随机过程的循环优化

Question

由于我将这段代码调用了1000多次,有没有办法优化这段代码需要1.73秒？

def generate():
    S0    = 0
    T     = 1.
    nt    = 100000
    lbd   = 500.
    mu    = 0
    sigma = 1.

    dt = T/nt
    St  = [S0]  * nt
    sqrtdt = np.sqrt(dt)
    dBt = np.random.normal(0, sqrtdt, nt)

    for k in xrange(1, nt):
        dSt = lbd * (mu - St[k-1]) * dt + sigma * dBt[k]
        St[k] = St[k-1] + dSt
    return St

Answer 1

您可以挤出更多的工作for-loop,但同时生成所有路径(假设您有足够的内存):

import numpy as np

def generate_orig(T=1., nt=100000, lbd=500., mu=0, sigma=1., S0=0):
    dt = T/nt
    St  = [S0]  * nt
    sqrtdt = np.sqrt(dt)
    dBt = np.random.normal(0, sqrtdt, nt)

    for k in xrange(1, nt):
        dSt = lbd * (mu - St[k-1]) * dt + sigma * dBt[k]
        St[k] = St[k-1] + dSt
    return St

def generate(T=1., nt=100000, lbd=500., mu=0, sigma=1., S0=0, npaths=1):
    dt = T/nt
    St  = np.full((nt, npaths), S0)
    sqrtdt = np.sqrt(dt)
    dBt = np.random.normal(0, sqrtdt, size=(nt, npaths))

    for k in xrange(1, nt):
        dSt = lbd * (mu - St[k-1]) * dt + sigma * dBt[k]
        St[k] = St[k-1] + dSt
    return St

这是100条路径的时间基准.

In [55]: %timeit [generate_orig() for i in xrange(100)]
1 loops, best of 3: 23.6 s per loop

In [56]: %timeit generate(npaths=100)
1 loops, best of 3: 1.97 s per loop

您也可以for-loop通过使用Cython 来提高性能.