won*_*jun 8 python ctypes numpy cython f2py
我可以将一维数组传递给python,如下所示.我想知道我是否可以通过使用ctypes,numpy将c ++双指针数组传递给python.
TEST.CPP:
#include <stdio.h>
extern "C" void cfun(const void * indatav, int rowcount, int colcount, void * outdatav);
void cfun(const void * indatav, int rowcount, int colcount, void * outdatav) {
//void cfun(const double * indata, int rowcount, int colcount, double * outdata) {
const double * indata = (double *) indatav;
double * outdata = (double *) outdatav;
int i;
puts("Here we go!");
for (i = 0; i < rowcount * colcount; ++i) {
outdata[i] = indata[i] * 4;
}
puts("Done!");
}
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test.py:
import numpy
import ctypes
indata = numpy.ones((5,6), dtype=numpy.double)
outdata = numpy.zeros((5,6), dtype=numpy.double)
lib = ctypes.cdll.LoadLibrary('./ctest.so')
fun = lib.cfun
# Here comes the fool part.
#fun(ctypes.c_void_p(indata.ctypes.data), ctypes.c_void_p(outdata.ctypes.data))
fun(ctypes.c_void_p(indata.ctypes.data), ctypes.c_int(5), ctypes.c_int(6),
ctypes.c_void_p(outdata.ctypes.data))
print 'indata: %s' % indata
print 'outdata: %s' % outdata
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这是一种方式.我没有看到使用numpy和双**的好方法.
test.cpp(Windows)
#include <stdio.h>
extern "C" __declspec(dllexport) void cfun(const double ** indata, int rowcount, int colcount, double ** outdata) {
for (int i = 0; i < rowcount; ++i) {
for (int j = 0; j < colcount; ++j) {
outdata[i][j] = indata[i][j] * 4;
}
}
}
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test.py
import numpy
import ctypes
# Allocate array of double*
indata = (ctypes.POINTER(ctypes.c_double) * 5)()
for i in range(5):
# Allocate arrays of double
indata[i] = (ctypes.c_double * 6)()
for j in range(6):
indata[i][j] = 1.0
outdata = (ctypes.POINTER(ctypes.c_double) * 5)()
for i in range(5):
outdata[i] = (ctypes.c_double * 6)()
for j in range(6):
outdata[i][j] = 1.0
lib = ctypes.cdll.LoadLibrary('test')
fun = lib.cfun
def dump(a,rows,cols):
for i in range(rows):
for j in range(cols):
print a[i][j],
print
dump(indata,5,6)
fun(ctypes.byref(indata),5,6,ctypes.byref(outdata))
dump(outdata,5,6)
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产量
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
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