Best way to isolate one coefficient of a multivariate polynomial in sympy

Question

I have a multivariate polynomial (which in the general case many many variables) whose coefficients list some data that I need to read off, but it doesn\'t seem like sympy g

Answer 1

The monomials of a Polynomial are listed in order that the generators appear (and that order is under the user's control):

>>> from sympy import Poly
>>> from sympy.abc import x, y, z
>>> Poly(x + 3*y**2, x, y).monoms()
[(1, 0), (0, 2)]
>>> Poly(x + 3*y**2, y, x).monoms()
[(2, 0), (0, 1)]

When querying to get the coefficients, either a monomial-tuple or an expression can be used:

>>> Poly(x + 3*y**2, x, y).coeff_monomial(y**2)
3
>>> Poly(x + 3*y**2, x, y).coeff_monomial((0, 2))
3

A dictionary of all coefficients for different monomials can be obtained in monomial-expression form by converting the Poly to an expression and then using the as_coefficients_dict method:

>>> p = Poly((x+2*y-z)**3)
>>> p.as_expr().as_coefficients_dict()
defaultdict(, {
x**3: 1, z**3: -1, y**3: 8, y**2*z: -12, x**2*z: -3,
x*z**2: 3, x**2*y: 6, y*z**2: 6, x*y**2: 12, x*y*z: -12})

Or, if you prefer the monomial-tuple form, you can use:

>>> [(m,p.coeff_monomial(m)) for m in p.monoms()]
[((3, 0, 0), 1), ((2, 1, 0), 6), ((2, 0, 1), -3), ((1, 2, 0), 12), ((1, 1, 1),
-12), ((1, 0, 2), 3), ((0, 3, 0), 8), ((0, 2, 1), -12), ((0, 1, 2), 6),
((0, 0, 3), -1)]

And that can be turned into a defaultdict that will give 0 for a non-existing monomial with:

>>> defaultdict(int, _)
defaultdict(, {(3, 0, 0): 1, (2, 1, 0): 6, (2, 0, 1): -3,
(1, 2, 0): 12, (1, 1, 1): -12, (1, 0, 2): 3, (0, 3, 0): 8, (0, 2, 1): -12,
(0, 1, 2): 6, (0, 0, 3): -1})