我有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,这应该没有效率。
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>());
}
}
}
}
有没有办法融合推力操作,以便排序并行?实际上,我正在寻找的是将循环迭代融合到一起的通用方法。
答案 0 :(得分: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较快的方法。
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2 here中所讨论的背对背稳定排序操作进行矢量化排序。
Thrust v1.8(可通过CUDA 7 RC获得,或通过直接下载thrust github repository包括support for nesting thrust algorithms,通过在自定义仿函数中包含推力算法调用传递给另一推力算法如果您使用thrust::for_each
操作来选择您需要执行的各个排序,则可以通过在传递给{{的仿函数中包含thrust::sort
操作来使用单推力算法调用来运行这些排序。 1}}。
以下是3种方法之间的完全比较:
在每种情况下,我们都会对每组1000个整数的16000组进行排序。
thrust::for_each
注意:
$ 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
$
更改为-arch=sm_20
-arch=sm_35 -rdc=true -lcudadevrt
大幅增加设备分配堆。cudaDeviceSetLimit
保留的内存量可能需要增加8倍。