I recently asked about trying to optimise a Python loop for a scientific application,并且收到我的an excellent, smart way of recoding it within NumPy which reduced execution time by a factor of around 100!在纯NumPy中重写for循环以减少执行时间
但是,B
值的计算实际上嵌套在其他几个循环中,因为它是在常规的位置网格中进行计算的。是否有类似智能的NumPy重写来缩短这个过程的时间?
我怀疑这个部分的性能增益不会很明显,并且其缺点大概是不可能向用户报告计算进度,结果不能写入输出文件直到计算结束,并且可能在一个巨大的步骤中这样做会产生内存影响?是否有可能绕过这些?你可以做
import numpy as np
import time
def reshape_vector(v):
b = np.empty((3,1))
for i in range(3):
b[i][0] = v[i]
return b
def unit_vectors(r):
return r/np.sqrt((r*r).sum(0))
def calculate_dipole(mu, r_i, mom_i):
relative = mu - r_i
r_unit = unit_vectors(relative)
A = 1e-7
num = A*(3*np.sum(mom_i*r_unit, 0)*r_unit - mom_i)
den = np.sqrt(np.sum(relative*relative, 0))**3
B = np.sum(num/den, 1)
return B
N = 20000 # number of dipoles
r_i = np.random.random((3,N)) # positions of dipoles
mom_i = np.random.random((3,N)) # moments of dipoles
a = np.random.random((3,3)) # three basis vectors for this crystal
n = [10,10,10] # points at which to evaluate sum
gamma_mu = 135.5 # a constant
t_start = time.clock()
for i in range(n[0]):
r_frac_x = np.float(i)/np.float(n[0])
r_test_x = r_frac_x * a[0]
for j in range(n[1]):
r_frac_y = np.float(j)/np.float(n[1])
r_test_y = r_frac_y * a[1]
for k in range(n[2]):
r_frac_z = np.float(k)/np.float(n[2])
r_test = r_test_x +r_test_y + r_frac_z * a[2]
r_test_fast = reshape_vector(r_test)
B = calculate_dipole(r_test_fast, r_i, mom_i)
omega = gamma_mu*np.sqrt(np.dot(B,B))
# write r_test, B and omega to a file
frac_done = np.float(i+1)/(n[0]+1)
t_elapsed = (time.clock()-t_start)
t_remain = (1-frac_done)*t_elapsed/frac_done
print frac_done*100,'% done in',t_elapsed/60.,'minutes...approximately',t_remain/60.,'minutes remaining'
我认为贾斯汀对配置文件的建议可能是明智的,但非常感谢......虽然我不确定我会使用它,但我认为试图理解这个例子可能是一种非常好的学习方式。 :) – Statto 2010-04-07 16:10:52