Parse a string of a linear equation into a vector of the coefficients (Python)

Question

I am looking for a Python solution that takes a string of a linear equation and outputs a vector of the coefficients.

To make things simple for the start, assume I have a set of equations:

• given in string form
• each with no more than 4 elements
• all are linear
• in all of them x appears only once

then I would like to get a vectorized representation where

• the first element is the x coefficient
• other elements are other coefficients in the equation (not evaluated, but as they are), as if they appeared on the other side of the equation as x
• zeros to complete the 4-d vector

I give here several input-output equations to get a sense of some of the challenges:

• '2*x+3=2+5' => [2, -3, 2, 5]
• '88/8=x' => [8, 88, 0, 0]
• '74=(35+18)+3*x' => [3, 74, -35, -18]
• '((4+4)*6)=x'] => [1/6, 4, 4, 0]
• '-X=(91.0+88.0)' => [-1, 91, 88, 0]
• 'X=(30.0/10.0)' => [10, 30, 0, 0]
• '0.16 + 0.41 = 2*x - 0.08' => [2, 0.16, 0.41, 0.08]
• '(0.25 + 0.37)*2 = x' => [1/2, 0.25, 0.37, 0]

I started coding a "brute force" solution that is highly rigorous and tedious, stumbled several times along the way, and figured there must be a nicer and more clever way to do this...

• I am using the sympy package, which makes things a bit easier. With sympify and formula.split and such I am able to extract the x coefficient and the result of the equation (although I really don't care about the result, but only the vector representation)
• I saw this and this but they are both in different languages, and not quite what I am looking for.

Sooo, anyone has any idea how to do it in Python?

Thanks! :)

Answer 1

This may get you going in the right direction:

>>> def peq(s):
...     from sympy import S
...     l, r = t = S(s, evaluate=False)
...     free = t.free_symbols
...     assert len(free) == 1
...     x = free.pop()
...     if r.has(x):
...         l, r = r, l
...     assert not r.has(x)
...     assert l.has(x)
...     assert not l.diff(x).free_symbols
...     v = []
...     v.append(l.coeff(x))
...     v.append(-(l.subs(x, 0)))
...     if not r.is_Add:
...         v.extend([r, S.Zero])
...     else:
...         assert r.is_Add and len(r.args) == 2
...         v.extend(r.args)
...     return v
>>> peq('2*x+3,-2+5/3')
[2,−3,−2,5/3]
>>> peq('2*x+3,-2')
[2,−3,−2,0]
>>> peq('x,-2')
[1,0,−2,0]