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Sympy .coeff_all() returned list is not readable by scipy

By : decky
Date : September 13 2020, 11:00 PM
it should still fix some issue Why did you just show a bit the error? Why not the full message, maybe even the full traceback!
code :
In [60]: sys = signal.lti(num_c, den_c)                                                                   
TypeError                                 Traceback (most recent call last)
<ipython-input-60-21f71ecd8884> in <module>
----> 1 sys = signal.lti(num_c, den_c)

/usr/local/lib/python3.6/dist-packages/scipy/signal/ltisys.py in __init__(self, *system, **kwargs)
    590         self._den = None
--> 592         self.num, self.den = normalize(*system)
    594     def __repr__(self):

/usr/local/lib/python3.6/dist-packages/scipy/signal/filter_design.py in normalize(b, a)
   1609     leading_zeros = 0
   1610     for col in num.T:
-> 1611         if np.allclose(col, 0, atol=1e-14):
   1612             leading_zeros += 1
   1613         else:

<__array_function__ internals> in allclose(*args, **kwargs)

/usr/local/lib/python3.6/dist-packages/numpy/core/numeric.py in allclose(a, b, rtol, atol, equal_nan)
   2170     """
-> 2171     res = all(isclose(a, b, rtol=rtol, atol=atol, equal_nan=equal_nan))
   2172     return bool(res)

<__array_function__ internals> in isclose(*args, **kwargs)

/usr/local/lib/python3.6/dist-packages/numpy/core/numeric.py in isclose(a, b, rtol, atol, equal_nan)
   2267     y = array(y, dtype=dt, copy=False, subok=True)
-> 2269     xfin = isfinite(x)
   2270     yfin = isfinite(y)
   2271     if all(xfin) and all(yfin):

TypeError: ufunc 'isfinite' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''
In [55]: num_c[0]                                                                                         
Out[55]: 500000.000000000

In [56]: type(_)                                                                                          
Out[56]: sympy.core.numbers.Float
In [61]: np.array(num_c)                                                                                  
Out[61]: array([500000.000000000], dtype=object)
In [63]: np.array(num_c).astype(float)                                                                    
Out[63]: array([500000.])
In [64]: sys = signal.lti(np.array(num_c).astype(float), np.array(den_c).astype(float))                   

In [65]: sys                                                                                              
array([1.00000000e+00, 1.02004008e+05]),
dt: None
sys = signal.lti([float(i) for i in num_c],[float(i) for i in den_c]) 

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Incompatability when upgrading Scipy cannot use SciPy function on SymPy matrix

By : ko.dash
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I don't know how it worked in your previous setup, but the process of converting SymPy matrices to NumPy arrays was explicit as early as 2012, per this answer, and SymPy has a utility function matrix2numpy for this purpose. So, in your context
code :
LA.eig(matrix2numpy(M, dtype=float))
LA.eig(np.array(M.tolist(), dtype=float))

Optimization in scipy from sympy

By : Purushottam E
Date : March 29 2020, 07:55 AM
around this issue The solution was, and I do not know why, to cast inputs to numpy.float:
code :
m_1 = np.float32(m_1)
m_2 = np.float32(m_2)
s_1 = np.float32(s_1)
s_2 = np.float32(s_2)

Sympy TypeError when using scipy.stats normal cdf

By : Tanmay Saraiya
Date : March 29 2020, 07:55 AM
Hope this helps Symbolic setup
If you are looking for symbolic solutions, use symbolic functions: e.g., SymPy's log not NumPy's log. The normal CDF is also available from SymPy's stats module as cdf(Normal("x", 0, 1)). The correct SymPy setup would be this:
code :
from sympy import Eq, Rational, Symbol, log
from sympy.stats import cdf, Normal
eqn = Eq((x-Rational('0.2'))/Rational('0.3'), cdf(Normal("x", 0, 1))(log(Rational(100, 110)) + x**2/2)/x)
Eq(10*x/3 - 2/3, (erf(sqrt(2)*(x**2/2 - log(11) + log(10))/2)/2 + 1/2)/x)
from scipy.stats import norm
from scipy.optimize import fsolve
import numpy as np

f = lambda x: (x-0.2)/0.3 - norm.cdf((np.log(100/110) + x**2/2)/x)
print(fsolve(f, 1))   # 1 is a random initial guess

How to pass sympy expressions to be used with scipy?

By : user3378436
Date : March 29 2020, 07:55 AM
With these it helps The provided code also gave me an error: TypeError: _lambdifygenerated() missing 1 required positional argument: 'y1'. Replacing the declaration ofy by y = DeferredVector('y') solved the problem, as suggested in this post. Supposing you are using recent versions of sympy and scipy.
I don't understand the strange error message about jac, as you entered it just fine. Maybe you are inadvertently using root from another package?
code :
from scipy.optimize import root as scipy_root
from sympy import Symbol, lambdify, DeferredVector

y = DeferredVector('y')

def solver():
    guess = [1,-1]
    sol = scipy_root(f, guess, method='hybr', jac=False) # Entrega el resultado
    return sol

Python optimization using sympy lambdify and scipy

By : Jahson
Date : March 29 2020, 07:55 AM
this one helps. As @Dair pointed out, sympy's lambdify in general requires more than one arguments, while scipy expects only one argument, a list (or an array) that contains all the values of each variable. Because my objective function is most conveniently defined using sympy, I need to find a way to get around this incompatibility of sympy and scipy.
@lhcgeneva pointed out the answer to a similar question. This answer does not conveniently handle a large number of independent variables, especially when the number of independent variables can change, requiring writing out the "vectorized" version of the objective function to be redefined. However, inspired by this post, I figured out the following solution using *tuple():
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