Get a value from solution set returned as finiteset by Sympy

A slightly more general solution is to simply convert the FiniteSet into a standard python list

>>> a=list(linsolve(lista,a1,a2))
>>> a
[(71/369, 7/41)]

Then you can extract elements using standard indexing — in this case a[0]. But then if you get multiple solutions, you can just pull out the one you want.

I found the sympy library way in this link http://docs.sympy.org/latest/tutorial/manipulation.html

Use .args atribute in the function or result object. If I have a function:

>>>func = Eq(u(x),−x+sin(x)) 
>>>func
u(x) = -x + sin(x)
>>>func.args[0] 
u(x)
>>>func.args[1]
-x+sin(x)

The same applies for a result that is a finite set type.

Weird. There's no description anywhere how to use the results of linsolve.

Even the specs only test the result set as a whole, without checking the elements separately.

Sequence unpacking

If you know that the system of equation admits at least one solution, you can use sequence unpacking with a trailing comma before the assignment :

>>> from sympy import linsolve, symbols, solve, Rational
>>> a1, a2 = symbols('a1 a2')
>>> equations = [-3*a1/10 - 3*a2/20 + Rational(1, 12), -3*a1/20 - 13*a2/105 + Rational(1, 20)]
>>> equations
[-3*a1/10 - 3*a2/20 + 1/12, -3*a1/20 - 13*a2/105 + 1/20]
>>> linsolve(equations, a1, a2)
{(71/369, 7/41)}
>>> solution, = linsolve(equations, a1, a2)
>>> solution
(71/369, 7/41)

This syntax will also work if there is an infinity of solution:

>>> solution, = linsolve([a1, a1], a1, a2)
>>> solution
(0, a2)

But it will fail if there are no solution:

>>> solution, = linsolve([a1 - 1, a1 - 2], a1, a2)
ValueError: not enough values to unpack (expected 1, got 0)

which might be the desired behaviour.

Iterate over the solutions:

Another possibility is to simply iterate over the solutions:

>>> for solution in linsolve(equations, a1, a2):
...     print(solution)
... 
(71/369, 7/41)

Nothing happens if there's no solution:

>>> for solution in linsolve([a1 - 1, a1 - 2], a1, a2):
...     print(solution)
... 

solve instead of linsolve

You could also use solve instead of linsolve, even though it's not recommended by the project because it can output different types:

>>> solve(equations, a1, a2)
{a1: 71/369, a2: 7/41}
>>> solve([a1 - 1, a1 - 2], a1, a2)
[]