Initializing Python dictionary using dict() vs {}

Question

I am witnessing a very weird behviour with the following Python code snippet.

list_of_dict = [{}]*2                  # create a list of 2 dict
list_of_dict[0]["add_to_dict_0"] = 1   # add a (key,val) pair to 1st dict

print list_of_dict[0]
print list_of_dict[1]

The output is:

{'add_to_dict_0': 1}

{'add_to_dict_0': 1}

Both the dictionaries got updated, although I (explicitly) updated only one. However, changing the initialization in Line 1 to the following:

list_of_dict = [dict() for _ in xrange(0,2)]

fixes the issue.

What is the reason for this behaviour? I am guessing that this is an artefact of the concept of name-object binding in Python (as opposed to variable-memory location binding in other languages like C/C++). My guess is that in the first case, both list_of_dict[0] and list_of_dict[1] are bound to the same dictionary object.

Is my suspicion correct? If yes, then why is the behaviour not replicated in the second case? What is the difference between dict_object = dict() and dict_object = {}?

Answer 1

There is no difference between dict_object = dict() and dict_object = {}

The problem here is [{}] * 2 will generate a list with 2 elements which are reference to a single object. [{} for x in range(2)] will generate a list with two elements which are reference two distinct objects.

Replacing {} with dict() will get the same result.

Answer 2

Both the dictionary in the list pointing to same object, Hence modifying one will will change others also

>>> list_of_dict = [{}]*2
>>> id(list_of_dict[0])
50030176
>>> id(list_of_dict[1])
50030176

When you created two empty dictionary by using dict(), dict() return different object every time

>>> d1 = dict()
>>> d2 = dict()
>>> id(d1)
50074656
>>> id(d2)
50074512