C\C++ 提升Python效率
本帖最后由 gapore 于 2018-12-7 10:37 编辑有没有大佬懂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import numpy as np
import scipy.linalg as linalg
N = 2328
Y = np.random.rand(N)
L = 200 # int in range(2,N/2)
K = N - L + 1
# Embedding
X = linalg.hankel(Y,np.zeros(K))
X = X[:-K+1,:]
# Singular value decomposition (SVD)
Xt = X.T
_,sig,Vt = linalg.svd(X)
Sig = linalg.diagsvd(sig,L,L)
r = np.linalg.matrix_rank(X)
Vt1 = Vt[:r,:]
Sig1 = Sig[:r,:r]
B1 = Sig1.dot(Vt1)
B1t = B1.T
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# coding = utf-8
import numpy as np
cimport cython
def FitLadReg(double[:] endog,double[:,:] exog):
exog_rank = np.linalg.matrix_rank(exog)
beta = np.ones(exog_rank)
xstar = exog
n_iter = 0
diff = 10
cycle = False
history = []
while n_iter < 1000 and diff > 1e-6 and not cycle:
n_iter += 1
beta0 = beta
xtx = np.dot(xstar.T, exog)
xty = np.dot(xstar.T, endog)
beta = np.dot(np.linalg.pinv(xtx), xty)
resid = endog - np.dot(exog, beta)
mask = np.abs(resid) < .000001
resid = ((resid >= 0) * 2 - 1) * .000001
resid = np.where(resid < 0, 0.5 * resid, 0.5 * resid)
resid = np.abs(resid)
xstar = exog / resid[:, np.newaxis]
diff = np.max(np.abs(beta - beta0))
history.append(beta)
if (n_iter >= 300) and (n_iter % 100 == 0):
# check for convergence circle, shouldn't happen
for ii in range(2, 10):
if np.all(beta == history[-ii]):
cycle = True
break
return history
编译后调用代码如下from FitLadReg import FitLadReg
A = np.zeros((r,L))
for i in range(L):
A[:,i] = FitLadReg(Xt[:,i],B1t)
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！！！
你用numba吧……
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