays use:

vec = list(v)

Changing one list changed the other list

What you did was a shallow copy. The difference between shallow and deep copying is only relevant for compound objects (objects that contain other objects, like lists or class instances):

A shallow copy constructs a new compound object and then (to the extent possible) inserts references into it to the objects found in the original.
A deep copy constructs a new compound object and then, recursively, inserts copies into it of the objects found in the original.

What you want to do it to create a whole new object so that one will not interfere with other. Here is how to do a deep copy.

import copy

a = [[1, 1], [2, 2], [3, 3]]
b = copy.deepcopy(a)
a[0][0] = 100
print(a)
print(b)

Changing an element in one list changes multiple lists

What matters is how you created your original mysolution list. As it seems, it contains four times the same list which is why changing it once will make it change in all four locations.

To initialize independent zero-filled lists like that, you can do the following:

mysolution = [[0] * 4 for i in range(4)]

List of lists changes reflected across sublists unexpectedly

When you write [x]*3 you get, essentially, the list [x, x, x]. That is, a list with 3 references to the same x. When you then modify this single x it is visible via all three references to it:

x = [1] * 4
xs = [x] * 3
print(f"id(x): {id(x)}")
# id(x): 140560897920048
print(
    f"id(xs[0]): {id(xs[0])}\n"
    f"id(xs[1]): {id(xs[1])}\n"
    f"id(xs[2]): {id(xs[2])}"
)
# id(xs[0]): 140560897920048
# id(xs[1]): 140560897920048
# id(xs[2]): 140560897920048

x[0] = 42
print(f"x: {x}")
# x: [42, 1, 1, 1]
print(f"xs: {xs}")
# xs: [[42, 1, 1, 1], [42, 1, 1, 1], [42, 1, 1, 1]]

To fix it, you need to make sure that you create a new list at each position. One way to do it is

[[1]*4 for _ in range(3)]

which will reevaluate [1]*4 each time instead of evaluating it once and making 3 references to 1 list.

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You might wonder why * can't make independent objects the way the list comprehension does. That's because the multiplication operator * operates on objects, without seeing expressions. When you use * to multiply [[1] * 4] by 3, * only sees the 1-element list [[1] * 4] evaluates to, not the [[1] * 4 expression text. * has no idea how to make copies of that element, no idea how to reevaluate [[1] * 4], and no idea you even want copies, and in general, there might not even be a way to copy the element.

The only option * has is to make new references to the existing sublist instead of trying to make new sublists. Anything else would be inconsistent or require major redesigning of fundamental language design decisions.

In contrast, a list comprehension reevaluates the element expression on every iteration. [[1] * 4 for n in range(3)] reevaluates [1] * 4 every time for the same reason [x**2 for x in range(3)] reevaluates x**2 every time. Every evaluation of [1] * 4 generates a new list, so the list comprehension does what you wanted.

Incidentally, [1] * 4 also doesn't copy the elements of [1], but that doesn't matter, since integers are immutable. You can't do something like 1.value = 2 and turn a 1 into a 2.

Changes made in Python list showing up in copy of original list

Use copy.deepcopy:

from copy import deepcopy

desc1 = deepcopy(desc)

list.copy only makes a shallow copy: a new list object, but its elements (here mutable sublists) will be references to the same objects as the elements of the original.

How do I clone a list so that it doesn't change unexpectedly after assignment?

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 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() 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)]

Changing list in dictionary changes all lists

from copy import deepcopy
empty = []
for x in range(2):
    empty.append([])
    for y in range(2):
        empty[x].append(False)

status = {k:[] for k in ["a", "b", "c"]}

status["a"] = deepcopy(empty)
status["b"] = deepcopy(empty)
status["c"] = deepcopy(empty)

print(status)
status["a"][0][0] = True
print(status)

you were close with the copy, but you actually neeed a deepcopy

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