CODEWITHSUNDEEP
numpysympy
manµ
manµ

Reputation: 47

Can I substitute IndexedBase with numpy array

Can I substitute a IndexedBase with a numpy array to get a new sympy expression? Something like:

import numpy as np
import sympy as sy

A = sy.IndexedBase('A', shape=(5))
k = sy.IndexedBase('k', shape=(5))
i = sy.symbols('i',cls=sy.Idx)

expr1 = sy.Sum(A[i]/sy.exp(sy.I * k[i]),(i,1,5))

kz = np.arange(1,5)

expr2 = expr1.subs(k[:],kz[:])

last line = error.

Thanks!

Upvotes: 0

Views: 200

Answers (2)

Taka
Taka

Reputation: 1

This might work for you!

import numpy as np
import sympy as sp
import inspect
i ,m= sp.symbols('i,m', integer=True)

A = sp.IndexedBase('A', shape=(m))
k = sp.IndexedBase('k', shape=(m))
#i = sp.symbols('i',cls=sy.Idx)

expr1 = sp.Sum(A[i]/sp.exp(sp.I * k[i]),(i,0,m-1))
print(expr1)
print(expr1.doit())

f_lam = sp.lambdify((A, k,m), expr1, 'numpy')

print(inspect.getsource(f_lam))

M=5
A2 = np.random.uniform(0,1, M)
kz = np.arange(0,M)

print(f_lam(A2,kz,M))

#(-0.08145603175240379+0.27923362018114933j)

Upvotes: 0

Francesco Bonazzi
Francesco Bonazzi

Reputation: 1957

Yes, you can:

In [6]: expr2 = expr1.replace(k, kz)

In [7]: expr2
Out[7]: 
  5                              
 ___                             
 ╲                               
  ╲    -ⅈ⋅[1, 2, 3, 4, 5][i]     
  ╱   ℯ                     ⋅A[i]
 ╱                               
 ‾‾‾                             
i = 1

Beware that the summation range is wrong, Python arrays have zero as offset.

Upvotes: 1

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