CODEWITHSUNDEEP
python-3.xsympy
bbalzano
bbalzano

Reputation: 33

How to use solutions created by sympy's linear equation solver

I am using sympy to solve a system of linear equations and I want to use one of the solutions in further calculation. First, I wanted to pull a specific solution from the generated dictionary for my system of linear equations. I tried dict['key'] as I believe you would normally but it does work. Below is the code I tried to run.

from sympy import symbols, solve

m2,x2a,x2b,x2c,m3,m4,m5,x5a,x5b,x5c = 
symbols('m2,x2a,x2b,x2c,m3,m4,m5,x5a,x5b,x5c')
solution = solve((100-m2-m3,
              .980*m3-.96*(.450*100),
              100*.300-m2*x2a,
              1-x2a-x2b-x2c,
              x2c*m2-.04*(.450*100),
              m2-m4-m5,
              m2*x2b-m4*.06-m5*x5b,
              .940*m4-.97*(x2a*m2),
              m5*x5a-.03*(x2a*m2),
              1-x5a-x5b-x5c),
            [m2,x2a,x2b,x2c,m3,m4,m5,x5a,x5b,x5c], dict=True)
print(solution[0]['m2'])

enter image description here

Upvotes: 1

Views: 87

Answers (1)

unutbu
unutbu

Reputation: 880547

Sometimes there can be more than one solution to an equation or system of equations. That is why sym.solve returns a list of dicts, not just a dict.

Notice that print(solution) begins with a bracket (indicating a list):

[{m2: ...}]

Therefore, to access the value of m2 for the first (and in this case only) solution, you would use

solution[0][m2]

In general, to loop through all solutions, you could use:

for s in solution:
    print(s[m2])

Notice also that the dict uses SymPy Symbols as keys, not strings.

If you ever run into a similar problem again, you can investigate the problem by inspecting the .keys attribute:

In [190]: list(solution[0].keys())
Out[190]: [m2, x2a, x2b, x2c, m3, m4, m5, x5a, x5b, x5c]

If the keys were strings, you would have seen ['m2', 'x2a', 'x2b', 'x2c', 'm3', 'm4', 'm5', 'x5a', 'x5b', 'x5c'].

Upvotes: 2

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