Why is [] faster than list()?


I recently compared the processing speeds of [] and list() and was surprised to discover that [] runs more than three times faster than list(). I ran the same test with {} and dict() and the results were practically identical: [] and {} both took around 0.128sec / million cycles, while list() and dict() took roughly 0.428sec / million cycles each.

Why is this? Do [] and {} (and probably () and '', too) immediately pass back a copies of some empty stock literal while their explicitly-named counterparts (list(), dict(), tuple(), str()) fully go about creating an object, whether or not they actually have elements?

I have no idea how these two methods differ but I’d love to find out.
I couldn’t find an answer in the docs or on SO, and searching for empty brackets turned out to be more problematic than I’d expected.

I got my timing results by calling timeit.timeit("[]") and timeit.timeit("list()"), and timeit.timeit("{}") and timeit.timeit("dict()"), to compare lists and dictionaries, respectively. I’m running Python 2.7.9.

I recently discovered “Why is if True slower than if 1?” that compares the performance of if True to if 1 and seems to touch on a similar literal-versus-global scenario; perhaps it’s worth considering as well.

Asked By: Augusta



Because list is a function to convert say a string to a list object, while [] is used to create a list off the bat. Try this (might make more sense to you):

x = "wham bam"
a = list(x)
>>> a
["w", "h", "a", "m", ...]


y = ["wham bam"]
>>> y
["wham bam"]

Gives you a actual list containing whatever you put in it.

Answered By: Torxed

Because [] and {} are literal syntax. Python can create bytecode just to create the list or dictionary objects:

>>> import dis
>>> dis.dis(compile('[]', '', 'eval'))
  1           0 BUILD_LIST               0
              3 RETURN_VALUE        
>>> dis.dis(compile('{}', '', 'eval'))
  1           0 BUILD_MAP                0
              3 RETURN_VALUE        

list() and dict() are separate objects. Their names need to be resolved, the stack has to be involved to push the arguments, the frame has to be stored to retrieve later, and a call has to be made. That all takes more time.

For the empty case, that means you have at the very least a LOAD_NAME (which has to search through the global namespace as well as the builtins module) followed by a CALL_FUNCTION, which has to preserve the current frame:

>>> dis.dis(compile('list()', '', 'eval'))
  1           0 LOAD_NAME                0 (list)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        
>>> dis.dis(compile('dict()', '', 'eval'))
  1           0 LOAD_NAME                0 (dict)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        

You can time the name lookup separately with timeit:

>>> import timeit
>>> timeit.timeit('list', number=10**7)
>>> timeit.timeit('dict', number=10**7)

The time discrepancy there is probably a dictionary hash collision. Subtract those times from the times for calling those objects, and compare the result against the times for using literals:

>>> timeit.timeit('[]', number=10**7)
>>> timeit.timeit('{}', number=10**7)
>>> timeit.timeit('list()', number=10**7)
>>> timeit.timeit('dict()', number=10**7)

So having to call the object takes an additional 1.00 - 0.31 - 0.30 == 0.39 seconds per 10 million calls.

You can avoid the global lookup cost by aliasing the global names as locals (using a timeit setup, everything you bind to a name is a local):

>>> timeit.timeit('_list', '_list = list', number=10**7)
>>> timeit.timeit('_dict', '_dict = dict', number=10**7)
>>> timeit.timeit('_list()', '_list = list', number=10**7)
>>> timeit.timeit('_dict()', '_dict = dict', number=10**7)

but you never can overcome that CALL_FUNCTION cost.

Answered By: Martijn Pieters

list() requires a global lookup and a function call but [] compiles to a single instruction. See:

Python 2.7.3
>>> import dis
>>> dis.dis(lambda: list())
  1           0 LOAD_GLOBAL              0 (list)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        
>>> dis.dis(lambda: [])
  1           0 BUILD_LIST               0
              3 RETURN_VALUE        
Answered By: Dan D.

The answers here are great, to the point and fully cover this question. I’ll drop a further step down from byte-code for those interested. I’m using the most recent repo of CPython; older versions behave similar in this regard but slight changes might be in place.

Here’s a break down of the execution for each of these, BUILD_LIST for [] and CALL_FUNCTION for list().

The BUILD_LIST instruction:

You should just view the horror:

PyObject *list =  PyList_New(oparg);
if (list == NULL)
    goto error;
while (--oparg >= 0) {
    PyObject *item = POP();
    PyList_SET_ITEM(list, oparg, item);

Terribly convoluted, I know. This is how simple it is:

  • Create a new list with PyList_New (this mainly allocates the memory for a new list object), oparg signalling the number of arguments on the stack. Straight to the point.
  • Check that nothing went wrong with if (list==NULL).
  • Add any arguments (in our case this isn’t executed) located on the stack with PyList_SET_ITEM (a macro).

No wonder it is fast! It’s custom-made for creating new lists, nothing else 🙂

The CALL_FUNCTION instruction:

Here’s the first thing you see when you peek at the code handling CALL_FUNCTION:

PyObject **sp, *res;
sp = stack_pointer;
res = call_function(&sp, oparg, NULL);
stack_pointer = sp;
if (res == NULL) {
    goto error;

Looks pretty harmless, right? Well, no, unfortunately not, call_function is not a straightforward guy that will call the function immediately, it can’t. Instead, it grabs the object from the stack, grabs all arguments of the stack and then switches based on the type of the object; is it a:

We’re calling the list type, the argument passed in to call_function is PyList_Type. CPython now has to call a generic function to handle any callable objects named _PyObject_FastCallKeywords, yay more function calls.

This function again makes some checks for certain function types (which I cannot understand why) and then, after creating a dict for kwargs if required, goes on to call _PyObject_FastCallDict.

_PyObject_FastCallDict finally gets us somewhere! After performing even more checks it grabs the tp_call slot from the type of the type we’ve passed in, that is, it grabs type.tp_call. It then proceeds to create a tuple out of of the arguments passed in with _PyStack_AsTuple and, finally, a call can finally be made!

tp_call, which matches type.__call__ takes over and finally creates the list object. It calls the lists __new__ which corresponds to PyType_GenericNew and allocates memory for it with PyType_GenericAlloc: This is actually the part where it catches up with PyList_New, finally. All the previous are necessary to handle objects in a generic fashion.

In the end, type_call calls list.__init__ and initializes the list with any available arguments, then we go on a returning back the way we came. 🙂

Finally, remmeber the LOAD_NAME, that’s another guy that contributes here.

It’s easy to see that, when dealing with our input, Python generally has to jump through hoops in order to actually find out the appropriate C function to do the job. It doesn’t have the curtesy of immediately calling it because it’s dynamic, someone might mask list (and boy do many people do) and another path must be taken.

This is where list() loses much: The exploring Python needs to do to find out what the heck it should do.

Literal syntax, on the other hand, means exactly one thing; it cannot be changed and always behaves in a pre-determined way.

Footnote: All function names are subject to change from one release to the other. The point still stands and most likely will stand in any future versions, it’s the dynamic look-up that slows things down.

Why is [] faster than list()?

The biggest reason is that Python treats list() just like a user-defined function, which means you can intercept it by aliasing something else to list and do something different (like use your own subclassed list or perhaps a deque).

It immediately creates a new instance of a builtin list with [].

My explanation seeks to give you the intuition for this.


[] is commonly known as literal syntax.

In the grammar, this is referred to as a “list display”. From the docs:

A list display is a possibly empty series of expressions enclosed in
square brackets:

list_display ::=  "[" [starred_list | comprehension] "]"

A list display yields a new list object, the contents being specified
by either a list of expressions or a comprehension. When a
comma-separated list of expressions is supplied, its elements are
evaluated from left to right and placed into the list object in that
order. When a comprehension is supplied, the list is constructed from
the elements resulting from the comprehension.

In short, this means that a builtin object of type list is created.

There is no circumventing this – which means Python can do it as quickly as it may.

On the other hand, list() can be intercepted from creating a builtin list using the builtin list constructor.

For example, say we want our lists to be created noisily:

class List(list):
    def __init__(self, iterable=None):
        if iterable is None:
        print('List initialized.')

We could then intercept the name list on the module level global scope, and then when we create a list, we actually create our subtyped list:

>>> list = List
>>> a_list = list()
List initialized.
>>> type(a_list)
<class '__main__.List'>

Similarly we could remove it from the global namespace

del list

and put it in the builtin namespace:

import builtins
builtins.list = List

And now:

>>> list_0 = list()
List initialized.
>>> type(list_0)
<class '__main__.List'>

And note that the list display creates a list unconditionally:

>>> list_1 = []
>>> type(list_1)
<class 'list'>

We probably only do this temporarily, so lets undo our changes – first remove the new List object from the builtins:

>>> del builtins.list
>>> builtins.list
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: module 'builtins' has no attribute 'list'
>>> list()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'list' is not defined

Oh, no, we lost track of the original.

Not to worry, we can still get list – it’s the type of a list literal:

>>> builtins.list = type([])
>>> list()


Why is [] faster than list()?

As we’ve seen – we can overwrite list – but we can’t intercept the creation of the literal type. When we use list we have to do the lookups to see if anything is there.

Then we have to call whatever callable we have looked up. From the grammar:

A call calls a callable object (e.g., a function) with a possibly
empty series of arguments:

call                 ::=  primary "(" [argument_list [","] | comprehension] ")"

We can see that it does the same thing for any name, not just list:

>>> import dis
>>> dis.dis('list()')
  1           0 LOAD_NAME                0 (list)
              2 CALL_FUNCTION            0
              4 RETURN_VALUE
>>> dis.dis('doesnotexist()')
  1           0 LOAD_NAME                0 (doesnotexist)
              2 CALL_FUNCTION            0
              4 RETURN_VALUE

For [] there is no function call at the Python bytecode level:

>>> dis.dis('[]')
  1           0 BUILD_LIST               0
              2 RETURN_VALUE

It simply goes straight to building the list without any lookups or calls at the bytecode level.


We have demonstrated that list can be intercepted with user code using the scoping rules, and that list() looks for a callable and then calls it.

Whereas [] is a list display, or a literal, and thus avoids the name lookup and function call.