Algorithm for inverting strings of algebraic expressions in Python

Question

Is there an easy way to make a function to invert the algorithm, for example, as follows:

>>> value = inverse("y = 2*x+3") >>> print(value) "x = (y-3)/2" 

If you cannot make the actual code for the function, please recommend me tools that will facilitate this task. The function will be used only for inverse algorithms with +, -, * and /

Answer 1

You should try SymPy for this:

 from sympy import solve from sympy.abc import x, y e = 2*x+3-y solve(e,x) #[y/2 - 3/2] solve(e,y) #[2*x + 3] 

Based on this, you can build your inverse() as (works for two variables):

 def inverse(string, left_string=None): from sympy import solve, Symbol, sympify string = '-' + string e = sympify(string.replace('=','+')) if left_string: ans = left_string + ' = ' + str(solve(e, sympify(left_string))[0]) else: left = sympify(string.split('=')[0].strip().replace('-','')) symbols = e.free_symbols symbols.remove( left ) right = list(symbols)[0] ans = str(right) + ' = ' + str(solve(e, right)[0]) return ans 

Examples:

 inverse(' x = 4*y/2') #'y = x/2' inverse(' y = 100/x + x**2') #'x = -y/(3*(sqrt(-y**3/27 + 2500) + 50)**(1/3)) - (sqrt(-y**3/27 + 2500) + 50)**(1/3)' inverse("screeny = (isox+isoy)*29/2.0344827586206895", "isoy") #'isoy = -isox + 0.0701545778834721*screeny' 

Answer 2

This is a bit long comment, but here is what I had in mind:

 import sympy def inverse(s): terms = [sympy.sympify(term) for term in s.split("=")] eqn = sympy.Eq(*terms) var_to_solve_for = min(terms[1].free_symbols) solns = sympy.solve(eqn, var_to_solve_for) output_eqs = [sympy.Eq(var_to_solve_for, soln) for soln in solns] return output_eqs 

After that we have

 >>> inverse("y = 2*x+3") [x == y/2 - 3/2] >>> inverse("x = 100/z + z**2") [z == -x/(3*(sqrt(-x**3/27 + 2500) + 50)**(1/3)) - (sqrt(-x**3/27 + 2500) + 50)**(1/3), z == -x/(3*(-1/2 - sqrt(3)*I/2)*(sqrt(-x**3/27 + 2500) + 50)**(1/3)) - (-1/2 - sqrt(3)*I/2)*(sqrt(-x**3/27 + 2500) + 50)**(1/3), z == -x/(3*(-1/2 + sqrt(3)*I/2)*(sqrt(-x**3/27 + 2500) + 50)**(1/3)) - (-1/2 + sqrt(3)*I/2)*(sqrt(-x**3/27 + 2500) + 50)**(1/3)] 

and etc.