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pythonnumpysympy
Alexander Tille
Alexander Tille

Reputation: 43

sympy: Lambdify doesn't convert sqrt expression

I try to lambdify some calculated eigenvalues but I get the following error.

File "<string>", line 1, in <lambda>
AttributeError: 'Symbol' object has no attribute 'sqrt'

To avoid a namespace clash (explained in this post What causes this error (AttributeError: 'Mul' object has no attribute 'cos') in Python?) I used the following import command instead of from sympy import * 

import sympy as sp
import numpy as np

def calculate_general_eigenvalues():
    Y, Z = sp.symbols("Y,Z")
    Rzy = sp.symbols("Rzy", positive=True)

    eigenvalues = [Y + Z,Rzy*Y + sp.sqrt(Rzy*Z)]

    print("eigenvalues of the system ")
    print(eigenvalues[0])
    print(eigenvalues[1])
    lam1 = sp.lambdify((Y,Z), eigenvalues[0] ,modules=['numpy'])
    lam2 = sp.lambdify((Y,Z), eigenvalues[1] ,modules=["numpy", {'sqrt': np.sqrt}])
    print(lam1(1,1))
    print(lam2(1,1))


    return (lam1,lam2)


l1,l2 = calculate_general_eigenvalues()

I also found a second hint here (Python: SymPy lambdify abs for use with NumPy) were they include the command lambdify(x, f(x), ["numpy", {'Abs': numpy.abs}]) but it doesn't work in my code as you can see

How can I solve my problem?

Upvotes: 1

Views: 1208

Answers (1)

user6655984
user6655984

Reputation:

The problem is not sqrt. Your second eigenvalue involves Rzy which you did not include among the variables when lambdifying it. Fix that, and it will work simply with modules="numpy". There is no need to remap sqrt: NumPy knows about sqrt, it just does not know how to apply it to a symbol such as Rzy.

lam1 = sp.lambdify((Y,Z), eigenvalues[0], modules='numpy')
lam2 = sp.lambdify((Y,Z,Rzy), eigenvalues[1], modules='numpy')
print(lam1(1,1))
print(lam2(1,1,1))

Upvotes: 2

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