Sympy TypeError 使用 solve 表示变量与其他变量的关系

Question

按照我在网上找到的一些例子，我可以这样做：

from sympy import var
from sympy import solve
Ldy, Ldz = var('Ldy Ldz')
g, x, y, z = var('g x y z')
xZ, yZ, zZ = var('xZ yZ zZ')
xdd, ydd, zdd = var('xdd ydd zdd')
E1 = z * xdd + (xZ - x) * (g + zdd)
E2 = z * ydd + (yZ - y) * (g + zdd) - Ldy
E3 = -y * xdd + x * ydd - zZ * (g + zdd) + Ldz
out = solve([E1, E2, E3], [xdd, ydd, Ldy])
print(type(xdd))
print("xdd = ", (out[xdd]).factor())

这会产生xdd = (g + zdd)*(x - xZ)/z.

现在，为我自己的方程做：

from sympy import symbols, solve
x, y, z, k12, k26, x0 = symbols("x, y, z, k12, k26, x0")
symbols = x, y, z, k12, k26, x0
eq1 = k12 * x**2 -y
eq2 = k26 * y**3 - z
eq3 = x * 2*y + 6*z - x0
out = solve([eq1, eq2, eq3], [x,y,z])
print("x = ", (out[x]).factor())

给而不是TypeError: list indices must be integers or slices, not Symbol.

我做错了什么？

Answer 1

问题是solve有多种返回类型：有时它返回一个列表，有时返回一个字典，有时返回一个字典列表。输出形式取决于要求解的方程的细节：变量的数量，解的数量。这意味着应该使用list=True或dict=True来强制从solve获得一致的输出。请注意，dict=True意味着输出是字典列表，因为可能存在多个解决方案——这里就是这种情况。在您的示例中：

out = solve([eq1, eq2, eq3], [x,y,z], dict=True)
for sol in out:
print("x = ", sol[x].factor())

指纹

x =  18**(1/3)*((3*x0 - sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(sqrt(6*k12*k26*x0 + 1) + 1)/(18*k12*x0)
x =  -18**(1/3)*((3*x0 + sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(sqrt(6*k12*k26*x0 + 1) - 1)/(18*k12*x0)
x =  -2**(1/3)*((3*x0 - sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) - 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 +1) + 1)/(36*k12*x0)
x =  -2**(1/3)*((3*x0 - sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) + 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 +1) + 1)/(36*k12*x0)
x =  2**(1/3)*((3*x0 + sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) - 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 + 1) - 1)/(36*k12*x0)
x =  2**(1/3)*((3*x0 + sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) + 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 + 1) - 1)/(36*k12*x0)

出于这个原因和其他原因，SymPy 开发人员建议使用 solveset 及其依赖项而不是solve。具体来说，nonlinsolve可以在这里使用：

out = nonlinsolve([eq1, eq2, eq3], [x,y,z])
for sol in out:
print("x = ", sol[x].factor())

哪些打印

x =  -18**(1/3)*((3*x0 + sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(sqrt(6*k12*k26*x0 + 1) - 1)/(18*k12*x0)
x =  18**(1/3)*((3*x0 - sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(sqrt(6*k12*k26*x0 + 1) + 1)/(18*k12*x0)
x =  2**(1/3)*((3*x0 + sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) - 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 + 1) - 1)/(36*k12*x0)
x =  2**(1/3)*((3*x0 + sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) + 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 + 1) - 1)/(36*k12*x0)
x =  -2**(1/3)*((3*x0 - sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) + 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 +1) + 1)/(36*k12*x0)
x =  -2**(1/3)*((3*x0 - sqrt(6*k12*k26*x0 + 1)/(k12*k26) + 1/(k12*k26))/k26)**(2/3)*(3**(2/3) - 3*3**(1/6)*I)*(sqrt(6*k12*k26*x0 +1) + 1)/(36*k12*x0)

求解集及其依赖项的返回类型始终是 SymPy 集。