python numpy`函数数组(lambda函数)

为了计算微分方程组的解,我定义了一个python类。 所以我定义了一个名为Rhs(right and side)的类,它应该表示dy/dt(I-th)的右边和右边。这个类包含一个浮点值(initial time,initial value,最后一次)和一个函数(函数数组)为了定义这个数组,我简单地定义了3个lambda函数,表示方程(I)并创建一个np.数组此函数的

func1 = lambda t,u : 10   * (u[1] - u[0])
func2 = lambda t,u : 28   * u[0] - u[1] - u[0] * u[2]  
func3 = lambda t,u : -8/3 * u[2] + u[0]*u[1]

然后以这种方式传递给rhs类:

func = np.array([func1,func2,func3])
y0 = np.array([1.,0.,0.])
problem3 = rhs.Rhs(func,0.0,100.0,y0,1000) 

Rhs等级如下:

    class Rhs:
    def __init__(self, fnum : np.ndarray , t0: np.float, tf: np.float, y0 : np.array, n: int , fanal = None ):
          self.func = fnum
          Rhs.solution  = fanal
          self.t0   = t0
          self.tf   = tf
          self.n    = n
          self.u0   = y0
   def createArray(self):
      '''   
         Create the Array time and f(time) 
         - the time array can be create now 
         - the solution array can be just initialize with the IV
      ''' 
      self.t = np.linspace(self.t0, self.tf, self.n )
      self.u = np.array([self.u0  for i in range(self.n) ])


      return self.t,self.u   

   def f(self,ti,ui):
      return self.func(ti,ui)     

   def Df(self,ti,ui):
      eps = 10e-6
      return ((self.func(ti,ui)+eps) - self.f(ti,ui))/eps

f

class Explicit:
     def __init__(self, dydt: rhs.Rhs, save : bool=True, _print: bool=False,  filename : str=None):
          self.dydt   = dydt
          self.dt     = (dydt.tf-dydt.t0)/dydt.n
          self._print = _print 

   def solve(self):
      self.time, self.u = self.dydt.createArray() 
      for i in range(len(self.time)-1): 
         self.u[i+1] = self.u[i] + self.dt*self.dydt.f(self.time[i],self.u[i])


      Explicit.solved = True 

      print('here')

      if self._print:
          with open(filename) as f:
              print('here')
              for i in range(len(self.u)):      
                f.write('%.4f %4f %4f %4f' %(self.time ,self.u[0,i], self.u[1], self.u[2]))


      if self.save:
         return self.time,self.u

这里的问题是:正确的方法是将向量u(形状=1000,3)传递给函数(为了使用3函数来应用于lambda函数系统中的3矢量索引……)我不理解为什么C++中没有这个问题。 请看这里: all the class hierarchy 我不知道用哪种方法计算这个东西

Traceback (most recent call last):
  File "drive.py", line 94, in <module>
    main()
  File "drive.py", line 63, in main
    fet,feu = fwdeuler_p1.solve()
  File "/home/marco/Programming/Python/Numeric/OdeSystem/euler.py", line 77, in solve
    self.u[i+1] = self.u[i] + self.dt*self.dydt.f(self.time[i],self.u[i])
  File "/home/marco/Programming/Python/Numeric/OdeSystem/rhs.py", line 44, in f
    return self.func(ti,ui)     
TypeError: 'numpy.ndarray' object is not callable

编辑 谢谢你的回复。。但不幸的是没有:( 这是错误消息:

drive.py:31: RuntimeWarning: overflow encountered in double_scalars
  func2 = lambda t,u : 28   * u[0] - u[1] - u[0] * u[2]
drive.py:32: RuntimeWarning: overflow encountered in double_scalars
  func3 = lambda t,u : -8/3 * u[2] + u[0]*u[1]
drive.py:31: RuntimeWarning: invalid value encountered in double_scalars
  func2 = lambda t,u : 28   * u[0] - u[1] - u[0] * u[2]
/home/marco/Programming/Python/Numeric/OdeSystem/euler.py:77: RuntimeWarning: invalid value encountered in add
  self.u[i+1] = self.u[i] + self.dt*self.dydt.f(self.time[i],self.u[i])

非常感谢你的帮助!你有别的解决办法吗?

u[0],u[1],u[2] 在方程(函数数组)中,顺便说一下,如果我增加步长(减少增量)