处理大型矩阵（NxM，1K <= N <= 20K＆amp; 10K＆lt; = M <= 200K），我经常需要通过Cython将Numpy矩阵传递给C ++来完成工作，这可以按预期工作没有复制。
然而，有时我需要在C ++中启动和预处理矩阵并将其传递给 Numpy（Python 3.6）。让我们假设矩阵是线性化的（因此大小为N * M，而它是1D矩阵 - col / row major在这里并不重要）。根据此处的信息：exposing C-computed arrays in Python without data copies＆amp;修改它以实现C ++兼容性，我能够通过C ++数组。
问题是如果我想使用标准向量而不是启动数组，我会得到分段错误。例如，考虑以下文件：
fast.h
#include <iostream>
#include <vector>
using std::cout; using std::endl; using std::vector;
int* doit(int length);
fast.cpp
#include "fast.h"
int* doit(int length) {
// Something really heavy
cout << "C++: doing it fast " << endl;
vector<int> WhyNot;
// Heavy stuff - like reading a big file and preprocessing it
for(int i=0; i<length; ++i)
WhyNot.push_back(i); // heavy stuff
cout << "C++: did it really fast" << endl;
return &WhyNot[0]; // or WhyNot.data()
}
faster.pyx
cimport numpy as np
import numpy as np
from libc.stdlib cimport free
from cpython cimport PyObject, Py_INCREF
np.import_array()
cdef extern from "fast.h":
int* doit(int length)
cdef class ArrayWrapper:
cdef void* data_ptr
cdef int size
cdef set_data(self, int size, void* data_ptr):
self.data_ptr = data_ptr
self.size = size
def __array__(self):
print ("Cython: __array__ called")
cdef np.npy_intp shape[1]
shape[0] = <np.npy_intp> self.size
ndarray = np.PyArray_SimpleNewFromData(1, shape,
np.NPY_INT, self.data_ptr)
print ("Cython: __array__ done")
return ndarray
def __dealloc__(self):
print("Cython: __dealloc__ called")
free(<void*>self.data_ptr)
print("Cython: __dealloc__ done")
def faster(length):
print("Cython: calling C++ function to do it")
cdef int *array = doit(length)
print("Cython: back from C++")
cdef np.ndarray ndarray
array_wrapper = ArrayWrapper()
array_wrapper.set_data(length, <void*> array)
print("Ctyhon: array wrapper set")
ndarray = np.array(array_wrapper, copy=False)
ndarray.base = <PyObject*> array_wrapper
Py_INCREF(array_wrapper)
print("Cython: all done - returning")
return ndarray
setup.py
from distutils.core import setup
from distutils.extension import Extension
from Cython.Distutils import build_ext
import numpy
ext_modules = [Extension(
"faster",
["faster.pyx", "fast.cpp"],
language='c++',
extra_compile_args=["-std=c++11"],
extra_link_args=["-std=c++11"]
)]
setup(
cmdclass = {'build_ext': build_ext},
ext_modules = ext_modules,
include_dirs=[numpy.get_include()]
)
如果你用
构建它python setup.py build_ext --inplace
并运行Python 3.6解释器，如果输入以下内容，则在尝试几次后会出现seg错误。
>>> from faster import faster
>>> a = faster(1000000)
Cython: calling C++ function to do it
C++: doing it fast
C++: did it really fast
Cython: back from C++
Ctyhon: array wrapper set
Cython: __array__ called
Cython: __array__ done
Cython: all done - returning
>>> a = faster(1000000)
Cython: calling C++ function to do it
C++: doing it fast
C++: did it really fast
Cython: back from C++
Ctyhon: array wrapper set
Cython: __array__ called
Cython: __array__ done
Cython: all done - returning
Cython: __dealloc__ called
Segmentation fault (core dumped)
需要注意的事项：
faster(1000000)并将结果设置为variable a以外的其他内容，则可以使用此功能。如果您输入较小的数字，例如faster(10)，您可以获得更详细的信息，例如：
Cython: calling C++ function to do it
C++: doing it fast
C++: did it really fast
Cython: back from C++
Ctyhon: array wrapper set
Cython: __array__ called
Cython: __array__ done
Cython: all done - returning
Cython: __dealloc__ called <--- Perhaps this happened too early or late?
*** Error in 'python': double free or corruption (fasttop): 0x0000000001365570 ***
======= Backtrace: =========
More info here ....
令人费解的是，为什么阵列不会发生这种情况？无论如何！
我经常使用矢量，并希望能够在这些场景中使用它们。
答案 0 :(得分：8)
我认为@ FlorianWeimer的答案提供了一个不错的解决方案（分配vector并将其传递到您的C ++函数中）但是应该可以从doit返回一个向量并避免副本通过使用移动构造函数。
from libcpp.vector cimport vector
cdef extern from "<utility>" namespace "std" nogil:
T move[T](T) # don't worry that this doesn't quite match the c++ signature
cdef extern from "fast.h":
vector[int] doit(int length)
# define ArrayWrapper as holding in a vector
cdef class ArrayWrapper:
cdef vector[int] vec
cdef Py_ssize_t shape[1]
cdef Py_ssize_t strides[1]
# constructor and destructor are fairly unimportant now since
# vec will be destroyed automatically.
cdef set_data(self, vector[int]& data):
self.vec = move(data)
# now implement the buffer protocol for the class
# which makes it generally useful to anything that expects an array
def __getbuffer__(self, Py_buffer *buffer, int flags):
# relevant documentation http://cython.readthedocs.io/en/latest/src/userguide/buffer.html#a-matrix-class
cdef Py_ssize_t itemsize = sizeof(self.vec[0])
self.shape[0] = self.vec.size()
self.strides[0] = sizeof(int)
buffer.buf = <char *>&(self.vec[0])
buffer.format = 'i'
buffer.internal = NULL
buffer.itemsize = itemsize
buffer.len = self.v.size() * itemsize # product(shape) * itemsize
buffer.ndim = 1
buffer.obj = self
buffer.readonly = 0
buffer.shape = self.shape
buffer.strides = self.strides
buffer.suboffsets = NULL
然后您应该可以将其用作：
cdef vector[int] array = doit(length)
cdef ArrayWrapper w
w.set_data(array) # "array" itself is invalid from here on
numpy_array = np.asarray(w)
答案 1 :(得分：5)
从doit返回时，WhyNot对象超出范围，并且数组元素被释放。这意味着&WhyNot[0]不再是有效指针。您需要将WhyNot对象存储在其他位置，可能位于调用者提供的位置。
执行此操作的一种方法是将doit拆分为三个函数doit_allocate，它分配向量并返回指向它的指针doit，如前所述（但接收的参数指向预分配向量, and doit_free`的指针，用于释放向量。
这样的事情：
vector<int> *
doit_allocate()
{
return new vector<int>;
}
int *
doit(vector<int> *WhyNot, int length)
{
// Something really heavy
cout << "C++: doing it fast " << endl;
// Heavy stuff - like reading a big file and preprocessing it
for(int i=0; i<length; ++i)
WhyNot->push_back(i); // heavy stuff
cout << "C++: did it really fast" << endl;
return WhyNot->front();
}
void
doit_free(vector<int> *WhyNot)
{
delete WhyNot;
}