C\C++ 提升Python效率

gapore

有没有大佬懂cython的啊？我有段程序想用cython加速一下，但是自己不懂c语言，胡乱编的效果有限

数据来源代码（Python）如下

1. import numpy as np
   
2. import scipy.linalg as linalg
   
3. 
   
4. N = 2328
   
5. Y = np.random.rand(N)
   
6. L = 200   # int in range(2,N/2)
   
7. K = N - L + 1
   
8. 
   
9. # Embedding
   
10. X = linalg.hankel(Y,np.zeros(K))
    
11. X = X[:-K+1,:]
    
12. 
    
13. # Singular value decomposition (SVD)
    
14. Xt = X.T
    
15. _,sig,Vt = linalg.svd(X)
    
16. Sig = linalg.diagsvd(sig,L,L)
    
17. r = np.linalg.matrix_rank(X)
    
18. Vt1 = Vt[:r,:]
    
19. Sig1 = Sig[:r,:r]
    
20. B1 = Sig1.dot(Vt1)
    
21. B1t = B1.T

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我想用最小一乘法求解矩阵 A ，使得 B1t * A = Xt ，编的python代码如下

1. # coding = utf-8
   
2. import numpy as np
   
3. cimport cython
   
4. 
   
5. def FitLadReg(double[:] endog,double[:,:] exog):
   
6.    
   
7.     exog_rank = np.linalg.matrix_rank(exog)
   
8.     beta = np.ones(exog_rank)
   
9.     xstar = exog
   
10. 
    
11.     n_iter = 0
    
12.     diff = 10
    
13.     cycle = False
    
14.     history = []
    
15.     while n_iter < 1000 and diff > 1e-6 and not cycle:
    
16.         n_iter += 1
    
17.         beta0 = beta
    
18.         xtx = np.dot(xstar.T, exog)   
    
19.         xty = np.dot(xstar.T, endog)
    
20.         beta = np.dot(np.linalg.pinv(xtx), xty)
    
21.         resid = endog - np.dot(exog, beta)
    
22. 
    
23.         mask = np.abs(resid) < .000001
    
24.         resid[mask] = ((resid[mask] >= 0) * 2 - 1) * .000001
    
25.         resid = np.where(resid < 0, 0.5 * resid, 0.5 * resid)
    
26.         resid = np.abs(resid)
    
27.         xstar = exog / resid[:, np.newaxis]
    
28.         diff = np.max(np.abs(beta - beta0))
    
29.         history.append(beta)
    
30. 
    
31.         if (n_iter >= 300) and (n_iter % 100 == 0):
    
32.             # check for convergence circle, shouldn't happen
    
33.             for ii in range(2, 10):
    
34.                 if np.all(beta == history[-ii]):
    
35.                     cycle = True
    
36.                     break
    
37. 
    
38.     return history

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编译后调用代码如下

1. from FitLadReg import FitLadReg
   
2. A = np.zeros((r,L))
   
3. for i in range(L):
   
4.     A[:,i] = FitLadReg(Xt[:,i],B1t)[0]

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FitLadReg函数cython编译前后效率差不多，请问要怎么改进函数代码呢？谢谢！！！

TCY

你用numba吧……