`
heipark
  • 浏览: 2094567 次
  • 性别: Icon_minigender_1
  • 来自: 北京
社区版块
存档分类
最新评论

Shared counter with Python's multiprocessing(转)

 
阅读更多

One of the methods of exchanging data between processes with the multiprocessing module is directly shared memory via multiprocessing.Value. As any method that's very general, it can sometimes be tricky to use. I've seen a variation of this question asked a couple of times on StackOverflow:

 I have some processes that do work, and I want them to increment some shared counter because [... some irrelevant reason ...] - how can this be done?

The wrong way

And surprisingly enough, some answers given to this question are wrong, since they usemultiprocessing.Value incorrectly, as follows:

import time
from multiprocessing import Process, Value

def func(val):
    for i in range(50):
        time.sleep(0.01)
        val.value += 1

if __name__ == '__main__':
    v = Value('i', 0)
    procs = [Process(target=func, args=(v,)) for i in range(10)]

    for p in procs: p.start()
    for p in procs: p.join()

    print v.value

This code is a demonstration of the problem, distilling only the usage of the shared counter. A "pool" of 10 processes is created to run the func function. All processes share a Value and increment it 50 times. You would expect this code to eventually print 500, but in all likeness it won't. Here's some output taken from 10 runs of that code:

> for i in {1..10}; do python sync_nolock_wrong.py; done
435
464
484
448
491
481
490
471
497
494

Why does this happen?

I must admit that the documentation of multiprocessing.Value can be a bit confusing here, especially for beginners. It states that by default, a lock is created to synchronize access to the value, so one may be falsely led to believe that it would be OK to modify this value in any way imaginable from multiple processes. But it's not.

Explanation - the default locking done by Value

This section is advanced and isn't strictly required for the overall flow of the post. If you just want to understand how to synchronize the counter correctly, feel free to skip it.

The locking done by multiprocessing.Value is very fine-grained. Value is a wrapper around a ctypesobject, which has an underlying value attribute representing the actual object in memory. All Value does is ensure that only a single process or thread may read or write this value attribute simultaneously. This is important, since (for some types, on some architectures) writes and reads may not be atomic. I.e. to actually fill up the object's memory, the CPU may need several instructions, and another process reading the same (shared) memory at the same time could see some intermediate, invalid state. The built-in lock of Value prevents this from happening.

However, when we do this:

val.value +=1

What Python actually performs is the following (disassembled bytecode with the dis module). I've annotated the locking done by Value in #<-- comments:

 0 LOAD_FAST                0 (val)
 3 DUP_TOP
                                     #<--- Value lock acquired
 4 LOAD_ATTR                0 (value)
                                     #<--- Value lock released
 7 LOAD_CONST               1 (1)
10 INPLACE_ADD
11 ROT_TWO
                                     #<--- Value lock acquired
12 STORE_ATTR               0 (value)
                                     #<--- Value lock released

So it's obvious that while process #1 is now at instruction 7 (LOAD_CONST), nothing prevents process #2 from also loading the (old) value attribute and be on instruction 7 too. Both processes will proceed incrementing their private copy and writing it back. The result: the actual value got incremented only once, not twice.

The right way

Fortunately, this problem is very easy to fix. A separate Lock is needed to guarantee the atomicity of modifications to the Value:

import time
from multiprocessing import Process, Value, Lock

def func(val, lock):
    for i in range(50):
        time.sleep(0.01)
        with lock:
            val.value += 1

if __name__ == '__main__':
    v = Value('i', 0)
    lock = Lock()
    procs = [Process(target=func, args=(v, lock)) for i in range(10)]

    for p in procs: p.start()
    for p in procs: p.join()

    print v.value

Now we get the expected result:

> for i in {1..10}; do python sync_lock_right.py; done
500
500
500
500
500
500
500
500
500
500

A value and a lock may appear like too much baggage to carry around at all times. So, we can create a simple "synchronized shared counter" object to encapsulate this functionality:

import time
from multiprocessing import Process, Value, Lock

class Counter(object):
    def __init__(self, initval=0):
        self.val = Value('i', initval)
        self.lock = Lock()

    def increment(self):
        with self.lock:
            self.val.value += 1

    def value(self):
        with self.lock:
            return self.val.value

def func(counter):
    for i in range(50):
        time.sleep(0.01)
        counter.increment()

if __name__ == '__main__':
    counter = Counter(0)
    procs = [Process(target=func, args=(counter,)) for i in range(10)]

    for p in procs: p.start()
    for p in procs: p.join()

    print counter.value()

Bonus: since we've now placed a more coarse-grained lock on the modification of the value, we may throw away Value with its fine-grained lock altogether, and just use multiprocessing.RawValue, that simply wraps a shared object without any locking.

 

转自:http://eli.thegreenplace.net/2012/01/04/shared-counter-with-pythons-multiprocessing

 

--end

分享到:
评论

相关推荐

Global site tag (gtag.js) - Google Analytics