Why lambdify never stops?

Question

x = symbols('x')
ch = 'exp(cos(cos(exp((sin(-0.06792841536110628))**(-6.045461643745118)))))'
f = lambdify(x, ch, "numpy")
print(float(f(2)))

It does not work, the programm is running and never ends(no error is issued). My goal is to avoid this kind of cases (among multiple cases) by doing a try/except but i can't as there is no error Why no error is released? How can i avoid these cases ?

Thanks for your helping me !

When I do `sympify(ch)` and then do a keyboard interrupt, the traceback shows a `mpmath` stack. It is trying to evaluate it numerically. Without symbols `ch` is just a numeric expression. What's the point to use `numpy` `lambdify`, muchless providing an argument like `2`? The equivalent in `numpy` produces a Runtimewarning and `nan`. — hpaulj, Apr 29 '21 at 16:16
Your inner most `sin` produces a negative value, and the power a complex value. `numpy` can only handle that if it starts with a complex argument `np.sin(-067+0j)`. But then the `exp` blows up, and so on... — hpaulj, Apr 29 '21 at 16:23

score 1 · Answer 1 · answered Apr 29 '21 at 16:31

lambdify is a lexical translator, converting a sympy expression to a python/numpy function.

Make a string with a symbol:

In [27]: ch = 'exp(cos(cos(exp((sin(x))**(-6.045461643745118)))))'

sympify(ch) has no problem, because it doesn't need to do any numeric calculation. So lambdify also works:

In [28]: f=lambdify(x,ch)
In [29]: f?
Signature: f(x)
Docstring:
Created with lambdify. Signature:

func(x)

Expression:

exp(cos(cos(exp((sin(x))**(-6.045461643745118)))))

Source code:

def _lambdifygenerated(x):
    return (exp(cos(cos(exp(sin(x)**(-6.045461643745118))))))

The equivalent mpmath:

def _lambdifygenerated(x):
    return (exp(cos(cos(exp(sin(x)**(mpf((1, 54452677612106279, -53, 56))))))))

And a working numeric evaluation:

In [33]: f(0j)
Out[33]: mpc(real='nan', imag='0.0')

score 0 · Answer 2 · answered Apr 29 '21 at 12:10

In general, I'm not sure you can. SymPy or NumPy will keep trying to compute the number until precision is exhausted. But you can create a function that will raise and error if numbers are out of bounds for your interest:

>>> from sympy import cos as _cos, I, exp
>>> def cos(x):
...     if abs(x) > 10**20: raise ValueError
...     return _cos(x)
>>> exp(cos(cos(exp(5*(1+I)))))
Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "<string>", line 2, in cos
ValueError
>>> f = lambda x: exp(cos(cos(exp(x))))
>>> f(sin(-0.06792841536110628)**-6.045461643745118)
Traceback (most recent call last):
  File "<string>", line 1, in <module>
  File "<string>", line 1, in <lambda>
  File "<string>", line 2, in cos
ValueError

But you have to think carefully about when you want to raise such an error. For example, SymPy has no trouble computing f(100) or f(100*I) if the non-error-catching cos is used. So think about when actually you want the error to rise.

Why lambdify never stops?

2 Answers2