python-如何求和与平方和与内存限制？

这是此问题的后续措施：How to do a sum of sums of the square of sum of sums?我在哪里寻求使用einsum的帮助(以实现更快的速度)并获得了一个很好的答案.我也有suggestion使用numba.我已经尝试了两者,似乎在达到...

这是此问题的后续措施：

How to do a sum of sums of the square of sum of sums?

我在哪里寻求使用einsum的帮助(以实现更快的速度)并获得了一个很好的答案.

我也有suggestion使用numba.我已经尝试了两者,似乎在达到某个点之后,numba的速度提高要好得多.

那么,如何在不遇到内存问题的情况下加快速度呢？

解决方法:

下面的解决方案提出了3种不同的方法来进行简单的和,以及4种不同的方法来进行平方和.

总和3种方法-循环,JIT循环,einsum(没有遇到内存问题)

求和平方和的4种方法-循环,JIT循环,扩展einsum,中间einsum

在这里,前三个不会遇到内存问题,而for循环和扩展的einsum都会遇到速度问题.这使JIT解决方案看起来是最好的.

import numpy as np
import time
from numba import jit

def fun1(Fu, Fv, Fx, Fy, P, B):
    Nu = Fu.shape[0]
    Nv = Fv.shape[0]
    Nx = Fx.shape[0]
    Ny = Fy.shape[0]
    Nk = Fu.shape[1]
    Nl = Fv.shape[1]
    I1 = np.zeros([Nu, Nv])
    for iu in range(Nu):
        for iv in range(Nv):
            for ix in range(Nx):
                for iy in range(Ny):
                    S = 0.
                    for ik in range(Nk):
                        for il in range(Nl):
                            S += Fu[iu,ik]*Fv[iv,il]*Fx[ix,ik]*Fy[iy,il]*P[ix,iy]*B[ik,il]
                    I1[iu, iv] += S
    return I1

def fun2(Fu, Fv, Fx, Fy, P, B):
    Nu = Fu.shape[0]
    Nv = Fv.shape[0]
    Nx = Fx.shape[0]
    Ny = Fy.shape[0]
    Nk = Fu.shape[1]
    Nl = Fv.shape[1]
    I2 = np.zeros([Nu, Nv])
    for iu in range(Nu):
        for iv in range(Nv):
            for ix in range(Nx):
                for iy in range(Ny):
                    S = 0.
                    for ik in range(Nk):
                        for il in range(Nl):
                            S += Fu[iu,ik]*Fv[iv,il]*Fx[ix,ik]*Fy[iy,il]*P[ix,iy]*B[ik,il]
                    I2[iu, iv] += S**2.
    return I2

if __name__ == '__main__':

    Nx = 30
    Ny = 40
    Nk = 50
    Nl = 60
    Nu = 70
    Nv = 8
    Fx = np.random.rand(Nx, Nk)
    Fy = np.random.rand(Ny, Nl)
    Fu = np.random.rand(Nu, Nk)
    Fv = np.random.rand(Nv, Nl)
    P = np.random.rand(Nx, Ny)
    B = np.random.rand(Nk, Nl)
    fjit1 = jit(fun1)
    fjit2 = jit(fun2)

    # For loop - becomes too slow so commented out
    # t = time.time()
    # I1 = fun1(Fu, Fv, Fx, Fy, P, B)
    # print 'fun1    :', time.time() - t

    # JIT compiled for loop - After a certain point beats einsum
    t = time.time()
    I1jit = fjit1(Fu, Fv, Fx, Fy, P, B)
    print 'jit1    :', time.time() - t

    # einsum great solution when no squaring is needed
    t = time.time()
    I1_ = np.einsum('uk, vl, xk, yl, xy, kl->uv', Fu, Fv, Fx, Fy, P, B)
    print '1 einsum:', time.time() - t

    # For loop - becomes too slow so commented out
    # t = time.time()
    # I2 = fun2(Fu, Fv, Fx, Fy, P, B)
    # print 'fun2    :', time.time() - t

    # JIT compiled for loop - After a certain point beats einsum
    t = time.time()
    I2jit = fjit2(Fu, Fv, Fx, Fy, P, B)
    print 'jit2    :', time.time() - t

    # Expanded einsum - As the size increases becomes very very slow
    # t = time.time()
    # I2_ = np.einsum('uk,vl,xk,yl,um,vn,xm,yn,kl,mn,xy->uv', Fu,Fv,Fx,Fy,Fu,Fv,Fx,Fy,B,B,P**2)
    # print '2 einsum:', time.time() - t

    # Intermediate einsum - As the sizes increase memory can become an issue
    t = time.time()
    temp = np.einsum('uk, vl, xk, yl, xy, kl->uvxy', Fu, Fv, Fx, Fy, P, B)
    I2__ = np.einsum('uvxy->uv', np.square(temp))
    print '2 einsum:', time.time() - t

    # print 'I1 == I1_   :', np.allclose(I1, I1_)
    print 'I1_ == Ijit1_:', np.allclose(I1_, I1jit)
    # print 'I2 == I2_   :', np.allclose(I2, I2_)
    print 'I2_ == Ijit2_:', np.allclose(I2__, I2jit)

评论：
请随时编辑/改善此答案.如果有人对并行化提出任何建议,那就太好了.