在共享内存中使用 numpy 数组进行多处理

Question

我想在共享内存中使用一个 numpy 数组来与多处理模块一起使用。困难在于像使用 numpy 数组一样使用它，而不仅仅是作为 ctypes 数组。

from multiprocessing import Process, Array
import scipy

def f(a):
    a[0] = -a[0]

if __name__ == '__main__':
    # Create the array
    N = int(10)
    unshared_arr = scipy.rand(N)
    arr = Array('d', unshared_arr)
    print "Originally, the first two elements of arr = %s"%(arr[:2])

    # Create, start, and finish the child processes
    p = Process(target=f, args=(arr,))
    p.start()
    p.join()

    # Printing out the changed values
    print "Now, the first two elements of arr = %s"%arr[:2]

这会产生如下输出：

Originally, the first two elements of arr = [0.3518653236697369, 0.517794725524976]
Now, the first two elements of arr = [-0.3518653236697369, 0.517794725524976]

可以以 ctypes 方式访问数组，例如arr[i] 有道理。但是，它不是一个numpy数组，我无法执行-1*arr或arr.sum()之类的操作。我想一个解决方案是将 ctypes 数组转换为 numpy 数组。但是（除了无法完成这项工作），我不相信它会再被分享。

对于必须成为常见问题的问题，似乎会​​有一个标准解决方案。

Answer 1

添加到@unutbu（不再可用）和@Henry Gomersall 的答案。您可以在需要时使用shared_arr.get_lock() 同步访问：

shared_arr = mp.Array(ctypes.c_double, N)
# ...
def f(i): # could be anything numpy accepts as an index such another numpy array
    with shared_arr.get_lock(): # synchronize access
        arr = np.frombuffer(shared_arr.get_obj()) # no data copying
        arr[i] = -arr[i]

示例

import ctypes
import logging
import multiprocessing as mp

from contextlib import closing

import numpy as np

info = mp.get_logger().info

def main():
    logger = mp.log_to_stderr()
    logger.setLevel(logging.INFO)

    # create shared array
    N, M = 100, 11
    shared_arr = mp.Array(ctypes.c_double, N)
    arr = tonumpyarray(shared_arr)

    # fill with random values
    arr[:] = np.random.uniform(size=N)
    arr_orig = arr.copy()

    # write to arr from different processes
    with closing(mp.Pool(initializer=init, initargs=(shared_arr,))) as p:
        # many processes access the same slice
        stop_f = N // 10
        p.map_async(f, [slice(stop_f)]*M)

        # many processes access different slices of the same array
        assert M % 2 # odd
        step = N // 10
        p.map_async(g, [slice(i, i + step) for i in range(stop_f, N, step)])
    p.join()
    assert np.allclose(((-1)**M)*tonumpyarray(shared_arr), arr_orig)

def init(shared_arr_):
    global shared_arr
    shared_arr = shared_arr_ # must be inherited, not passed as an argument

def tonumpyarray(mp_arr):
    return np.frombuffer(mp_arr.get_obj())

def f(i):
    """synchronized."""
    with shared_arr.get_lock(): # synchronize access
        g(i)

def g(i):
    """no synchronization."""
    info("start %s" % (i,))
    arr = tonumpyarray(shared_arr)
    arr[i] = -1 * arr[i]
    info("end   %s" % (i,))

if __name__ == '__main__':
    mp.freeze_support()
    main()

如果您不需要同步访问或创建自己的锁，则不需要mp.Array()。在这种情况下，您可以使用 mp.sharedctypes.RawArray。

Answer 2

虽然已经给出的答案很好，但只要满足两个条件，这个问题就有一个更简单的解决方案：

1. 您使用的是 POSIX 兼容 操作系统（例如 Linux、Mac OSX）；和
2. 您的子进程需要对共享数组的只读访问权限。

在这种情况下，您不需要显式地使变量共享，因为子进程将使用 fork 创建。分叉的孩子自动共享父母的内存空间。在 Python 多处理的上下文中，这意味着它共享所有 模块级 变量；请注意，对于您显式传递给子进程或您在 multiprocessing.Pool 上调用的函数的参数，这 不成立。

一个简单的例子：

import multiprocessing
import numpy as np

# will hold the (implicitly mem-shared) data
data_array = None

# child worker function
def job_handler(num):
    # built-in id() returns unique memory ID of a variable
    return id(data_array), np.sum(data_array)

def launch_jobs(data, num_jobs=5, num_worker=4):
    global data_array
    data_array = data

    pool = multiprocessing.Pool(num_worker)
    return pool.map(job_handler, range(num_jobs))

# create some random data and execute the child jobs
mem_ids, sumvals = zip(*launch_jobs(np.random.rand(10)))

# this will print 'True' on POSIX OS, since the data was shared
print(np.all(np.asarray(mem_ids) == id(data_array)))

Answer 3

我编写了一个小型 python 模块，它使用 POSIX 共享内存在 python 解释器之间共享 numpy 数组。也许你会发现它很方便。

https://pypi.python.org/pypi/SharedArray

它是这样工作的：

import numpy as np
import SharedArray as sa

# Create an array in shared memory
a = sa.create("test1", 10)

# Attach it as a different array. This can be done from another
# python interpreter as long as it runs on the same computer.
b = sa.attach("test1")

# See how they are actually sharing the same memory block
a[0] = 42
print(b[0])

# Destroying a does not affect b.
del a
print(b[0])

# See how "test1" is still present in shared memory even though we
# destroyed the array a.
sa.list()

# Now destroy the array "test1" from memory.
sa.delete("test1")

# The array b is not affected, but once you destroy it then the
# data are lost.
print(b[0])

Answer 4

您可以使用sharedmem 模块：https://bitbucket.org/cleemesser/numpy-sharedmem

这是您的原始代码，这次使用行为类似于 NumPy 数组的共享内存（注意最后一条调用 NumPy sum() 函数的附加语句）：

from multiprocessing import Process
import sharedmem
import scipy

def f(a):
    a[0] = -a[0]

if __name__ == '__main__':
    # Create the array
    N = int(10)
    unshared_arr = scipy.rand(N)
    arr = sharedmem.empty(N)
    arr[:] = unshared_arr.copy()
    print "Originally, the first two elements of arr = %s"%(arr[:2])

    # Create, start, and finish the child process
    p = Process(target=f, args=(arr,))
    p.start()
    p.join()

    # Print out the changed values
    print "Now, the first two elements of arr = %s"%arr[:2]

    # Perform some NumPy operation
    print arr.sum()

Answer 5

Array 对象有一个与之关联的get_obj() 方法，该方法返回呈现缓冲区接口的 ctypes 数组。我认为以下应该可行...

from multiprocessing import Process, Array
import scipy
import numpy

def f(a):
    a[0] = -a[0]

if __name__ == '__main__':
    # Create the array
    N = int(10)
    unshared_arr = scipy.rand(N)
    a = Array('d', unshared_arr)
    print "Originally, the first two elements of arr = %s"%(a[:2])

    # Create, start, and finish the child process
    p = Process(target=f, args=(a,))
    p.start()
    p.join()

    # Print out the changed values
    print "Now, the first two elements of arr = %s"%a[:2]

    b = numpy.frombuffer(a.get_obj())

    b[0] = 10.0
    print a[0]

运行时，这会打印出 a 的第一个元素现在是 10.0，显示 a 和 b 只是同一内存中的两个视图。

为了确保它仍然是多处理器安全的，我相信您将不得不使用存在于 Array 对象 a 上的 acquire 和 release 方法，以及它的内置锁来使确保所有内容都可以安全访问（尽管我不是多处理器模块方面的专家）。