Changing a list derived from a reference list also changes the reference list?

[MauriceMeilleur]

a = 1
b = [(a, a), (a + 1, a + 1)]

d = 1
g = b
print('original:', b, 'new, pending:', g)
c = [0, 1]
for n in c:
    g[n] = (g[n][0] - d, g[n][1] - d)
print('original:', b, 'new:', g)

returns:

original: [(1, 1), (2, 2)] new, pending: [(1, 1), (2, 2)]
original: [(0, 0), (1, 1)] new: [(0, 0), (1, 1)]

What is going on here? Why does changing g also change b?

[gferreira]

hello @MauriceMeilleur,

assigning a list to a new variable does not create a new list, just a new reference to the same list. (this also applies to dictionaries; both are mutable object types.)

there are different ways to create a copy of a list:

A = [1, 2, 3]
B = A.copy() # or list(A) # or A[:]
B.append(4)
print(A)
print(B)

cheers!

[MauriceMeilleur]

@gferreira Thanks, Gustavo! I don't know how I've managed to go nearly four years without knowing this—I feel like I've used this method in the past to restore a reference. Lesson learned.

[frederik]

g is just a reference to the same object. You need to create a copy to get a new object with the same content.

import copy

a = 1
b = [(a, a), (a + 1, a + 1)]
print(id(b))

g = b
print(id(g))

g = copy.deepcopy(b)
print(id(g))