How to get coefficients in sympy expression/ Python

Question

How to get coefficients in sympy expression

b = sympy.Symbol('b')
a = sympy.Symbol('a')
x = sympy.Symbol('x')

y = 2*x**3 + a*x**2 +b*x

The result should be :

coeff for x**3 =2 coeff for x**2 = a coeff for x = b

Does this answer your question? [How to extract all coefficients in sympy](https://stackoverflow.com/questions/22955888/how-to-extract-all-coefficients-in-sympy) — Lomtrur, Mar 05 '20 at 14:31

score 2 · Accepted Answer · answered Mar 05 '20 at 15:38

It is surprisingly awkward to get the expression form of the terms in the polynomial but you can do it like this:

In [28]: b = sympy.Symbol('b') 
    ...: a = sympy.Symbol('a') 
    ...: x = sympy.Symbol('x') 
    ...:  
    ...: y = 2*x**3 + a*x**2 +b*x

In [29]: p = Poly(y, x)

In [30]: p
Out[30]: Poly(2*x**3 + a*x**2 + b*x, x, domain='ZZ[a,b]')

In [31]: {x**m[0]:p.coeff_monomial(m) for m in p.monoms()}
Out[31]: 
⎧       2      3   ⎫
⎨x: b, x : a, x : 2⎬
⎩                  ⎭

smichr · Answer 2 · 2020-03-05T19:41:37.063

Although getting the coefficients is easy

>>> y = 2*x**3 + a*x**2 + b
>>> p = Poly(y, x)
>>> p.coeffs()
[2, a, b]
>>> p.all_coeffs()
[2, a, 0, b]

getting them linked up to their monomial is the unimplemented/ackward part. @OscarBenjamin showed how to do this with the monomials of the Poly. You can also work with the expression without converting it to a Poly:

>>> dict(i.as_independent(x)[::-1] for i in Add.make_args(y))
{1: b, x**3: 2, x**2: a}

It would be a nice "battery" for SymPy to include a mapping keyword on Poly.coeffs that would allow this dictionary to be returned. Alternatively, Expr.as_coefficients_dict could be given an optional variable keyword that would allow the coefficients to be returned which are indepenedent of any symbols in variable.

This simple dict-comprehension will fail if there are uncollected terms like a*x + b*x in the expression. A more robust little function is posted here.

How to get coefficients in sympy expression/ Python

2 Answers2