List Of Lists Changes Reflected Across Sublists Unexpectedly

I needed to create a list of lists in Python, so I typed the following: my_list = [[1] * 4] * 3 The list looked like this: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]] Then I cha

Solution 1:

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

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

---

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.

Solution 2:

size = 3matrix_surprise = [[0] * size] * size
matrix = [[0]*size for _ in range(size)]

Live visualization using Python Tutor:

Solution 3:

Actually, this is exactly what you would expect. Let's decompose what is happening here:

You write

lst = [[1] * 4] * 3

This is equivalent to:

lst1 = [1]*4lst = [lst1]*3

This means lst is a list with 3 elements all pointing to lst1. This means the two following lines are equivalent:

lst[0][0] = 5
lst1[0] = 5

As lst[0] is nothing but lst1.

To obtain the desired behavior, you can use a list comprehension:

lst = [ [1]*4 for n in range(3) ]

In this case, the expression is re-evaluated for each n, leading to a different list.

Solution 4:

[[1] * 4] * 3

or even:

[[1, 1, 1, 1]] * 3

Creates a list that references the internal [1,1,1,1] 3 times - not three copies of the inner list, so any time you modify the list (in any position), you'll see the change three times.

It's the same as this example:

>>> inner = [1,1,1,1]
>>> outer = [inner]*3
>>> outer
[[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]
>>> inner[0] = 5
>>> outer
[[5, 1, 1, 1], [5, 1, 1, 1], [5, 1, 1, 1]]

where it's probably a little less surprising.

Solution 5:

my_list = [[1]*4] * 3 creates one list object [1,1,1,1] in memory and copies its reference 3 times over. This is equivalent to obj = [1,1,1,1]; my_list = [obj]*3. Any modification to obj will be reflected at three places, wherever obj is referenced in the list. The right statement would be:

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

or

my_list = [[1for __ in range(4)] for _ in range(3)]

Important thing to note here is that the * operator is mostly used to create a list of literals. Although 1 is immutable, obj = [1]*4 will still create a list of 1 repeated 4 times over to form [1,1,1,1]. But if any reference to an immutable object is made, the object is overwritten with a new one.

This means if we do obj[1] = 42, then obj will become [1,42,1,1]not[42,42,42,42] as some may assume. This can also be verified:

>>>my_list = [1]*4>>>my_list
[1, 1, 1, 1]

>>>id(my_list[0])
4522139440
>>>id(my_list[1])  # Same as my_list[0]
4522139440

---

>>>my_list[1] = 42# Since my_list[1] is immutable, this operation overwrites my_list[1] with a new object changing its id.>>>my_list
[1, 42, 1, 1]

>>>id(my_list[0])
4522139440
>>>id(my_list[1])  # id changed
4522140752
>>>id(my_list[2])  # id still same as my_list[0], still referring to value `1`.
4522139440