[] and {} vs list() and dict(), which is better?

Question

In terms of speed, it's no competition for empty lists/dicts:

>>> from timeit import timeit
>>> timeit("[]")
0.040084982867934334
>>> timeit("list()")
0.17704233359267718
>>> timeit("{}")
0.033620194745424214
>>> timeit("dict()")
0.1821558326547077

and for non-empty:

>>> timeit("[1,2,3]")
0.24316302770330367
>>> timeit("list((1,2,3))")
0.44744206316727286
>>> timeit("list(foo)", setup="foo=(1,2,3)")
0.446036018543964
>>> timeit("{'a':1, 'b':2, 'c':3}")
0.20868602015059423
>>> timeit("dict(a=1, b=2, c=3)")
0.47635635255323905
>>> timeit("dict(bar)", setup="bar=[('a', 1), ('b', 2), ('c', 3)]")
0.9028228448029267

Also, using the bracket notation lets you use list and dictionary comprehensions, which may be reason enough.

In my opinion [] and {} are the most pythonic and readable ways to create empty lists/dicts.

Be wary of set()'s though, for example:

this_set = {5}
some_other_set = {}

Can be confusing. The first creates a set with one element, the second creates an empty dict and not a set.

The dict literal might be a tiny bit faster as its bytecode is shorter:

In [1]: import dis
In [2]: a = lambda: {}
In [3]: b = lambda: dict()

In [4]: dis.dis(a)
  1           0 BUILD_MAP                0
              3 RETURN_VALUE

In [5]: dis.dis(b)
  1           0 LOAD_GLOBAL              0 (dict)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE

Same applies to the list vs []

Be careful list() and [] works differently:

>>> def a(p):
...     print(id(p))
... 
>>> for r in range(3):
...     a([])
... 
139969725291904
139969725291904
139969725291904
>>> for r in range(3):
...     a(list())
... 
139969725367296
139969725367552
139969725367616

list() always creates a new object on the heap, but [] can reuse memory cells in many situations.

IMHO, using list() and dict() makes your Python look like C. Ugh.

In the case of difference between [] and list(), there is a pitfall that I haven't seen anyone else point out. If you use a dictionary as a member of the list, the two will give entirely different results:

In [1]: foo_dict = {"1":"foo", "2":"bar"}

In [2]: [foo_dict]
Out [2]: [{'1': 'foo', '2': 'bar'}]

In [3]: list(foo_dict)
Out [3]: ['1', '2']