Testing for an empty dictionary in Python

[John Nagle]

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

is expensive for large dictionaries, and makes loops O(N^2).

John Nagle

[Ryan Ginstrom]

On Behalf Of John Nagle

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

is expensive for large dictionaries, and makes loops O(N^2).

I believe that the following is fairly efficient:

>>def dict_is_empty(D ):

for k in D:
return False
return True

>>dict_is_empty (dict(a=1))

False

>>dict_is_empty ({})

True

Regards,
Ryan Ginstrom

[D'Arcy J.M. Cain]

On Sun, 23 Mar 2008 08:53:02 -0700
John Nagle <na***@animats. comwrote:

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

is expensive for large dictionaries, and makes loops O(N^2).

Try this:

if dict:

It should be faster as it only checks whether or not there are
any members and does not run keys() or len() on the dictionary. Of
course, you should test it.

--
D'Arcy J.M. Cain <da***@druid.ne t | Democracy is three wolves
http://www.druid.net/darcy/ | and a sheep voting on
[phone] (DoD#0082) (eNTP) | what's for dinner.

[Brian Lane]

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John Nagle wrote:

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

is expensive for large dictionaries, and makes loops O(N^2).

John Nagle

if dict:
...

:)

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[Paddy]

On Mar 23, 3:53 pm, John Nagle <na...@animats. comwrote:

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

is expensive for large dictionaries, and makes loops O(N^2).

John Nagle

As others have stated, if <container object>: is false for built-in
container types such as dicts, lists, sets, tuples,...
Its nice to make any of your owh container types follow the same
convention too.

- Paddy.

[Paul Rubin]

John Nagle <na***@animats. comwrites:

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

I like to think len(dict) is constant time but I haven't checked the code.
Same for bool(dict) (which is what you get when you run "if dict: ...").

[Paddy]

On Mar 23, 4:14 pm, Paddy <paddy3...@goog lemail.comwrote :

On Mar 23, 3:53 pm, John Nagle <na...@animats. comwrote:

What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() 0) :

is expensive for large dictionaries, and makes loops O(N^2).

John Nagle

As others have stated, if <container object>: is false for built-in
container types such as dicts, lists, sets, tuples,...
Its nice to make any of your own container types follow the same
convention too.

- Paddy.

I missed out *empty* didn't I.
Its false for empty container types.

- Paddy.

[John Nagle]

Brian Lane wrote:

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Hash: SHA1

John Nagle wrote:

> What's the cheapest way to test for an empty dictionary in Python?

if len(dict.keys() ) :

is expensive for large dictionaries, and makes loops O(N^2).

John Nagle

if dict:

Cute.

I'd already figured out that

len(dict)

works, which is probably better than

len(dict.keys() 0

which requires creating a list.

John Nagle

[Arnaud Delobelle]

On Mar 23, 5:45*pm, Paul Rubin <http://phr...@NOSPAM.i nvalidwrote:

John Nagle <na...@animats. comwrites:

* *What's the cheapest way to test for an empty dictionary in Python?

* *if len(dict.keys() 0) :

I like to think len(dict) is constant time but I haven't checked the code.
Same for bool(dict) (which is what you get when you run "if dict: ...").

It has to be constant time as it is a lower bound for inserts (which
average to constant time).

--
Arnaud

[Steven D'Aprano]

On Sun, 23 Mar 2008 18:56:51 -0700, John Machin wrote:

>Python knows the truth value of built-in types like dicts without
actually converting them to bools, or for that matter calling __len__
or __nonzero__ on them.

What the 2.5.1 interpreter does is call PyObject_IsTrue , which checks to
see if the built_in or extension type is a mapping (not just a dict)
with a length method and if so calls it; similarly for sequences:

else if (v->ob_type->tp_as_mappin g != NULL &&
v->ob_type->tp_as_mappin g->mp_length != NULL)
res = (*v->ob_type->tp_as_mappin g->mp_length)(v );
else if (v->ob_type->tp_as_sequen ce != NULL &&
v->ob_type->tp_as_sequen ce->sq_length != NULL)
res = (*v->ob_type->tp_as_sequen ce->sq_length)(v );

Was that what you meant by "without ... calling __len__"?

What I meant was that the interpreter *didn't* do was lookup the name
'__len__' via the normal method resolution procedure. There's no need to
search the dict's __dict__ attribute looking for an attribute named
__len__, or indeed any sort of search at all. It's a direct pointer
lookup to get to mp_length.

I didn't mean to imply that Python magically knew whether a dict was
empty without actually checking to see if it were empty. Apologies for
being less than clear.

Also, since I frequently criticize others for confusing implementation
details with Python the language, I should criticize myself as well. What
I described is specific to the CPython implementation -- there's no
requirement for other Python versions to do the same thing.

--
Steven