Sympy似乎随着数字的增加而崩溃

2 回答

• 2

  这是一个浮点精度问题 . solve 默认情况下将解决方案插入到原始等式中并对其进行评估（使用浮点算法）以便对错误解决方案进行排序 . 您可以通过设置 check=False 来禁用此功能 . 例如，对于Hugh Bothwell的代码

  for fv in range(1870, 1875, 1):
      sols = sp.solve(eq.subs({FV:fv}), check=False)
      print("{}: {}".format(fv, sols))
  

  这使

  1870: [66.6116466112007]
  1871: [66.6386438584579]
  1872: [66.6656266802551]
  1873: [66.6925950919998]
  1874: [66.7195491090752]
  

  回复于 2024-05-12T16:10:39+08:00
• 1

  我没有答案，但我确实有一个更简单的演示案例;-)

  import sympy as sp
  
  FV, n = sp.symbols("FV n")
  eq = sp.Eq(FV, sp.S("500 * 1.02 ** n"))
  
  # see where it breaks
  for fv in range(1870, 1875, 1):
      sols = sp.solve(eq.subs({FV:fv}))
      print("{}: {}".format(fv, sols))
  

  哪个产生

  1870: [66.6116466112007]
  1871: [66.6386438584579]
  1872: []
  1873: []
  1874: []
  

  猜测这是准确度下降到足以找不到 n 的可验证解决方案的地方？

  ---

  此外，虽然我在这里做了一个相当广泛的重写，你可能会觉得有用 . 它与您的代码几乎完全相同，但是以更加松散耦合的方式 .

  import sympy as sp
  
  class Equation:
      def __init__(self, label, equality_str, eq="=="):
          self.label = label
          # parse the equality
          lhs, rhs = equality_str.split(eq)
          self.equality = sp.Eq(sp.sympify(lhs), sp.sympify(rhs))
          # index free variables by name
          self.vars = {var.name: var for var in self.equality.free_symbols}
  
      def prompt_for_values(self):
          # show variables to be entered
          var_names = sorted(self.vars, key=str.lower)
          print("\nFree variables are: " + ", ".join(var_names))
          print("Enter a value for all but one (press Enter to skip):")
          # prompt for values by name
          var_values = {}
          for name in var_names:
              value = input("Value of {}: ".format(name)).strip()
              if value:
                  var_values[name] = sp.sympify(value)
          # convert names to Sympy variable references
          return {self.vars[name]:value for name,value in var_values.items()}
  
      def solve(self):
          values = self.prompt_for_values()
          solutions = sp.solve(self.equality.subs(values))
          # remove complex answers
          solutions = [sol.evalf() for sol in solutions if sol.is_real]
          return solutions
  
      def __str__(self):
          return str(self.equality)
  
  # Define some equations!
  equations = [
      Equation("Time value of money (discrete)",   "FV == PV * (1 + i) ** n"),
      Equation("Time value of money (continuous)", "FV == PV * exp(i * n)"  )
  ]
  
  # Create menu
  menu_lo = 1
  menu_hi = len(equations) + 1
  menu_prompt = "\n".join(
      [""]
      + ["{}: {}".format(i, eq.label) for i, eq in enumerate(equations, 1)]
      + ["{}: Exit".format(menu_hi)]
      + ["? "]
  )
  
  def get_int(prompt, lo=None, hi=None):
      while True:
          try:
              value = int(input(prompt))
              if (lo is None or lo <= value) and (hi is None or value <= hi):
                  return value
          except ValueError:
              pass
  
  def main():
      while True:
          choice = get_int(menu_prompt, menu_lo, menu_hi)
          if choice == menu_hi:
              print("Goodbye!")
              break
          else:
              solutions = equations[choice - 1].solve()
              num = len(solutions)
              if num == 0:
                  print("No solutions found")
              elif num == 1:
                  print("1 solution found: " + str(solutions[0]))
              else:
                  print("{} solutions found:".format(num))
                  for sol in solutions:
                      print(sol)
  
  if __name__ == "__main__":
      main()
  

  回复于 2024-05-12T16:10:39+08:00