Eliminate a variable to relate two functions in Python using SymPy

Question

I have two equations that are parametrized by a variable "t". These look like:

X = p(t) 
Y = q(t)

where p and q are polynomials in t. I want to use Python's SymPy library to eliminate the t variable and express Y = F(X) for some function X. I have tried using solve() in SymPy but this is not working too well. I know that Maple and Mathematica both have eliminate() functions that can accomplish this, but I wanted to know if Python might have a general function that does this.

Answer 1

I refer to this example from the 'Scope' section of https://reference.wolfram.com/language/ref/Eliminate.html.

Eliminate[2 x + 3 y + 4 z == 1 && 9 x + 8 y + 7 z == 2, z]

>>> from sympy import *
>>> var('x y z')
(x, y, z)
>>> solve(2*x+3*y+4*z-1, z)
[-x/2 - 3*y/4 + 1/4]
>>> solve(9*x+8*y+7*z-2, z)
[-9*x/7 - 8*y/7 + 2/7]
>>> (-9*x/7 - 8*y/7 + Rational(2,7))-(-x/2 - 3*y/4 + Rational(1,4)).simplify()
-11*x/14 - 11*y/28 + 1/28
>>> 28*((-9*x/7 - 8*y/7 + Rational(2,7))-(-x/2 - 3*y/4 + Rational(1,4)).simplify())
-22*x - 11*y + 1

• Solve each equation for z.
• Subtract one expression for z from the other.
• Note only that numeric fractions need to be coded — I've used Rational because I forget other methods — so that fractional arithmetic is used.
• I multiply through to get rid of the denominators.
• This approach will work only for the elimination of a single variable. I haven't considered the second and subsequent examples.

I hope this is useful.

Answer 2

Here is a lightly tested simple routine

def eliminate(eqs, z):
    """return eqs with parameter z eliminated from each equation; the first
    element in the returned list will be the definition of z that was used
    to eliminate z from the other equations.

    Examples
    ========

    >>> eqs = [Eq(2*x + 3*y + 4*z, 1),
    ...        Eq(9*x + 8*y + 7*z, 2)]
    >>> eliminate(eqs, z)
    [Eq(z, -x/2 - 3*y/4 + 1/4), Eq(11*x/2 + 11*y/4 + 7/4, 2)]
    >>> Eq(y,solve(_[1], y)[0])
    Eq(y, -2*x + 1/11)
    """
    from sympy.solvers.solveset import linsolve
    Z = Dummy()
    rv = []
    for i, e in enumerate(eqs):
        if z not in e.free_symbols:
            continue
        e = e.subs(z, Z)
        if z in e.free_symbols:
            break
        try:
            s = linsolve([e], Z)
            if s:
                zi = list(s)[0][0]
                rv.append(Eq(z, zi))
                rv.extend([eqs[j].subs(z, zi)
                    for j in range(len(eqs)) if j != i])
                return rv
        except ValueError:
            continue
    raise ValueError('only a linear parameter can be eliminated')

There is a more complex routine at this issue.