有没有一种方法可以对带有参数的 sympy 函数进行绘图和数值积分？

Question

我正在 Jupyter 笔记本中进行一些符号计算，并希望在继续其他符号计算之前绘制函数并计算数值积分。

我尝试对函数进行羔羊化，但如果我实际重新输入函数定义，我只能得到绘图和积分。有没有办法直接用符号形式进行这些操作？如果没有，有没有办法正确地对具有固定输入参数 a 和 b 的函数进行 lambda 化？

from sympy import *
from sympy.utilities.lambdify import lambdify 
import matplotlib.pyplot as plt
import numpy as np
import scipy.integrate as int
x,a,b = symbols('x a b')
u = log(exp(-b*x**2/2) + 1/a)
t = np.linspace(-5,5,100)
# Attempt to lambdify u:
f = (lambda x,a=1,b=3: lambdify((x,a,b),u, modules = ['numpy','sympy']))
y = [f(t[i]) for i in range(len(t))]
plt.plot(t,y)
int.quad(f,-5,5)
# Gives error:
Traceback (most recent call last):

  File "C:\Users\...\AppData\Local\Continuum\anaconda3\lib\site-packages\IPython\core\interactiveshell.py", line 3296, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)

  File "<ipython-input-86-75a94a575357>", line 2, in <module>
    y = [f(t[i]) for i in range(len(t))]

  File "<ipython-input-86-75a94a575357>", line 2, in <listcomp>
    y = [f(t[i]) for i in range(len(t))]

  File "<ipython-input-86-75a94a575357>", line 1, in <lambda>
    f = (lambda x,a=1,b=3: lambdify((x,a,b),u, modules = ['numpy','sympy']))

  File "C:\Users\...\AppData\Local\Continuum\anaconda3\lib\site-packages\sympy\utilities\lambdify.py", line 476, in lambdify
    c = compile(funcstr, filename, 'exec')

  File "<lambdifygenerated-10>", line 1
    def _lambdifygenerated(-5.0, 1, 3):
                           ^
SyntaxError: invalid syntax

# But explicitly typing u works:
f = (lambda x, a=1, b=3: log(a + exp(b*x**2/2)))
y = [f(t[i]) for i in range(len(t))]
plt.plot(t,y)
int.quad(f,-5,5)
(126.10732269388095,1.0767576643095748𝑒−07)

Answer 1

Lambdify 创建一个函数：您调用它一次来获取该函数，然后多次调用 that 函数。你可以这样做：

In [11]: a = Symbol('a')                                                                                                          

In [12]: b = Symbol('b')                                                                                                          

In [13]: u = log(exp(-b*x**2/2) + 1/a)                                                                                            

In [14]: flambda = lambdify((x,a,b),u, modules = ['numpy','sympy'])                                                               

In [16]: flambda(1, 1, 3)                                                                                                         
Out[16]: 0.20141327798275246

In [17]: f = lambda x,a=1,b=3: flambda(x, a, b)                                                                                   

In [18]: f(1)                                                                                                                     
Out[18]: 0.20141327798275246