将sympy表达式转换为numpy数组的函数

Question

我有一个同意写的ODE系统:

from sympy.parsing.sympy_parser import parse_expr

xs = symbols('x1 x2')
ks = symbols('k1 k2')
strs = ['-k1 * x1**2 + k2 * x2', 'k1 * x1**2 - k2 * x2']
syms = [parse_expr(item) for item in strs]

我想这转换成向量量值函数,接受的x值,第k值的1D numpy的阵列的一维阵列numpy的,返回在那些点评估的方程的一个一维阵列numpy的.签名看起来像这样:

import numpy as np
x = np.array([3.5, 1.5])
k = np.array([4, 2])
xdot = my_odes(x, k)

我想要这样的东西的原因是给这个功能scipy.integrate.odeint,所以它需要快.

尝试1:潜艇

当然,我可以写一个包装器subs:

def my_odes(x, k):
    all_dict = dict(zip(xs, x))
    all_dict.update(dict(zip(ks, k)))
    return np.array([sym.subs(all_dict) for sym in syms])

但这是超级慢的,特别是对于我的真实系统,它更大,并且运行了很多次.我需要将此操作编译为C代码.

尝试2:theano

我可以通过sympy与theano的整合来接近:

from sympy.printing.theanocode import theano_function

f = theano_function(xs + ks, syms)

def my_odes(x, k):
    return np.array(f(*np.concatenate([x,k]))))

这会编译每个表达式,但输入和输出的所有打包和解包都会减慢它的速度.返回的函数theano_function接受numpy数组作为参数,但每个符号需要一个数组,而不是每个符号一个元素.这是相同的行为functify,并ufunctify为好.我不需要广播行为; 我需要它将数组的每个元素解释为一个不同的符号.

尝试3:DeferredVector

如果我使用DeferredVector我可以创建一个接受numpy数组的函数,但我无法将其编译为C代码或返回一个numpy数组而不自行打包它.

import numpy as np
import sympy as sp
from sympy import DeferredVector

x = sp.DeferredVector('x')
k =  sp.DeferredVector('k')
deferred_syms = [s.subs({'x1':x[0], 'x2':x[1], 'k1':k[0], 'k2':k[1]}) for s in syms]
f = [lambdify([x,k], s) for s in deferred_syms]

def my_odes(x, k):
    return np.array([f_i(x, k) for f_i in f])

使用DeferredVector我不需要解压缩输入,但我仍然需要打包输出.此外,我可以使用lambdify,但版本ufuncify和theano_function版本消亡,因此没有生成快速C代码.

from sympy.utilities.autowrap import ufuncify
f = [ufuncify([x,k], s) for s in deferred_syms] # error

from sympy.printing.theanocode import theano_function
f = theano_function([x,k], deferred_syms) # error

Answer 1

您可以使用sympy功能lambdify.例如,

from sympy import symbols, lambdify
from sympy.parsing.sympy_parser import parse_expr
import numpy as np

xs = symbols('x1 x2')
ks = symbols('k1 k2')
strs = ['-k1 * x1**2 + k2 * x2', 'k1 * x1**2 - k2 * x2']
syms = [parse_expr(item) for item in strs]

# Convert each expression in syms to a function with signature f(x1, x2, k1, k2):
funcs = [lambdify(xs + ks, f) for f in syms]


# This is not exactly the same as the `my_odes` in the question.
# `t` is included so this can be used with `scipy.integrate.odeint`.
# The value returned by `sym.subs` is wrapped in a call to `float`
# to ensure that the function returns python floats and not sympy Floats.
def my_odes(x, t, k):
    all_dict = dict(zip(xs, x))
    all_dict.update(dict(zip(ks, k)))
    return np.array([float(sym.subs(all_dict)) for sym in syms])

def lambdified_odes(x, t, k):
    x1, x2 = x
    k1, k2 = k
    xdot = [f(x1, x2, k1, k2) for f in funcs]
    return xdot


if __name__ == "__main__":
    from scipy.integrate import odeint

    k1 = 0.5
    k2 = 1.0
    init = [1.0, 0.0]
    t = np.linspace(0, 1, 6)
    sola = odeint(lambdified_odes, init, t, args=((k1, k2),))
    solb = odeint(my_odes, init, t, args=((k1, k2),))
    print(np.allclose(sola, solb))

True 在脚本运行时打印.

速度要快得多(注意时序结果单位的变化):

In [79]: t = np.linspace(0, 10, 1001)

In [80]: %timeit sol = odeint(my_odes, init, t, args=((k1, k2),))
1 loops, best of 3: 239 ms per loop

In [81]: %timeit sol = odeint(lambdified_odes, init, t, args=((k1, k2),))
1000 loops, best of 3: 610 µs per loop