[Numpy-discussion] Numpy's policy for releasing memory
Tue Nov 13 06:31:34 CST 2012
I've been using psutil, pmap (linux command), and resource in various
capacities, all on cpython. When I wasn't seeing memory freed when I
expected, I got to wondering if maybe numpy was maintaining pools of
buffers for reuse or something like that. It sounds like that's not the
case, though, so I'm following up other possibilities.
On Tue, Nov 13, 2012 at 1:08 PM, Olivier Delalleau <firstname.lastname@example.org> wrote:
> How are you monitoring memory usage?
> Personally I've been using psutil and it seems to work well, although I've
> used it only on Windows and not in applications with large numpy arrays, so
> I can't tell whether it would work you.
> Also, keep in mind that:
> - The "auto-delete object when it goes out of scope" behavior is specific
> to the CPython implementation and not part of the Python standard, so if
> you're actually using a different implementation you may see a different
> - CPython deals with small objects in a special way, not actually
> releasing allocated memory. For more info:
> -=- Olivier
> 2012/11/13 Austin Bingham <email@example.com>
>> OK, if numpy is just subject to Python's behavior then what I'm seeing
>> must be due to the vagaries of Python. I've noticed that things like
>> removing a particular line of code or reordering seemingly unrelated calls
>> (unrelated to the memory issue, that is) can affect when memory is reported
>> as free. I'll just assume that everything is in order and carry on. Thanks!
>> On Tue, Nov 13, 2012 at 9:41 AM, Nathaniel Smith <firstname.lastname@example.org> wrote:
>>> On Tue, Nov 13, 2012 at 8:26 AM, Austin Bingham
>>> <email@example.com> wrote:
>>> > I'm trying to understand how numpy decides when to release memory and
>>> > whether it's possible to exert any control over that. The situation is
>>> > I'm profiling memory usage on a system in which a great deal of the
>>> > memory is tied up in ndarrays. Since numpy manages ndarray memory on
>>> its own
>>> > (i.e. without the python gc, or so it seems), I'm finding that I can't
>>> > much to convince numpy to release memory when things get tight. For
>>> > object, for example, I can explicitly run gc.collect().
>>> > So, in an effort to at least understand the system better, can anyone
>>> > me how/when numpy decides to release memory? And is there any way via
>>> > the Python or C-API to explicitly request release? Thanks.
>>> Numpy array memory is released when the corresponding Python objects
>>> are deleted, so it exactly follows Python's rules. You can't
>>> explicitly request release, because by definition, if memory is not
>>> released, then it means that it's still accessible somehow, so
>>> releasing it could create segfaults. Perhaps you have stray references
>>> sitting around that you have forgotten to clear -- that's a common
>>> cause of memory leaks in Python. gc.get_referrers() can be useful to
>>> debug such things.
>>> Some things to note:
>>> - Numpy uses malloc() instead of going through the Python low-level
>>> memory allocation layer (which itself is a wrapper around malloc with
>>> various optimizations for small objects). This is really only relevant
>>> because it might create some artifacts depending on how your memory
>>> profiler gathers data.
>>> - gc.collect() doesn't do that much in Python... it only matters if
>>> you have circular references. Mostly Python releases the memory
>>> associated with objects as soon as the object becomes unreferenced.
>>> You could try avoiding circular references, and then gc.collect()
>>> won't even do anything.
>>> - If you have multiple views of the same memory in numpy, then they
>>> share the same underlying memory, so that memory won't be released
>>> until all of the views objects are released. (The one thing to watch
>>> out for is you can do something like 'huge_array = np.zeros((2,
>>> 10000000)); tiny_array = a[:, 100]' and now since tiny_array is a view
>>> onto huge_array, so long as a reference to tiny_array exists the full
>>> big memory allocation will remain.)
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