sympy - Bisection方法 - TypeError: unable determine truth value of Relational.

Question

我试图用Sympy实现Bisection方法，但我遇到了这个错误。

if f[a]*f[c] > 0:  # Opposite sign a and c
File "C:\Users\maico\AppData\Local\Programs\Python\Python38\lib\site-packages\sympy\core\relational.py", line 376, in __nonzero__
    raise TypeError("cannot determine truth value of Relational")
TypeError: cannot determine truth value of Relational

这是我的代码

from sympy import plot_implicit, latex, lambdify, Float
from sympy.abc import x
from sympy.parsing.latex import parse_latex

eq = input("Latex equation: ")    
raw_equation = eq.replace("=0", "").replace("= 0", "").replace("e", "E")
equation = parse_latex("y = " + raw_equation)
f = lambdify(x, parse_latex(self.__raw_equation), 'numpy')

# Bisection
a = 0 # start interval
b = 1 # end interval

eps = a - b
r = None
nlimit = 8
for n in range(nlimit):
    c = (a + b) / 2
    fd = {a: f(a), b: f(b), c: f(c)}
    solved = False
    for v in [a, b, c]:
        if fd[v] == 0:
            r = "Iterations: {} - Result: {}".format(n + 1, v)
            solved = True
            break
    if solved:
        break
    if fd[a]*fd[c] > 0:  # Opposite sign a and c <-- ERROR
        b = c
    else:  # Opposite sign b and c
        a = c
...

你可以用下面的公式来测试一下 xe^x-1=0

你能帮帮我吗？

Answer 1

有了这些入口

from sympy import plot_implicit, latex, lambdify, Float
from sympy import symbols
from sympy.abc import x
from sympy.parsing.latex import parse_latex

E = symbols('E')
equation = E**x*x - 1
f = lambdify(x, equation, 'numpy')
print(f.__doc__)
....

我得到的是:

1450:~/mypy$ python3 stack61370217.py 
Created with lambdify. Signature:

func(x)

Expression:

E**x*x - 1

Source code:

def _lambdifygenerated(x):
    return (E**x*x - 1)


Imported modules:


Traceback (most recent call last):
  File "stack61370217.py", line 29, in <module>
    if fd[a]*fd[c] > 0:  # Opposite sign a and c <-- ERROR
  File "/usr/local/lib/python3.6/dist-packages/sympy/core/relational.py", line 376, in __nonzero__
    raise TypeError("cannot determine truth value of Relational")
TypeError: cannot determine truth value of Relational

如果换成

from sympy import E

羔羊的表达方式变成了

def _lambdifygenerated(x):
    return (x*exp(x) - 1)

它的生产 c 0.00390625

你没有显示定义 E. 我不知道latex解析是否会产生。 我没有安装足够的包来运行它。 在任何情况下，那个符号 E 传播到 numpy 表达方式：

In [86]: E=symbols('E')                                                                                

In [87]: def f(x): 
    ...:     return E**x*x-1 
    ...:                                                                                               

In [88]: f(10)                                                                                         
Out[88]: 
    10    
10⋅E   - 1

In [89]: f(10)*f(20)>0                                                                                 
Out[89]: 
⎛    10    ⎞ ⎛    20    ⎞    
⎝10⋅E   - 1⎠⋅⎝20⋅E   - 1⎠ > 0

In [90]: if f(10)*f(20)>0: pass                                                                        
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-90-14ce5e03fbac> in <module>
----> 1 if f(10)*f(20)>0: pass

/usr/local/lib/python3.6/dist-packages/sympy/core/relational.py in __nonzero__(self)
    374 
    375     def __nonzero__(self):
--> 376         raise TypeError("cannot determine truth value of Relational")
    377 
    378     __bool__ = __nonzero__

TypeError: cannot determine truth value of Relational

如果是 E 是同情心的定义 e这被翻译成 exp(x)一切都很顺利。