c++ - 为什么 numpy 的 cumsum 比手动 C++ 的循环快得多?
问题描述
我试图理解为什么 numpy 的cumsum函数比手动 c++ 循环快得多。我正在使用 pybind 提供从 C++ 代码到 python 的映射。
数字优先:
[nav] In [24]: x = np.arange(100_000, dtype=np.float32)
[nav] In [25]: %timeit np.cumsum(x)
295 µs ± 34.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
[ins] In [26]: %timeit derived.cumsum(x)
9.26 ms ± 137 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
现在如何derived.cumsum(x)
获得?
template <typename Sequence,
typename = std::enable_if_t<std::is_rvalue_reference_v<Sequence&&>>>
inline py::array_t<typename Sequence::value_type> as_pyarray(Sequence&& seq) {
auto size = seq.size();
auto data = seq.data();
auto seq_ptr = std::make_unique<Sequence>(std::move(seq));
auto capsule = py::capsule(seq_ptr.get(), [](void* p) {
std::unique_ptr<Sequence>(reinterpret_cast<Sequence*>(p));
});
seq_ptr.release();
return py::array(size, data, capsule);
}
std::vector<float> cumsum(const std::vector<float>& nums) {
std::vector<float> result(nums.size());
float sum = 0;
for (size_t i = 0; i < nums.size(); ++i) {
sum += nums[i];
result[i] = sum;
}
return result;
}
PYBIND11_MODULE(derived, m) {
m.def("cumsum", [](const std::vector<float>& nums) {
auto result = cumsum(nums);
return as_pyarray(std::move(result));
});
}
as_pyarray
是从这个pybind 问题中获得的,它确保在 python 和 c++ 之间进行交互时零拷贝。
请注意,C++ 代码是使用-O3 -DNDEBUG -fPIC
.
解决方案
问题在于 cumsum 的论点。const std::vector<float>&
仍在做一个相当昂贵的副本。直接py::array_t<float>
作为论据解决了这个问题。
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