在PyPy下使用__slots__

Question

I have this simple code that helped me to measure how classes with __slots__ perform (taken from here ):

import timeit

def test_slots():
    class Obj(object):
        __slots__ = ('i', 'l')

        def __init__(self, i):
            self.i = i
            self.l = []

    for i in xrange(1000):
        Obj(i)

print timeit.Timer('test_slots()', 'from __main__ import test_slots').timeit(10000)

If I run it via python2.7 - I would get something around 6 seconds - ok, it's really faster (and also more memory-efficient) than without slots.

But, if I run the code under PyPy (using 2.2.1 - 64bit for Mac OS/X), it starts to use 100% CPU and "never" returns (waited for minutes - no result).

What is going on? Should I use __slots__ under PyPy?

Here's what happens if I pass different number to timeit() :

timeit(10) - 0.067s
timeit(100) - 0.5s
timeit(1000) - 19.5s
timeit(10000) - ? (probably more than a Game of Thrones episode)

Thanks in advance.

---

Note that the same behavior is observed if I use namedtuple s:

import collections
import timeit

def test_namedtuples():
    Obj = collections.namedtuple('Obj', 'i l')

    for i in xrange(1000):
      Obj(i, [])

print timeit.Timer('test_namedtuples()', 'from __main__ import test_namedtuples').timeit(10000)

Answer 1

In each of the 10,000 or so iterations of the timeit code, the class is recreated from scratch. Creating classes is probably not a well-optimized operation in PyPy; even worse, doing so will probably discard all of the optimizations that the JIT learned about the previous incarnation of the class. PyPy tends to be slow until the JIT has warmed up, so doing things that require it to warm up repeatedly will kill your performance.

The solution here is, of course, to simply move the class definition outside of the code being benchmarked.

Answer 2

要直接回答标题中的问题： __slots__对于PyPy的性能毫无意义（但不会影响）。