Memoryview规范-Cython

Question

我有一个给定memoryview向量的函数，我想计算该向量的范数。到目前为止，我通过将memoryview转换为Numpy数组并通过np.sqrt(V.dot(V))计算范数来实现这一点。现在，出于速度原因，我想放弃该步骤，但是该程序在使用以下实现时有时会失败。

cdef do_something(np.double_t[::1] M_mem):
    cdef:
        int i
        np.double_t norm_mv = 0
        np.double_t norm_np = 0
        np.ndarray[np.double_t, ndim=1] V = np.copy(np.asarray(M_mem))

    # Original implementation -- working
    norm_np = np.sqrt(V.dot(V))

    # My failed try with memoryview -- not working
    for i in range(M_mem.shape[0]):
        norm_mv += M_mem[i]**2
    norm_mv = np.sqrt(norm_mv)

    # norm_mv != norm_np

我怀疑原因是Floating Point Arithmetic妨碍了足够大的向量。是否存在一种数值稳定的方法来计算Cython memoryview的范数？

更新

检查后，发现舍入错误可能毫无意义。相反，发生了一件非常奇怪的事情。我的实际功能如下：

@cython.boundscheck(False)
@cython.cdivision(True)
@cython.wraparound(False)
cdef np.double_t[:,::1] GS_coefficients(np.double_t[:,::1] M_mem):
    cdef:
        int n, i, k
        int N_E = M_mem.shape[1]
        np.ndarray[np.double_t, ndim=2] W = np.asarray(M_mem)
        np.ndarray[np.double_t, ndim=2] V = np.copy(W)
        np.double_t[:,::1] G = np.eye(N_E, dtype=np.float64)
        np.longdouble_t norm  = 0 # np.sqrt(V[:,0].dot(V[:,0]))
    for i in range(M_mem.shape[0]):
        norm += M_mem[i,0]**2
    norm = sqrt(norm)
    print("npx: ", np.sqrt(V[:,0].dot(V[:,0]))) # line 1
    print("cp: ", norm) # line 2
    V[:,0] /= norm
    G[0,0] /= norm
    for n in range(1, N_E):
        for i in range(0, n):
            G[n,i] = - (V[:,i].dot(W[:,n]))
            V[:,n] += G[n,i] * V[:,i]
        norm = np.sqrt(V[:,n].dot(V[:,n]))
        V[:,n] /= norm
        for i in range(n+1):
            G[n,i] /= norm
    return G

我插入了print语句来检查与norm的结果“相等”。现在的问题是，上面的代码可以正常工作。但是，当我注释掉第一个打印语句（第1行）时，代码可以正常运行该函数，但在程序执行后不久就会失败。那里发生了什么事？这不只是一个print语句，甚至在操作上也不会影响其他任何内容吗？

更新2

以下是我尝试的一个最小，完整和可验证的示例：

DEF N_E_cpt = 4

cimport cython
cimport numpy as np
import numpy as np
from libc.math cimport sqrt

@cython.boundscheck(False)
@cython.cdivision(True)
@cython.wraparound(False)
cdef np.double_t[:,::1] GS_coefficients(np.double_t[:,::1] M_mem):
    """Writes the coefficients, that the Gram-Schmidt procedure
    provides in a Matrix and retruns it."""
    cdef:
        int n, i, k
        int N_E = M_mem.shape[1]
        np.ndarray[np.double_t, ndim=2] W = np.asarray(M_mem)
        np.ndarray[np.double_t, ndim=2] V = np.copy(W)
        np.double_t[:,::1] G = np.eye(N_E, dtype=np.float64)
        np.longdouble_t norm  = 0 # np.sqrt(V[:,0].dot(V[:,0]))
    for i in range(M_mem.shape[0]):
        norm += M_mem[i,0]**2
    norm = sqrt(norm)
    print("npx: ", np.sqrt(V[:,0].dot(V[:,0]))) # line 1
    print("cp: ", norm) # line 2
    V[:,0] /= norm
    G[0,0] /= norm
    for n in range(1, N_E):
        for i in range(0, n):
            G[n,i] = - (V[:,i].dot(W[:,n]))
            V[:,n] += G[n,i] * V[:,i]
        norm = np.sqrt(V[:,n].dot(V[:,n]))
        V[:,n] /= norm
        for i in range(n+1):
            G[n,i] /= norm
    return G

@cython.boundscheck(False)
@cython.cdivision(True)
@cython.wraparound(False)
cdef np.double_t[:,::1] G_mat(np.double_t[:,::1] M_mem):
    """Calls GS_coefficients and uses the coefficients to calculate
    the entries of the transformation matrix G_ij"""
    cdef:
        np.double_t[:,::1] G_mem = GS_coefficients(M_mem)
        int N_E = G_mem.shape[1]
        np.double_t carr[N_E_cpt][N_E_cpt]
        np.double_t[:,::1] G = carr
        int n, i, j

    # delete lower triangle in G
    G[...] = G_mem
    for i in range(N_E_cpt):
        for j in range(0, i):
            G[i,j] = 0.

    for n in range(1, N_E):
        for i in range(0, n):
            for j in range(0, i+1):
                G[n,j] += G_mem[n,i] * G[i,j]
    return G


def run_test():
    cdef:
        np.double_t[:,::1] A_mem
        np.double_t[:,::1] G
        np.ndarray[np.double_t, ndim=2] A = np.random.rand(400**2, N)
        int N = 4

    A_mem = A
    G = G_mat(A_mem)
    X = np.zeros((400**2, N))
    for i in range(0, N):
        for j in range(0,i+1):
            X[:,i] += G[i,j] * A[:,j]
    print(X)
    print("\n", X.T.dot(X))

run_test()

我认为没有必要了解该代码的功能。对我来说，真正的奥秘在于，为什么print声明会有所作为。

因此，此代码应该采用的是将一组非正交向量作为矩阵向量写成列向量，然后返回一个正交化矩阵，对矩阵向量进行正交化，如下所示：

因此A_ {orthonormal}等效于代码中的X矩阵。当您将正交矩阵的转置矩阵与正交矩阵本身相乘时，您将获得单位矩阵，只要print语句＃line1在其中，就可以得到单位矩阵。删除后，还会出现非对角线的条目，这意味着矩阵甚至不是正交的。为什么？

Answer 1

至少有一个错字

for i in range(M_mem.shape[0]):
    norm += M_mem[i]**2

->

for i in range(M_mem.shape[0]):
    norm_mv += M_mem[i]**2

否则，我推荐以下更惯用的版本：

import numpy as np
cimport numpy as np
from libc.math cimport sqrt

def do_something(double[::1] M_mem):
    cdef:
        int i
        double norm_mv = 0
        double norm_np = 0
        double[::1] V = np.copy(np.asarray(M_mem))

    # Original implementation -- working
    norm_np = np.sqrt(np.dot(V, V))

    # My failed try with memoryview -- not working
    for i in range(M_mem.shape[0]):
        norm_mv += M_mem[i]**2
    norm_mv = sqrt(norm_mv)

    # norm_mv != norm_np
    return norm_np, norm_mv

import 和 cimport numpy并使用libc.math中的标量数学函数而不是NumPy版本。您仍然可以通过用@cython.boundscheck(False)装饰例程来加快代码速度（然后需要cimport cython）。