我有一个推力device_vector分为100个块(但在GPU内存上完全连续),我想删除每个块的最后5个元素,而不必重新分配新的device_vector来复制它。
// Layout in memory before (number of elements in each contiguous subblock listed):
// [ 95 | 5 ][ 95 | 5 ][ 95 | 5 ]........
// Layout in memory after cutting out the last 5 of each chunk (number of elements listed)
// [ 95 ][ 95 ][ 95 ].........
thrust::device_vector v;
// call some function on v;
// so elements 95-99, 195-99, 295-299, etc are removed (assuming 0-based indexing)
如何正确实现此功能?我希望避免在GPU内存中分配新的向量以保存转换。我知道有Thrust模板函数来处理这些类型的操作,但是我很难将它们串在一起。 Thrust提供的东西能做到吗?
答案 0 :(得分:1)
不分配缓冲区mem意味着您必须保留复制顺序,这不能与完全利用GPU硬件并行。
这是使用Thrust和缓冲存储器执行此操作的版本。
它需要Thrust 1.6.0+,因为lambda表达式functor用于迭代器。
#include "thrust/device_vector.h"
#include "thrust/iterator/counting_iterator.h"
#include "thrust/iterator/permutation_iterator.h"
#include "thrust/iterator/transform_iterator.h"
#include "thrust/copy.h"
#include "thrust/functional.h"
using namespace thrust::placeholders;
int main()
{
const int oldChunk = 100, newChunk = 95;
const int size = 10000;
thrust::device_vector<float> v(
thrust::counting_iterator<float>(0),
thrust::counting_iterator<float>(0) + oldChunk * size);
thrust::device_vector<float> buf(newChunk * size);
thrust::copy(
thrust::make_permutation_iterator(
v.begin(),
thrust::make_transform_iterator(
thrust::counting_iterator<int>(0),
_1 / newChunk * oldChunk + _1 % newChunk)),
thrust::make_permutation_iterator(
v.begin(),
thrust::make_transform_iterator(
thrust::counting_iterator<int>(0),
_1 / newChunk * oldChunk + _1 % newChunk))
+ buf.size(),
buf.begin());
return 0;
}
我认为由于使用了mod运算符%
,上述版本可能无法达到最高性能。为了获得更高的性能,您可以考虑使用cuBLAS函数cublas_geam()
float alpha = 1;
float beta = 0;
cublasSgeam(handle, CUBLAS_OP_N, CUBLAS_OP_N,
newChunk, size,
&alpha,
thrust::raw_pointer_cast(&v[0]), oldChunk,
&beta,
thrust::raw_pointer_cast(&v[0]), oldChunk,
thrust::raw_pointer_cast(&buf[0]), newChunk);