python - 这个python函数可以向量化吗？

Question

我一直在研究这个函数，它生成我正在开发的模拟代码所需的一些参数，但在提高其性能方面遇到了瓶颈。

分析代码表明这是主要的瓶颈，因此我可以对其进行任何改进，无论多么微小都会很棒。

我想尝试向量化此函数的一部分，但不确定是否可行。

主要的挑战是存储在我的数组 params 中的参数取决于 params 的索引。我看到的唯一直接的解决方案是使用 np.ndenumerate，但这似乎很慢。

是否可以对存储在数组中的值取决于它们的存储位置的这种类型的操作进行矢量化？或者创建一个只给我数组索引的元组的生成器会更聪明/更快吗？

import numpy as np
from scipy.sparse import linalg as LA

def get_params(num_bonds, energies):
    """
    Returns the interaction parameters of different pairs of atoms.

    Parameters
    ----------
    num_bonds : ndarray, shape = (M, 20)
        Sparse array containing the number of nearest neighbor bonds for 
        different pairs of atoms (denoted by their column) and next-
        nearest neighbor bonds. Columns 0-9 contain nearest neighbors, 
        10-19 contain next-nearest neighbors

    energies : ndarray, shape = (M, )
        Energy vector corresponding to each atomic system stored in each 
        row of num_bonds.
    """

    # -- Compute the bond energies
    x = LA.lsqr(num_bonds, energies, show=False)[0]

    params = np.zeros([4, 4, 4, 4, 4, 4, 4, 4, 4])

    nn = {(0,0): x[0], (1,1): x[1], (2,2): x[2], (3,3): x[3], (0,1): x[4],
          (1,0): x[4], (0,2): x[5], (2,0): x[5], (0,3): x[6], (3,0): x[6],
          (1,2): x[7], (2,1): x[7], (1,3): x[8], (3,1): x[8], (2,3): x[9],
          (3,2): x[9]}

    nnn = {(0,0): x[10], (1,1): x[11], (2,2): x[12], (3,3): x[13], (0,1): x[14],
           (1,0): x[14], (0,2): x[15], (2,0): x[15], (0,3): x[16], (3,0): x[16],
           (1,2): x[17], (2,1): x[17], (1,3): x[18], (3,1): x[18], (2,3): x[19],
           (3,2): x[19]}

    """
    params contains the energy contribution of each site due to its
    local environment. The shape is given by the number of possible atom
    types and the number of sites in the lattice.
    """
    for (i,j,k,l,m,jj,kk,ll,mm), val in np.ndenumerate(params):

        params[i,j,k,l,m,jj,kk,ll,mm] = nn[(i,j)] + nn[(i,k)] + nn[(i,l)] + \
                                        nn[(i,m)] + nnn[(i,jj)] + \
                                        nnn[(i,kk)] + nnn[(i,ll)] + nnn[(i,mm)]

return np.ascontiguousarray(params)

Answer 1

这是使用 broadcasted 的矢量化方法总结-

# Gather the elements sorted by the keys in (row,col) order of a dense 
# 2D array for both nn and nnn
sidx0 = np.ravel_multi_index(np.array(nn.keys()).T,(4,4)).argsort()
a0 = np.array(nn.values())[sidx0].reshape(4,4)

sidx1 = np.ravel_multi_index(np.array(nnn.keys()).T,(4,4)).argsort()
a1 = np.array(nnn.values())[sidx1].reshape(4,4)

# Perform the summations keep the first axis aligned for nn and nnn parts
parte0 = a0[:,:,None,None,None] + a0[:,None,:,None,None] + \
     a0[:,None,None,:,None] + a0[:,None,None,None,:]

parte1 = a1[:,:,None,None,None] + a1[:,None,:,None,None] + \
     a1[:,None,None,:,None] + a1[:,None,None,None,:]    

# Finally add up sums from nn and nnn for final output    
out = parte0[...,None,None,None,None] + parte1[:,None,None,None,None]

运行时测试

函数定义-

def vectorized_approach(nn,nnn):
    sidx0 = np.ravel_multi_index(np.array(nn.keys()).T,(4,4)).argsort()
    a0 = np.array(nn.values())[sidx0].reshape(4,4)    
    sidx1 = np.ravel_multi_index(np.array(nnn.keys()).T,(4,4)).argsort()
    a1 = np.array(nnn.values())[sidx1].reshape(4,4)
    parte0 = a0[:,:,None,None,None] + a0[:,None,:,None,None] + \
         a0[:,None,None,:,None] + a0[:,None,None,None,:]    
    parte1 = a1[:,:,None,None,None] + a1[:,None,:,None,None] + \
         a1[:,None,None,:,None] + a1[:,None,None,None,:]
    return parte0[...,None,None,None,None] + parte1[:,None,None,None,None]

def original_approach(nn,nnn):
    params = np.zeros([4, 4, 4, 4, 4, 4, 4, 4, 4])
    for (i,j,k,l,m,jj,kk,ll,mm), val in np.ndenumerate(params):    
        params[i,j,k,l,m,jj,kk,ll,mm] = nn[(i,j)] + nn[(i,k)] + nn[(i,l)] + \
                                        nn[(i,m)] + nnn[(i,jj)] + \
                                        nnn[(i,kk)] + nnn[(i,ll)] + nnn[(i,mm)]
    return params

设置输入-

# Setup inputs
x = np.random.rand(30)
nn = {(0,0): x[0], (1,1): x[1], (2,2): x[2], (3,3): x[3], (0,1): x[4],
      (1,0): x[4], (0,2): x[5], (2,0): x[5], (0,3): x[6], (3,0): x[6],
      (1,2): x[7], (2,1): x[7], (1,3): x[8], (3,1): x[8], (2,3): x[9],
      (3,2): x[9]}

nnn = {(0,0): x[10], (1,1): x[11], (2,2): x[12], (3,3): x[13], (0,1): x[14],
       (1,0): x[14], (0,2): x[15], (2,0): x[15], (0,3): x[16], (3,0): x[16],
       (1,2): x[17], (2,1): x[17], (1,3): x[18], (3,1): x[18], (2,3): x[19],
       (3,2): x[19]}

时间 -

In [98]: np.allclose(original_approach(nn,nnn),vectorized_approach(nn,nnn))
Out[98]: True

In [99]: %timeit original_approach(nn,nnn)
1 loops, best of 3: 884 ms per loop

In [100]: %timeit vectorized_approach(nn,nnn)
1000 loops, best of 3: 708 µs per loop

欢迎使用 1000x+ 加速!

---

对于具有此类外部产品的通用数量的系统，这是一个遍历这些维度的通用解决方案 -

m,n = a0.shape # size of output array along each axis
N = 4  # Order of system
out = a0.copy()
for i in range(1,N):
    out = out[...,None] + a0.reshape((m,)+(1,)*i+(n,))

for i in range(N):
    out = out[...,None] + a1.reshape((m,)+(1,)*(i+n)+(n,))