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我有一个函数，它给出了一个 memoryview 向量，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)). 现在，出于速度原因，我想摆脱该步骤，但该程序在某些时候会因以下实现而失败。

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

我怀疑这是因为浮点运算妨碍了足够大的向量。是否有一种数值稳定的方法来计算 Cython 内存视图的范数？

更新

检查后发现舍入误差可能没有意义。相反，发生了一件非常奇怪的事情。我的实际功能如下所示：

@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语句来检查结果“有多相等” norm。奇怪的是，现在一切正常，就像上面的代码一样。但是，当我注释掉第一个打印语句（第 1 行）时，代码运行良好，但在程序中不久就失败了。那里发生了什么？这不只是一个print不应该影响其他任何操作的声明吗？

更新 2

这是我尝试一个最小的、完整的和可验证的例子：

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作用。对我来说，真正的奥秘就是为什么这种print说法会有所作为。

因此，这段代码应该采取的是将一组非正交向量写成矩阵 A 中的列向量，并返回一个正交化矩阵，该矩阵对向量集进行正交归一化，如下所示：

代码风格中的正交归一化公式

所以 A_{orthonormal} 等价于代码中的 X 矩阵。当您将正交矩阵的转置与正交矩阵本身相乘时，您会得到单位矩阵，只要print语句 # line1 在那里，您就会得到统一矩阵。一旦你删除它，你也会得到非对角线条目，这意味着矩阵甚至不是正交的。为什么？

4

1 回答 1

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，我推荐下面更惯用的版本：

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 并使用标量数学函数 fromlibc.math而不是 NumPy 版本。@cython.boundscheck(False)您仍然可以通过使用（您需要）装饰例程来加快代码速度cimport cython。

于 2018-10-12T12:50:28.940 回答