Cython:创建没有NumPy数组的memoryview吗? [英] Cython: Create memoryview without NumPy array?

问题描述

由于我发现内存视图方便且快速，因此我尝试避免在cython中创建NumPy数组，并使用给定数组的视图.但是，有时无法避免的是，不更改现有阵列而是创建一个新阵列.在上层函数中，这不是很明显，但是在通常称为子例程中，它是明显的.考虑以下功能

#@cython.profile(False)
@cython.boundscheck(False)
@cython.wraparound(False)
@cython.nonecheck(False)
cdef double [:] vec_eq(double [:] v1, int [:] v2, int cond):
    ''' Function output corresponds to v1[v2 == cond]'''
    cdef unsigned int n = v1.shape[0]
    cdef unsigned int n_ = 0
    # Size of array to create
    cdef size_t i
    for i in range(n):
        if v2[i] == cond:
            n_ += 1
    # Create array for selection
    cdef double [:] s = np.empty(n_, dtype=np_float) # Slow line
    # Copy selection to new array
    n_ = 0
    for i in range(n):
        if v2[i] == cond:
            s[n_] = v1[i]
            n_ += 1
    return s

分析告诉我，这里有一定的速度

我所能做的就是调整函数，有时导致，例如，计算出该向量的均值，有时是求和.所以我可以重写它，以便求和或取平均值.但是没有一种方法可以直接以很少的开销直接创建内存视图，动态地定义大小.类似于首先使用malloc创建交流缓冲，然后在函数将缓冲转换为视图的最后，传递指针并大步前进. >

修改1: 也许对于简单的情况，调整功能e. G.这样是可以接受的方法.我只添加了一个参数并求和/取平均值.这样，我不必创建数组，并且可以轻松处理内部函数malloc.这不会更快，是吗?

# ...
cdef double vec_eq(double [:] v1, int [:] v2, int cond, opt=0):
    # additional option argument
    ''' Function output corresponds to v1[v2 == cond].sum() / .mean()'''
    cdef unsigned int n = v1.shape[0]
    cdef int n_ = 0
    # Size of array to create
    cdef Py_ssize_t i
    for i in prange(n, nogil=True):
        if v2[i] == cond:
            n_ += 1
    # Create array for selection
    cdef double s = 0
    cdef double * v3 = <double *> malloc(sizeof(double) * n_)
    if v3 == NULL:
        abort()
    # Copy selection to new array
    n_ = 0
    for i in range(n):
        if v2[i] == cond:
            v3[n_] = v1[i]
            n_ += 1
    # Do further computation here, according to option
    # Option 0 for the sum
    if opt == 0:
        for i in prange(n_, nogil=True):
            s += v3[i]
        free(v3)
        return s
    # Option 1 for the mean
    else:
        for i in prange(n_, nogil=True):
            s += v3[i]
        free(v3)
        return s / n_
    # Since in the end there is always only a single double value, 
    # the memory can be freed right here

解决方案

不知道如何处理cpython数组，因此我最终通过自制的内存视图"( 解决方案

Didn't know, how to deal with cpython arrays, so I solved this finally by a self made 'memory view', as proposed by fabrizioM. Wouldn't have thought that this would work. Creating a new np.array in a tight loop is pretty expensive, so this gave me a significant speed up. Since I only need a 1 dimensional array, I didn't even had to bother with strides. But even for a higher dimensional arrays, I think this could go well.

cdef class Vector:
    cdef double *data
    cdef public int n_ax0

    def __init__(Vector self, int n_ax0):
        self.data = <double*> malloc (sizeof(double) * n_ax0)
        self.n_ax0 = n_ax0

    def __dealloc__(Vector self):
        free(self.data)

...
#@cython.profile(False)
@cython.boundscheck(False)
cdef Vector my_vec_func(double [:, ::1] a, int [:] v, int cond, int opt):
    # function returning a Vector, which can be hopefully freed by del Vector
    cdef int vecsize
    cdef size_t i
    # defs..
    # more stuff...
    vecsize = n
    cdef Vector v = Vector(vecsize)

    for i in range(vecsize):
        # computation
        v[i] = ...

    return v

...
vec = my_vec_func(...
ptr_to_data = vec.data
length_of_vec = vec.n_ax0

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