ami*_*mir 3 sorting cuda thrust
我有一个5000x500矩阵,我想用cuda分别对每行进行排序.我可以使用arrayfire,但这只是关于thrust :: sort的for循环,这应该不高效.
https://github.com/arrayfire/arrayfire/blob/devel/src/backend/cuda/kernel/sort.hpp
for(dim_type w = 0; w < val.dims[3]; w++) {
dim_type valW = w * val.strides[3];
for(dim_type z = 0; z < val.dims[2]; z++) {
dim_type valWZ = valW + z * val.strides[2];
for(dim_type y = 0; y < val.dims[1]; y++) {
dim_type valOffset = valWZ + y * val.strides[1];
if(isAscending) {
thrust::sort(val_ptr + valOffset, val_ptr + valOffset + val.dims[0]);
} else {
thrust::sort(val_ptr + valOffset, val_ptr + valOffset + val.dims[0],
thrust::greater<T>());
}
}
}
}
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有没有办法融合推力操作,以便排序并行?实际上,我正在寻找的是融合循环迭代的通用方法.
Rob*_*lla 13
我可以想到两种可能性,其中一种可能是由@JaredHoberock提出的.我不知道在推力中融合for循环迭代的一般方法,但第二种方法是更通用的方法.我的猜测是,在这种情况下,第一种方法将是两种方法中更快的方法.
使用矢量化排序.如果要由嵌套for循环排序的区域不重叠,则可以使用此处讨论的2个背靠背稳定排序操作进行矢量化排序.
推力v1.8(可与CUDA 7 RC一起使用,或通过直接从推力github存储库下载包括支持嵌套推力算法,通过在自定义仿函数中包含推力算法调用传递给另一个推力算法.如果使用该thrust::for_each
操作选择您需要执行的各种排序,您可以使用单个推力算法调用来运行这些排序,方法是将thrust::sort
操作包含在您传递给的仿函数中thrust::for_each
.
这是3种方法之间的完全比较:
在每种情况下,我们分别对每组1000个整数的16000组进行排序.
$ cat t617.cu
#include <thrust/device_vector.h>
#include <thrust/device_ptr.h>
#include <thrust/host_vector.h>
#include <thrust/sort.h>
#include <thrust/execution_policy.h>
#include <thrust/generate.h>
#include <thrust/equal.h>
#include <thrust/sequence.h>
#include <thrust/for_each.h>
#include <iostream>
#include <stdlib.h>
#define NSORTS 16000
#define DSIZE 1000
int my_mod_start = 0;
int my_mod(){
return (my_mod_start++)/DSIZE;
}
bool validate(thrust::device_vector<int> &d1, thrust::device_vector<int> &d2){
return thrust::equal(d1.begin(), d1.end(), d2.begin());
}
struct sort_functor
{
thrust::device_ptr<int> data;
int dsize;
__host__ __device__
void operator()(int start_idx)
{
thrust::sort(thrust::device, data+(dsize*start_idx), data+(dsize*(start_idx+1)));
}
};
#include <time.h>
#include <sys/time.h>
#define USECPSEC 1000000ULL
unsigned long long dtime_usec(unsigned long long start){
timeval tv;
gettimeofday(&tv, 0);
return ((tv.tv_sec*USECPSEC)+tv.tv_usec)-start;
}
int main(){
cudaDeviceSetLimit(cudaLimitMallocHeapSize, (16*DSIZE*NSORTS));
thrust::host_vector<int> h_data(DSIZE*NSORTS);
thrust::generate(h_data.begin(), h_data.end(), rand);
thrust::device_vector<int> d_data = h_data;
// first time a loop
thrust::device_vector<int> d_result1 = d_data;
thrust::device_ptr<int> r1ptr = thrust::device_pointer_cast<int>(d_result1.data());
unsigned long long mytime = dtime_usec(0);
for (int i = 0; i < NSORTS; i++)
thrust::sort(r1ptr+(i*DSIZE), r1ptr+((i+1)*DSIZE));
cudaDeviceSynchronize();
mytime = dtime_usec(mytime);
std::cout << "loop time: " << mytime/(float)USECPSEC << "s" << std::endl;
//vectorized sort
thrust::device_vector<int> d_result2 = d_data;
thrust::host_vector<int> h_segments(DSIZE*NSORTS);
thrust::generate(h_segments.begin(), h_segments.end(), my_mod);
thrust::device_vector<int> d_segments = h_segments;
mytime = dtime_usec(0);
thrust::stable_sort_by_key(d_result2.begin(), d_result2.end(), d_segments.begin());
thrust::stable_sort_by_key(d_segments.begin(), d_segments.end(), d_result2.begin());
cudaDeviceSynchronize();
mytime = dtime_usec(mytime);
std::cout << "vectorized time: " << mytime/(float)USECPSEC << "s" << std::endl;
if (!validate(d_result1, d_result2)) std::cout << "mismatch 1!" << std::endl;
//nested sort
thrust::device_vector<int> d_result3 = d_data;
sort_functor f = {d_result3.data(), DSIZE};
thrust::device_vector<int> idxs(NSORTS);
thrust::sequence(idxs.begin(), idxs.end());
mytime = dtime_usec(0);
thrust::for_each(idxs.begin(), idxs.end(), f);
cudaDeviceSynchronize();
mytime = dtime_usec(mytime);
std::cout << "nested time: " << mytime/(float)USECPSEC << "s" << std::endl;
if (!validate(d_result1, d_result3)) std::cout << "mismatch 2!" << std::endl;
return 0;
}
$ nvcc -arch=sm_20 -std=c++11 -o t617 t617.cu
$ ./t617
loop time: 8.51577s
vectorized time: 0.068802s
nested time: 0.567959s
$
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笔记:
-arch=sm_20
为-arch=sm_35 -rdc=true -lcudadevrt
cudaDeviceSetLimit
.cudaDeviceSetLimit
可能需要增加预留的内存量,可能需要增加8倍.