Python 当表达式包含许多符号时,Sympy-积分速度较慢
假设我有下面的表达式,我想对变量Python 当表达式包含许多符号时,Sympy-积分速度较慢,python,sympy,symbolic-math,Python,Sympy,Symbolic Math,假设我有下面的表达式,我想对变量z从0到L进行积分 import sympy as sp mdot, D, R, alpha, beta, xi, mu0, q, cp, Tin, L = sp.symbols("\dot{m}, D, R, alpha, beta, xi, mu_0, q, c_p, T_in, L", real=True, positive=True, constant=True) z = sp.symbols("z", real=True, positive=True)
z
从0到L
进行积分
import sympy as sp
mdot, D, R, alpha, beta, xi, mu0, q, cp, Tin, L = sp.symbols("\dot{m}, D, R, alpha, beta, xi, mu_0, q, c_p, T_in, L", real=True, positive=True, constant=True)
z = sp.symbols("z", real=True, positive=True)
n = sp.Symbol("n", real=True)
firstexpr = 8 * mdot**2 * R / (sp.pi**2 * D**5) * (alpha + beta * (sp.pi * D * mu0 / (4 * mdot))**xi * (q * z / (mdot * cp) + Tin)**(n * xi)) * (q * z / (mdot * cp) + Tin)
res1 = sp.integrate(firstexpr, (z, 0, L), conds="none")
这将花费永远的时间:我不得不在电脑上10分钟后停止计算,而没有得到答案
如果我重写表达式,使其只包含最少数量的常量符号,并对其进行积分,最后替换原始符号,情况会显著改善:
a = 8 * mdot**2 * R / (sp.pi**2 * D**5)
b = beta * (sp.pi * D * mu0 / (4 * mdot))**xi
c = q / (mdot * cp)
_a, _b, _c = sp.symbols("a, b, c", real=True, positive=True, constant=True)
secondexpr = _a * (alpha + _b * (_c * z + Tin)**(n * xi)) * (_c * z + Tin)
res2 = sp.integrate(secondexpr, (z, 0, L), conds="none")
sp.simplify(res2.subs([(_a, a), (_b, b), (_c, c)]))
为什么在第一种情况下,sympy要花很长时间?在创作我的符号时,我是否错过了一些假设