Почему LEN (<a list object>) so slow?

Question

I'm running the following code in an ipython session:

# This call is slow, but that is expected. (It loads 3 GB of data.) 
In [3]: arc, arc_sub, upls, go = foo_mod.ready_set() 

# This call is also slow, as `upls` is huge. 
In [4]: upls = list(upls) 

# This call is slow in meatspace, but `%timeit` doesn't notice! 
In [5]: %timeit -n1 -r1 len(upls) 
1 loops, best of 1: 954 ns per loop 

%timeit is straight-up lying here. With or without the %timeit, the command takes upwards of 10s to actually run. Only the first time, however; subsequent calls to len are quick.

Even time.time() sings a similar tune:

In [5]: import time 

In [6]: s = time.time(); len_ = len(upls); e = time.time() 

In [7]: e - s 
Out[7]: 7.104873657226562e-05 

But it took seconds in the real world for In [6] to actually complete. I just don't seem to be able to capture where the actual time is being spent!

There's nothing terribly unusual about the list, aside from it's huge: it's a real list; it holds ~¼ billion bson.ObjectId objects. (Prior to the list() call, it's a set object; that call is also slow, but that makes sense; list(<set instance>) is O(n), and my set is huge.)

Edit re GC

If I run gc.set_debug(gc.DEBUG_STATS) just prior to ready_set, which itself is a slow call, I see tons of GC cycles. This is expected. gen3 grows:

gc: objects in each generation: 702 701 3289802 
gc: done, 0.0000s elapsed. 
gc: collecting generation 0... 
gc: objects in each generation: 702 1402 3289802 
gc: done, 0.0000s elapsed. 
gc: collecting generation 0... 
gc: objects in each generation: 702 2103 3289802 

Unfortunately the console outputs make this runtime of this impossibly slow. If I instead delay the gc.set_debug call until just after ready_set, I don't see any GC cycles, but gc.get_count() claims the generations are tiny:

In [6]: gc.get_count() 
Out[6]: (43, 1, 193) 

In [7]: len(upls) 
Out[7]: 125636395 

(but why/how is get_count less objects than what's in the list?; they're definitely all unique, since they just went through a set…) The fact that involving gc in the code makes len speedy leads me to believe I'm paused for a collect-the-world.

(Versions, just in case:

Python 2.7.6 (default, Mar 22 2014, 22:59:56) 
IPython 3.2.0 -- An enhanced Interactive Python. 

)

Answer 1

I will summarize the comments to your question to the answer.

As everyone said (and you pointed it out), Python's list object knows its size and it returns just the stored number:

static Py_ssize_t 
list_length(PyListObject *a) 
{ 
    return Py_SIZE(a); 
} 

Где Py_SIZEis defined:

Py_SIZE (о)

Этот макрос используется для доступа к члену ob_size объекта Python Он расширяется.: (((PyVarObject*)(o))->ob_size)

Поэтому я могу c onclude он не должен делать никаких вычислений. Единственный подозреваемый - это объект, который вы пытаетесь преобразовать в список. Но если вы клянетесь, это действительно list без каких-либо поддельных объектов, имитирующих его метод с некоторыми ленивыми вычислениями - это не так.

Так что я предполагаю, что все методы timeit действительно показывают точное время, проведенное для вызова функции len.

И единственный процесс, связанный с потерей времени, является .. Мусороуборочный комбайн. В конце ваших измерений он обнаруживает, что никто не использует такую ​​большую часть данных и начинает освобождать память. Конечно, требуется несколько секунд.