Let's say I have the following code:
class Foo:
pass
foo_list = [Foo() for _ in range(10)]
How can I proceed to create a copy of foo_list without references to both the list and the objects contained therein being passed to the next list?
What you're doing is just a shallow copy shallow) .
DOCS :
The difference between shallow and deep copying is only relevant for compound objects (objects that contain other objects, like lists or class instances)
Translation:
The difference between shallow copy and deep copy is only relevant for compound objects, * is your case , (objects that contain others objects, such as lists or instances of classes)
import copy
y = [[1,2,3], [4,5,6]]
x = copy.deepcopy(y)
x[0].append(10)
print(y) # [[1, 2, 3], [4, 5, 6]]
print(x) # [[1, 2, 3, 10], [4, 5, 6]]
print(id(y[0])) # 140313159068680
print(id(x[0])) # 140313158999816
x and y , and everything that belongs to them, are 'considered' different / independent objects, references to their internal objects are different and so it is possible to operate on each one without affecting the other contained in the other list.
y = [[1,2,3], [4,5,6]]
x = y[:]
x[0].append(10)
print(y) # [[1, 2, 3, 10], [4, 5, 6]]
print(x) # [[1, 2, 3, 10], [4, 5, 6]]
print(id(y[0])) # 139853165128712
print(id(x[0])) # 139853165128712
Here, the internal objects of x and y have exactly the same references, so they literally share the same objects / values.
Using your example to demonstrate the differences:
from copy import deepcopy
class Foo: pass
foo = Foo()
foo.bar = 10
shallow_copies = [foo for _ in range(10)] # todos os foo partilham as mesmas referencias internas
deep_copies = [deepcopy(foo) for _ in range(10)] # copias independentes, referencias diferentes
shallow_copies[0].bar = 100 # mudar valor da propriedade do primeiro foo
deep_copies[0].bar = 100 # mudar valor da propriedade do primeiro foo
print([f.bar for f in shallow_copies]) # [100, 100, 100, 100, 100, 100, 100, 100, 100, 100]
print([f.bar for f in deep_copies]) # [100, 10, 10, 10, 10, 10, 10, 10, 10, 10]
print(all(f.bar is shallow_copies[0].bar for f in shallow_copies[1:])) # True , todos os foo tem bar com a mesma ref
print(all(f.bar is deep_copies[0].bar for f in deep_copies[1:])) # False