How to extract all coefficients in sympy

You can get a coefficient of a specific term by using coeff();

x, a = symbols("x, a")
expr = 3 + x + x**2 + a*x*2
expr.coeff(x)
# 2*a + 1

Here I want to extract all the coefficients of x, x**2 (and so on), like;

# for example
expr.coefficients(x)
# want {1: 3, x: (2*a + 1), x**2: 1}

There is a method as_coefficients_dict(), but it seems this doesn’t work in the way I want;

expr.as_coefficients_dict()
# {1: 3, x: 1, x**2: 1, a*x: 2}
expr.collect(x).as_coefficients_dict()
# {1: 3, x**2: 1, x*(2*a + 1): 1}

Answers:

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Method 1

all_coeffs() can be sometime better than using coeffs() for a Poly.
The difference lies in output of these both. coeffs() returns a list containing all coefficients which has a value and ignores those whose coefficient is 0 whereas all_coeffs() returns all coefficients including those whose coefficient is zero.

>>> a = Poly(x**3 + a*x**2 - b, x)
>>> a.coeffs()
[1, a, -b]

>>> a.all_coeffs()
[1, a, 0, -b]

Method 2

The easiest way is to use Poly

>>> a = Poly(expr, x)
>>> a.coeffs()
[1, 2*a + 1, 3]

Method 3

Collection of coefficients can be handled with Poly and then separation of the monomials into dependent and independent parts can be handled with Expr.as_independent:

def codict(expr, *x):
  collected = Poly(expr, *x).as_expr()
  i, d = collected.as_independent(*x, as_Add=True)
  rv = dict(i.as_independent(*x, as_Mul=True)[::-1] for i in Add.make_args(d))
  if i:
      assert 1 not in rv
      rv.update({S.One: i})
  return rv

>>> var('a x z y')
(a, x, z, y)
>>> expr = 3 + x + x**2 + a*x*2
>>> codict(expr, x)
{x**2: 1, x: 2*a + 1, 1: 3}
>>> codict(expr+y+z, x)
{x**2: 1, x: 2*a + 1, 1: y + z + 3}
>>> codict(expr+y+z, x,y)
{y: 1, x**2: 1, x: 2*a + 1, 1: z + 3}
>>> codict(expr+y+z, x,y,z)
{y: 1, z: 1, x**2: 1, x: 2*a + 1, 1: 3}

Method 4

One thing you can do is use a dictionary comprehension like so:

dict = {x**p: expr.collect(x).coeff(x**p) for p in range(1,n)}

where n is the highest power+1. In this case n=3. So you would have the list [1,2]

This would give

dict = {x: (2*a+1), x**2: 1}

Then you can add in the single term with

dict[1] = 3

So

 dict = {1:3,x:(2*a+1),x**2:1}

You may also try:

a = list(reversed(expr.collect(x).as_ordered_terms()))
dict = {x**p: a[p],coeff(x**p) for p in range(1,n)}
dict[1] = a[0] # Would only apply if there is single term such as the 3 in the example

where n is the highest power + 1.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0