我无法弄清楚为什么以下代码(chi2距离)在使用OMP编译时需要更长的时间。在question之后我释放了GIL,但仍然没有任何改进。
np::ndarray additive_chi2_kernel(const np::ndarray& _h0,
const np::ndarray& _h1) {
auto dtype = np::dtype::get_builtin<float>();
auto h0 = _h0.astype(dtype);
auto h1 = _h1.astype(dtype);
enter code here
assert (h0.get_nd() == 2 && h1.get_nd() == 2);
float* ptr_h0 = reinterpret_cast<float*>(h0.get_data());
float* ptr_h1 = reinterpret_cast<float*>(h1.get_data());
int M0 = h0.shape(0);
int M1 = h1.shape(0);
int N = h1.shape(1);
float* result = new float[M0*M1]();
auto save_state = PyEval_SaveThread();
#pragma omp parallel
for(int m0 = 0; m0 < M0; ++m0) {
for(int m1 = 0; m1 < M1; ++m1) {
float error = 0;
for(int i = 0; i < N; ++i) {
float sum = ptr_h0[m0*N+i] + ptr_h1[m1*N+i];
if (sum != 0){
float diff = ptr_h0[m0*N+i] - ptr_h1[m1*N+i];
error += (diff*diff/sum);
}
}
result[m0*M1+m1] = error;
}
}
PyEval_RestoreThread(save_state);
np::ndarray D = np::from_data(result,np::dtype::get_builtin<float>(),
py::make_tuple(M0,M1),py::make_tuple(sizeof(float)*M1,
sizeof(float)), py::object());
return D;
}
Boost包装是:
BOOST_PYTHON_MODULE(vgic) {
// Initialize numpy
PyEval_InitThreads();
np::initialize();
py::def("additive_chi2_kernel",additive_chi2_kernel);
}
编译器标志:-std=c++11 -Wall -fopenmp -O3 -fPIC
所有线程都应该处于活动状态
PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
20706 memecs 20 0 3980080 818004 42880 R 3179 0.6 7:34.04 python
但运行时间高于单个核心。对于M0 = 100,M1 = 1000,N = 1000,我得到
OMP: 2.214 seconds
SINGLE-CORE: 1.175 seconds.
可能出现的问题?
答案 0 :(得分:3)
您应该使用#pragma omp parallel for
来划分线程之间的外部循环。 #pragma omp parallel
只生成一组线程,但每个线程都在计算所有内容。