Python/Sympy：求解具有不同值的方程

Question

我正在尝试解决这三个方程：

a1 + a2 = b1
a2 + a3 = b2
b1 + b2 = c1

我为三个随机选择的变量生成值（不允许组合 b1、b2 和 c1），所以我可能有 a1 = 5、a3 = 10 和 c1 = 100，所以我可以用 sympy 求解该方程.

我的问题是我似乎无法将随机输入传输到代码的 sympy 部分并对其进行循环。

a1, a2, a3, b1, b2, c1 = symbols('a1 a2 a3 b1 b2 c1')
solve([a1.subs(a1, 5) + a2 - b1, a2 + a3.subs (a3, 10) - b2, b1 + b2 - c1.subs (c1, 100)], (a1, a2, a3, b1, b2, c1))

当我假设我在示例中提到的 a1、a3 和 c1 时，这有效，但我事先随机选择了这些变量。我尝试根据选择的变量创建 for 循环，但是三个变量的可能组合太多，所以我放弃了。

Answer 1

如果我正确理解了您的要求，您可以使用 Sympy 的 linsolve 函数先以符号方式求解方程，然后再用数字替换。关键步骤是告诉linsolve 您要求解哪些变量。我建议使用sets 将您正在求解的变量与您要为其插入值的变量分开。你可以像这样声明你的变量：

all_vars = symbols('a1 a2 a3 b1 b2 c1')
# randomly choose which variables will have values
plug_in_for_vars = ...
solve_for_vars = tuple(set(all_vars) - set(plug_in_for_vars))

然后你可以定义你的方程式

equations = [a1 + a2 - b1, a2 + a3 - b2, b1 + b2 - c1]

并将它们和选择的变量传递给linsolve

solution = linsolve(equations, solve_for_vars)

然后你可以插入选择的值。

solution.subs({variable: value(variable) for variable in plug_in_for_vars})

当然，value(variable) 是您为确定要插入的数值所做的任何事情的代理。

Answer 2

只有 19 种不同的方法可以做到这一点，所以只是象征性地这样做以查看所有可能性并不算太糟糕：

from sympy import symbols, subsets, solve, Dict
v = a1, a2, a3, b1, b2, c1 = symbols('a1 a2 a3 b1 b2 c1')
eqs = (a1 + a2 - b1, a2 + a3 - b2, b1 + b2 - c1)
# we will not assign b1,b2,c1 so we will never solve for a1,a2,a3
ignore = set(v[:3])
for i in subsets(v, 3):
    if set(i) != ignore:
        subs = set(v)-set(i)
        sol = solve(eqs,i,dict=True)[0]
        print('replace',subs,'solve for',set(sol),':\n\t', sol)

输出是

replace {b2, c1, a3} solve for {a2, a1, b1} :
    {a1: a3 - 2*b2 + c1, a2: -a3 + b2, b1: -b2 + c1}
replace {c1, a3, b1} solve for {a2, b2, a1} :
    {a1: a3 + 2*b1 - c1, a2: -a3 - b1 + c1, b2: -b1 + c1}
replace {b2, a3, b1} solve for {a2, c1, a1} :
    {a1: a3 + b1 - b2, a2: -a3 + b2, c1: b1 + b2}
replace {a2, b2, c1} solve for {a1, a3, b1} :
    {a1: -a2 - b2 + c1, b1: -b2 + c1, a3: -a2 + b2}
replace {a2, c1, b1} solve for {b2, a1, a3} :
    {a1: -a2 + b1, a3: -a2 - b1 + c1, b2: -b1 + c1}
replace {a2, b2, b1} solve for {c1, a1, a3} :
    {a1: -a2 + b1, a3: -a2 + b2, c1: b1 + b2}
replace {a2, c1, a3} solve for {b2, a1, b1} :
    {a1: -2*a2 - a3 + c1, b1: -a2 - a3 + c1, b2: a2 + a3}
replace {a2, b2, a3} solve for {a1, b1} :
    {a1: -a2 - b2 + c1, b1: -b2 + c1}
replace {a2, a3, b1} solve for {b2, a1, c1} :
    {a1: -a2 + b1, b2: a2 + a3, c1: a2 + a3 + b1}
replace {a1, b2, c1} solve for {a2, a3, b1} :
    {a2: -a1 - b2 + c1, a3: a1 + 2*b2 - c1, b1: -b2 + c1}
replace {a1, c1, b1} solve for {a2, b2, a3} :
    {a2: -a1 + b1, a3: a1 - 2*b1 + c1, b2: -b1 + c1}
replace {b2, a1, b1} solve for {a2, c1, a3} :
    {a2: -a1 + b1, a3: a1 - b1 + b2, c1: b1 + b2}
replace {a1, c1, a3} solve for {a2, b2, b1} :
    {a2: -a1/2 - a3/2 + c1/2, b1: a1/2 - a3/2 + c1/2, b2: -a1/2 + a3/2 + c1/2}
replace {b2, a1, a3} solve for {a2, c1, b1} :
    {a2: -a3 + b2, b1: a1 - a3 + b2, c1: a1 - a3 + 2*b2}
replace {a1, a3, b1} solve for {a2, b2, c1} :
    {a2: -a1 + b1, b2: -a1 + a3 + b1, c1: -a1 + a3 + 2*b1}
replace {a1, c1, a2} solve for {b2, a3, b1} :
    {a3: -a1 - 2*a2 + c1, b1: a1 + a2, b2: -a1 - a2 + c1}
replace {a2, b2, a1} solve for {c1, a3, b1} :
    {b1: a1 + a2, c1: a1 + a2 + b2, a3: -a2 + b2}
replace {a2, a1, b1} solve for {b2, a3} :
    {a3: -a2 - b1 + c1, b2: -b1 + c1}
replace {a2, a1, a3} solve for {b2, c1, b1} :
    {b1: a1 + a2, b2: a2 + a3, c1: a1 + 2*a2 + a3}

如果您想查看值，可以执行Dict(sol).subs(dict(zip(subs, (1,2,3)))) 其中1,2,3 是您要使用的值：

Dict(sol).subs(dict(zip(subs, (1,2,3))))
{b1: 3, b2: 4, c1: 7}