References

Strings are immutable so Python saves memory and points to the same memory location:

stringreference

Lists are mutable so they refer to different location, but the immutable elements inside list will indeed point to the same locations.

listreference

Aliasing

Since variables refer to objects, if we assign one variable to another, both variables refer to the same object:

a = [81, 82, 83]
b = a
print(a is b)

>>> True

listreference

Because the same list has two different names, a and b, we say that it is aliased. Changes made with one alias affect the other.

Although this behavior can be useful, it is sometimes unexpected or undesirable. In general, it is safer to avoid aliasing when you are working with mutable objects. Of course, for immutable objects, there’s no problem. That’s why Python is free to alias strings and integers when it sees an opportunity to economize.

Cloning

a = [1,2,3]
b = a 
a is b
>>> True

b = a[:]
a == b
>>> False

Repeating

origlist = [45, 76, 34, 55]
newlist = [origlist] * 3        # keep reference as the original list
anotherlist = origlist * 3        # creates new list based on the original list

print(newlist)
print(anotherlist)

origlist[1] = 99

print(newlist)                 # Note the change
print(anotherlist)             # Note the lack of change

>>> [[45, 76, 34, 55], [45, 76, 34, 55], [45, 76, 34, 55]]
>>> [45, 76, 34, 55, 45, 76, 34, 55, 45, 76, 34, 55]
>>> [[45, 99, 34, 55], [45, 99, 34, 55], [45, 99, 34, 55]]
>>> [45, 76, 34, 55, 45, 76, 34, 55, 45, 76, 34, 55]

Append versus Concatenate

With append, the original list is simply modified. On the other hand, with concatenation, an entirely new list is created, which makes sense considering its behavior if assigned to a different variable. Append mutates and does not return anything. Cannot be used for reassignment.

origlist = [1,2,3,4]
origlist = origlist + ["cat"]

origlist.append(3)