1/cosh的Sympy符号FFT（x）

Question

我正在尝试通过通过错误地计算出$ 1/cosh（x）$的符号FFT，

# %%
import sympy as sp 
import numpy as np 
import sympy.abc as spa

g = sp.fourier_transform(1/sp.cosh(spa.x), spa.x, spa.k)
print(g)
sp.plot(g)

但请继续遇到

> --------------------------------------------------------------------------- TypeError                                 Traceback (most recent call
> last)
> ~\AppData\Local\Programs\Python\Python38-32\lib\site-packages\sympy\plotting\experimental_lambdify.py
> in __call__(self, args)
>     175             #The result can be sympy.Float. Hence wrap it with complex type.
> --> 176             result = complex(self.lambda_func(args))
>     177             if abs(result.imag) > 1e-7 * abs(result):
> 
> ~\AppData\Local\Programs\Python\Python38-32\lib\site-packages\sympy\plotting\experimental_lambdify.py
> in __call__(self, *args, **kwargs)
>     271     def __call__(self, *args, **kwargs):
> --> 272         return self.lambda_func(*args, **kwargs)
>     273 
> 
> <string> in <lambda>(x0)

任何想法，这意味着什么？

Answer 1

在isympy会话中：

In [40]: g = fourier_transform(1/cosh(x),x,k)

In [41]: g
Out[41]: 
                ⎛   1         ⎞
FourierTransform⎜───────, x, k⎟
                ⎝cosh(x)      ⎠

In [42]: plot(g)
/usr/local/lib/python3.8/dist-packages/sympy/plotting/experimental_lambdify.py:188: UserWarning: The evaluation of the expression is problematic. We are trying a failback method that may still work. Please report this as a bug.
  return self.__call__(args)
-----------------------------------------------------------------------
TypeError                             Traceback (most recent call last)
File /usr/local/lib/python3.8/dist-packages/sympy/plotting/experimental_lambdify.py:176, in lambdify.__call__(self, args)
    174 try:
    175     #The result can be sympy.Float. Hence wrap it with complex type.
--> 176     result = complex(self.lambda_func(args))
    177     if abs(result.imag) > 1e-7 * abs(result):

File /usr/local/lib/python3.8/dist-packages/sympy/plotting/experimental_lambdify.py:272, in Lambdifier.__call__(self, *args, **kwargs)
    271 def __call__(self, *args, **kwargs):
--> 272     return self.lambda_func(*args, **kwargs)

File <string>:1, in <lambda>(x0)
....
File /usr/local/lib/python3.8/dist-packages/sympy/core/expr.py:345, in Expr.__float__(self)
    343 if result.is_number and result.as_real_imag()[1]:
    344     raise TypeError("Cannot convert complex to float")
--> 345 raise TypeError("Cannot convert expression to float")

TypeError: Cannot convert expression to float

我要抱怨不完整的追溯，但很长。您仍然离开了最后一个块，一个带有错误的错误

 Cannot convert expression to float

> code> print（g）;你为什么不展示呢？请显示尽可能多的信息。它可以帮助我们中那些无法复制您的代码的人（目前）。

---

我以前从未使用过这个问题，但是即使我使用subs用数字替换变量，它仍然不会产生数字值。所以没有什么可绘的。

In [52]: g.subs({x:2,k:3}).evalf()
Out[52]: 
                ⎛   1         ⎞
FourierTransform⎜───────, 2, 3⎟
                ⎝cosh(2)      ⎠