OpenCV Python接口与ctypes库之间的互操作性
我正在使用OpenCV 2.3中的Python接口。我有一个用C编写的库,它需要OpenCV对象作为参数,比如OpenCV Python接口与ctypes库之间的互操作性,python,opencv,ctypes,Python,Opencv,Ctypes,我正在使用OpenCV 2.3中的Python接口。我有一个用C编写的库,它需要OpenCV对象作为参数,比如IplImage。像这样: void MyFunction(IplImage* image); 我希望从我的Python代码中调用这个函数。我试过: library.MyFunction(image) 但这给了我一个错误: ArgumentError: argument 1: <type 'exceptions.TypeError'>: Don't know how to
IplImage
。像这样:
void MyFunction(IplImage* image);
我希望从我的Python代码中调用这个函数。我试过:
library.MyFunction(image)
但这给了我一个错误:
ArgumentError: argument 1: <type 'exceptions.TypeError'>: Don't know how to convert parameter 1
如何执行此操作?
cv.iplimage
是由定义的CPython对象。它包装了您需要的指针。有几种方法可以找到这个指针。我将使用CPythonid
返回对象基址这一事实
import cv, cv2
from ctypes import *
您可以使用c\u void\u p
而不是定义IplImage
。我在下面定义它是为了证明指针是正确的
class IplROI(Structure):
pass
class IplTileInfo(Structure):
pass
class IplImage(Structure):
pass
IplImage._fields_ = [
('nSize', c_int),
('ID', c_int),
('nChannels', c_int),
('alphaChannel', c_int),
('depth', c_int),
('colorModel', c_char * 4),
('channelSeq', c_char * 4),
('dataOrder', c_int),
('origin', c_int),
('align', c_int),
('width', c_int),
('height', c_int),
('roi', POINTER(IplROI)),
('maskROI', POINTER(IplImage)),
('imageId', c_void_p),
('tileInfo', POINTER(IplTileInfo)),
('imageSize', c_int),
('imageData', c_char_p),
('widthStep', c_int),
('BorderMode', c_int * 4),
('BorderConst', c_int * 4),
('imageDataOrigin', c_char_p)]
CPython对象:
class iplimage_t(Structure):
_fields_ = [('ob_refcnt', c_ssize_t),
('ob_type', py_object),
('a', POINTER(IplImage)),
('data', py_object),
('offset', c_size_t)]
将示例加载为iplimage
:
data = cv2.imread('lena.jpg') # 512 x 512
step = data.dtype.itemsize * 3 * data.shape[1]
size = data.shape[1], data.shape[0]
img = cv.CreateImageHeader(size, cv.IPL_DEPTH_8U, 3)
cv.SetData(img, data, step)
在CPython中,img
的id
是它的基址。使用ctypes,您可以直接访问此对象的a
字段,即所需的IplImage*
>>> ipl = iplimage_t.from_address(id(img))
>>> a = ipl.a.contents
>>> a.nChannels
3
>>> a.depth
8
>>> a.colorModel
'RGB'
>>> a.width
512
>>> a.height
512
>>> a.imageSize
786432
cv.iplimage
是由定义的CPython对象。它包装了您需要的指针。有几种方法可以找到这个指针。我将使用CPythonid
返回对象基址这一事实
import cv, cv2
from ctypes import *
您可以使用c\u void\u p
而不是定义IplImage
。我在下面定义它是为了证明指针是正确的
class IplROI(Structure):
pass
class IplTileInfo(Structure):
pass
class IplImage(Structure):
pass
IplImage._fields_ = [
('nSize', c_int),
('ID', c_int),
('nChannels', c_int),
('alphaChannel', c_int),
('depth', c_int),
('colorModel', c_char * 4),
('channelSeq', c_char * 4),
('dataOrder', c_int),
('origin', c_int),
('align', c_int),
('width', c_int),
('height', c_int),
('roi', POINTER(IplROI)),
('maskROI', POINTER(IplImage)),
('imageId', c_void_p),
('tileInfo', POINTER(IplTileInfo)),
('imageSize', c_int),
('imageData', c_char_p),
('widthStep', c_int),
('BorderMode', c_int * 4),
('BorderConst', c_int * 4),
('imageDataOrigin', c_char_p)]
CPython对象:
class iplimage_t(Structure):
_fields_ = [('ob_refcnt', c_ssize_t),
('ob_type', py_object),
('a', POINTER(IplImage)),
('data', py_object),
('offset', c_size_t)]
将示例加载为iplimage
:
data = cv2.imread('lena.jpg') # 512 x 512
step = data.dtype.itemsize * 3 * data.shape[1]
size = data.shape[1], data.shape[0]
img = cv.CreateImageHeader(size, cv.IPL_DEPTH_8U, 3)
cv.SetData(img, data, step)
在CPython中,img
的id
是它的基址。使用ctypes,您可以直接访问此对象的a
字段,即所需的IplImage*
>>> ipl = iplimage_t.from_address(id(img))
>>> a = ipl.a.contents
>>> a.nChannels
3
>>> a.depth
8
>>> a.colorModel
'RGB'
>>> a.width
512
>>> a.height
512
>>> a.imageSize
786432
在Ubuntu14.04的cv2中,图像不会以C风格的IplImage返回。它作为numpy.ndarray()返回。使用my_array.data_as(C_void_p)将ndarray()值传递给C函数相对容易。出现此错误
AttributeError:'numpy.ndarray'对象没有属性“data_as”
对于Python2.7.12,在Ubuntu14.04的cv2中,图像不会作为C风格的IplImage返回。它作为numpy.ndarray()返回。使用my_array.data_as(C_void_p)将ndarray()值传递给C函数相对容易。出现此错误AttributeError:'numpy.ndarray'对象没有属性“data_as”
,适用于Python 2.7.12