Python 如何在 SymPy 中求解线性方程组？
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How can I solve system of linear equations in SymPy?
提问by Aniket Vij
Sorry, I am pretty new to sympy and python in general.
抱歉，总的来说，我对 sympy 和 python 还很陌生。
I want to solve the following underdetermined linear system of equations:
我想解决以下欠定线性方程组：
x + y + z = 1
x + y + 2z = 3
采纳答案by Amit Kumar
SymPy recently got a new Linear system solver: linsolvein sympy.solvers.solveset, you can use that as follows:
SymPy 最近获得了一个新的线性系统求解器：linsolve在 中sympy.solvers.solveset，您可以按如下方式使用它：
In [38]: from sympy import *
In [39]: from sympy.solvers.solveset import linsolve
In [40]: x, y, z = symbols('x, y, z')
List of Equations Form:
方程列表形式：
In [41]: linsolve([x + y + z - 1, x + y + 2*z - 3 ], (x, y, z))
Out[41]: {(-y - 1, y, 2)}
Augmented Matrix Form:
增强矩阵形式：
In [59]: linsolve(Matrix(([1, 1, 1, 1], [1, 1, 2, 3])), (x, y, z))
Out[59]: {(-y - 1, y, 2)}
A*x = b Form
A*x = b 形式
In [59]: M = Matrix(((1, 1, 1, 1), (1, 1, 2, 3)))
In [60]: system = A, b = M[:, :-1], M[:, -1]
In [61]: linsolve(system, x, y, z)
Out[61]: {(-y - 1, y, 2)}
Note: Order of solution corresponds the order of given symbols.
注意：解的顺序对应于给定符号的顺序。
回答by Scott
You can solve in matrix form Ax=b(in this case an underdetermined system but we can use solve_linear_system):
您可以以矩阵形式求解Ax=b（在本例中为欠定系统，但我们可以使用solve_linear_system）：
from sympy import Matrix, solve_linear_system
x, y, z = symbols('x, y, z')
A = Matrix(( (1, 1, 1, 1), (1, 1, 2, 3) ))
solve_linear_system(A, x, y, z)
{x: -y - 1, z: 2}
Or rewrite as (my editing, not sympy):
或重写为（我的编辑，而不是同情）：
[x]= [-1] [-1]
[y]= y[1] + [0]
[z]= [0] [2]
In the case of a square Awe could define band use A.LUsolve(b).
在正方形的情况下，A我们可以定义b和使用A.LUsolve(b).
回答by PaulDong
In addition to the great answers given by @AMiT Kumar and @Scott, SymPy 1.0 has added even further functionalities. For the underdetermined linear system of equations, I tried below and get it to work without going deeper into sympy.solvers.solveset. That being said, do go there if curiosity leads you.
除了@AMiT Kumar 和@Scott 给出的出色答案之外，SymPy 1.0 还添加了更多功能。对于欠定线性方程组，我在下面尝试并使其工作，而无需深入研究sympy.solvers.solveset. 话虽如此，如果好奇心引导你，那就去那里吧。
from sympy import *
x, y, z = symbols('x, y, z')
eq1 = x + y + z
eq2 = x + y + 2*z
solve([eq1-1, eq2-3], (x, y,z))
That gives me {z: 2, x: -y - 1}.
Again, great package, SymPy developers!
这给了我{z: 2, x: -y - 1}。再次，伟大的包，SymPy 开发人员！
回答by Aziz Alto
Another example on matrix linear system equations, lets assume we are solving for this system:
矩阵线性系统方程的另一个例子，假设我们正在求解这个系统：
In SymPywe could do something like:
在SymPy我们可以做这样的事情：
>>> import sympy as sy
... sy.init_printing()
>>> a, b, c, d = sy.symbols('a b c d')
... A = sy.Matrix([[a-b, b+c],[3*d + c, 2*a - 4*d]])
... A
? a - b b + c ?
? ?
?c + 3?d 2?a - 4?d?
>>> B = sy.Matrix([[8, 1],[7, 6]])
... B
?8 1?
? ?
?7 6?
>>> A - B
? a - b - 8 b + c - 1 ?
? ?
?c + 3?d - 7 2?a - 4?d - 6?
>>> sy.solve(A - B, (a, b, c, d))
{a: 5, b: -3, c: 4, d: 1}
回答by Ali80
import sympy as sp
x, y, z = sp.symbols('x, y, z')
eq1 = sp.Eq(x + y + z, 1) # x + y + z = 1
eq2 = sp.Eq(x + y + 2 * z, 3) # x + y + 2z = 3
ans = sp.solve((eq1, eq2), (x, y, z))
this is similar to @PaulDong answer with some minor changes
这类似于@PaulDong 的回答，但有一些细微的变化
- its a good practice getting used to not using
import *(numpy has many similar functions) - defining equations with
sp.Eq()results in cleaner code later on
- 习惯不使用是一个好习惯
import *（numpy 有很多类似的功能） - 定义方程，
sp.Eq()稍后会产生更清晰的代码
