相关疑难解决方法(0)

假设我有一个大内存numpy数组,我有一个函数func,它接受这个巨大的数组作为输入(连同一些其他参数).func具有不同参数可以并行运行.例如:

def func(arr, param):
    # do stuff to arr, param

# build array arr

pool = Pool(processes = 6)
results = [pool.apply_async(func, [arr, param]) for param in all_params]
output = [res.get() for res in results]

如果我使用多处理库,那么这个巨型数组将被多次复制到不同的进程中.

有没有办法让不同的进程共享同一个数组？此数组对象是只读的,永远不会被修改.

更复杂的是,如果arr不是一个数组,而是一个任意的python对象,有没有办法分享它？

[EDITED]

我读了答案,但我仍然有点困惑.由于fork()是copy-on-write,因此在python多处理库中生成新进程时不应调用任何额外的成本.但是下面的代码表明存在巨大的开销:

from multiprocessing import Pool, Manager
import numpy as np; 
import time

def f(arr):
    return len(arr)

t = time.time()
arr = np.arange(10000000)
print "construct array = ", time.time() - t;


pool = Pool(processes = 6)

t = …

我想在共享内存中使用numpy数组与多处理模块一起使用.困难是使用它像一个numpy数组,而不仅仅是一个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 = …

我有一个关于如何尽可能快地计算numpy距离的问题,

def getR1(VVm,VVs,HHm,HHs):
    t0=time.time()
    R=VVs.flatten()[numpy.newaxis,:]-VVm.flatten()[:,numpy.newaxis]
    R*=R
    R1=HHs.flatten()[numpy.newaxis,:]-HHm.flatten()[:,numpy.newaxis]
    R1*=R1
    R+=R1
    del R1
    print "R1\t",time.time()-t0, R.shape, #11.7576191425 (108225, 10500) 
    print numpy.max(R) #4176.26290975
    # uses 17.5Gb ram
    return R


def getR2(VVm,VVs,HHm,HHs):
    t0=time.time()
    precomputed_flat = numpy.column_stack((VVs.flatten(), HHs.flatten()))
    measured_flat = numpy.column_stack((VVm.flatten(), HHm.flatten()))
    deltas = precomputed_flat[None,:,:] - measured_flat[:, None, :]
    #print time.time()-t0, deltas.shape # 5.861109972 (108225, 10500, 2)
    R = numpy.einsum('ijk,ijk->ij', deltas, deltas)
    print "R2\t",time.time()-t0,R.shape, #14.5291359425 (108225, 10500)
    print numpy.max(R) #4176.26290975
    # uses 26Gb ram
    return R


def getR3(VVm,VVs,HHm,HHs):
    from numpy.core.umath_tests import inner1d
    t0=time.time()
    precomputed_flat = …