我在用Python编写一个薛定谔方程的模拟，但在使用Numba库中的 @njit来优化函数时遇到了问题。

这是代码。我只是想求解一个离散微分方程。

import numpy as np
from scipy.integrate import solve_ivp as ivp
from numba import njit

m = 1/2             #Un. Geom.
h_bar = 1           #Unidades geometrizadas
infty = 5000        #Esto sirve como infinito


# =============================================================================
# # Parámetros
# =============================================================================

z0 = 0
zf = 1

omega = 1
t0 = 0
tf = 1
a = 1
b = 3

m = 50


# =============================================================================
# # Funciones
# =============================================================================

def f(z):
    return 2**0.5*np.sin(2*np.pi*z)

def Ll(a,b,t): 
    return a + b*t

def dLl(b):
    return b

def Ls(a,b,t):
    return a + b*np.sin(omega*t)

def dLs(b,t):
    return omega*b*np.cos(omega*t)


@njit
def res(t,phi):
    global h
    dphi = np.zeros_like(z, dtype = complex)       #recordar que para dphi0 y dphi-1 son ceros

    for i in range(0,m):
        if i != 0 and i != m-1:
            #print(i,len(dphi))
            #print('phi_i= ', phi[i], 'phi_{i-1}= ', phi[i-1], 'phi_{i+1}= ',phi[i+1],'||',dphi[i])
            dphi[i] = 1/h**2*1j/Ll(a,b,t)**2*(phi[i+1]-2*phi[i]+phi[i-1]) + 1/2/h*i*dLl(b)/Ll(a,b,t)*(phi[i+1] - phi[i-1])
            # dphi[i] = 1/h**2*1j/L**2*(phi[i+1]-2*phi[i]+phi[i-1]) + 1/2/h*i*dLs(b,t)/Ls(a,b,t)(phi[i+1] - phi[i-1]) #esta es para el sin

        else: 
            if i == 0: 
                print(i,phi[i])

                dphi[i] = 1/h**2*1j/Ll(a,b,t)**2*(phi[i+1]-2*phi[i]+ 0) + 1/2/h*i*dLl(b)/Ll(a,b,t)*(phi[i+1] - 0)

            if  i == m-1:
                print(i,phi[i])

                dphi[i] = 1/h**2*1j/Ll(a,b,t)**2*( 0 -2*phi[i]+phi[i-1]) + 1/2/h*i*dLl(b)/Ll(a,b,t)*( 0 - phi[i-1])


    print(dphi)
    return dphi   

# =============================================================================
# # Resolución numérica
#  Esto se hace con ivp
# =============================================================================

h = 1/m
z = np.linspace(z0+h,zf-h,m)
z = np.transpose(z)
print('tamaño z  =  ', len(z))

y0 = f(z) 
t = np.linspace(t0,tf,m)

sol = ivp(res,(t0,tf), y0, t_eval=t)
print(sol.y)

我的函数在没有 @njit的情况下可以正常工作，但当我取消注释时，得到的结果是 "Cannot determine Numba type of ”。我一直在找答案，唯一能想到的就是移除Numba，但计算时间很长（31 s）。

在对所有被调用的函数实现 @njit之后，我得到这个信息：

TypingError: No implementation of function Function(<function zeros_like at 0x000001EAA78FB160>) found for signature:

zeros_like(readonly array(float64, 1d, C), dtype=Literal[str](complex128))

There are 2 candidate implementations:
      - Of which 2 did not match due to:
      Overload of function 'zeros_like': File: numba\core\typing\npydecl.py: Line 525.
        With argument(s): '(readonly array(float64, 1d, C), dtype=unicode_type)':
       No match.

During: resolving callee type: Function(<function zeros_like at 0x000001EAA78FB160>)
During: typing of call at C:\...\untitled0.py (66)

我一直在查阅Numba的文档，据说函数np.zeros_like只包含包中的前两个参数（我现在只用这两个），但我仍然收到这个信息。我试着在函数中引入z，但这也不起作用。

解决方案

正如你在问题的编辑中所述，首先需要用 njit 给函数 L1 和 dL1 进行装饰。

使用numba 0.62.1时我得到一个不同的错误：

There are 2 candidate implementations:
  - Of which 2 did not match due to:
  Overload in function 'ol_np_zeros_like': File: numba/np/arrayobj.py: Line 4539.
    With argument(s): '(readonly array(float64, 1d, C), dtype=Function(<class 'complex'>))':
   Rejected as the implementation raised a specific error:
     TypingError: Failed in nopython mode pipeline (step: nopython frontend)
   No implementation of function Function(<built-in function empty_like>) found for signature:

    >>> empty_like(readonly array(float64, 1d, C), dtype=Function(<class 'complex'>))

无论哪种情况，错误都指向对 numpy.zeros_like 的使用，特别是 dtype 参数。

这可以通过把第 dphi = np.zeros_like(z, dtype = complex) 行中的 complex 改为 "complex" 或 np.complex128 来修复。