scipy.integrate.ode.set_solout 工作吗? [英] Does scipy.integrate.ode.set_solout work?

问题描述

集成例程的 scipy.integrate.ode 接口提供了一种在任何步骤违反约束时停止集成的方法，set_solout.但是，即使在最简单的示例中，我也无法使用此方法.这是一次尝试:

The scipy.integrate.ode interface to integration routines provides a method for stopping the integration if a constraint is violated at any step, set_solout. However, I cannot get this method to work, even in the simplest examples. Here's one attempt:

import numpy as np
from scipy.integrate import ode

def f(t, y):
    """Exponential decay."""
    return -y

def solout(t, y):
    if y[0] < 0.5:
        return -1
    else:
        return 0

y_initial = 1
t_initial = 0

r = ode(f).set_integrator('dopri5') # Integrator that supports solout
r.set_initial_value(y_initial, t_initial)
r.set_solout(solout)

# Integrate until t = 5, but stop when solout constraint violated
r.integrate(5)

# The time when solout should have terminated integration:
intersection_time = np.log(2)

当 t = log(2) = 0.693... 时，集成应该被 solout 停止，但反而愉快地继续到 t = 5，当 >y = 0.007.

The integration should have been stopped by solout when t = log(2) = 0.693..., but instead happily continues until t = 5, when y = 0.007.

这是 scipy 中的错误，还是我没有正确使用 set_solout?

Is this a bug in scipy, or am I not using set_solout correctly?

推荐答案

事实证明你需要在调用 set_initial_value 之前调用 set_solout .(我通过研究 scipy 测试套件中的 set_solout 测试.)因此，在我的问题代码中颠倒两次调用的顺序会产生正确的结果.

It turns out you need to call set_solout before calling set_initial_value. (I figured this out by studying the set_solout tests in the scipy test suite.) So, reversing the order of the two calls in my question code produces the correct result.

即使这种行为是正确的，也应该在 set_solout 的文档中提及.我已经在 GitHub 上发布了 SciPy 的问题.

Even if this behavior is correct, it ought to be mentioned in the documentation for set_solout. I've posted an issue with SciPy on GitHub.

更新:此问题已在 SciPy 0.17.0 中修复；即使在set_initial_value之后调用set_solout也会起作用，并且问题代码会产生正确的结果.

UPDATE: This issue is fixed in SciPy 0.17.0; set_solout will work even if called after set_initial_value, and the question code will produce the correct result.

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