在任一幼崽或推力,我們可以排序的.w
「鑰匙」只是,那種做一個鍵值,其中值只是一個線性遞增指數:
0, 1, 2, 3, ...
然後我們可以使用合成重新排列索引序列以便在一個步驟中重新排列原始的float4
陣列(有效地按.w
排序)。這將允許您保留基數分選速度(無論是幼體還是推力),並且也可能相當有效,因爲float4
數量只需移動/重新排列一次,而不是在分類操作過程中連續移動。
這裏是一個完整的推力示例,在32M元素上,演示了一個「普通」推力排序,使用函數來指定排序.w
元素(sort_f4_w
),接着是上述方法。在這種情況下,我的特定的設置(Fedora的20,CUDA 7,Quadro5000),第二個方法似乎是約5倍快:
$ cat t686.cu
#include <iostream>
#include <vector_types.h>
#include <stdlib.h>
#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/sort.h>
#include <thrust/iterator/transform_iterator.h>
#include <thrust/iterator/permutation_iterator.h>
#include <thrust/sequence.h>
#include <thrust/copy.h>
#include <thrust/equal.h>
#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;
}
#define DSIZE (32*1048576)
struct sort_f4_w
{
__host__ __device__
bool operator()(const float4 &a, const float4 &b) const {
return (a.w < b.w);}
};
// functor to extract the .w element from a float4
struct f4_to_fw : public thrust::unary_function<float4, float>
{
__host__ __device__
float operator()(const float4 &a) const {
return a.w;}
};
// functor to extract the .x element from a float4
struct f4_to_fx : public thrust::unary_function<float4, float>
{
__host__ __device__
float operator()(const float4 &a) const {
return a.x;}
};
bool validate(thrust::device_vector<float4> &d1, thrust::device_vector<float4> &d2){
return thrust::equal(thrust::make_transform_iterator(d1.begin(), f4_to_fx()), thrust::make_transform_iterator(d1.end(), f4_to_fx()), thrust::make_transform_iterator(d2.begin(), f4_to_fx()));
}
int main(){
unsigned long long t1_time, t2_time;
float4 *mydata = new float4[DSIZE];
for (int i = 0; i < DSIZE; i++){
mydata[i].x = i;
mydata[i].y = i;
mydata[i].z = i;
mydata[i].w = rand()/(float)RAND_MAX;}
thrust::host_vector<float4> h_data(mydata, mydata+DSIZE);
// do once as a warm-up run, then report timings on second run
for (int i = 0; i < 2; i++){
thrust::device_vector<float4> d_data1 = h_data;
thrust::device_vector<float4> d_data2 = h_data;
// first time sort using typical thrust approach
t1_time = dtime_usec(0);
thrust::sort(d_data1.begin(), d_data1.end(), sort_f4_w());
cudaDeviceSynchronize();
t1_time = dtime_usec(t1_time);
// now extract keys and create index values, sort, then rearrange
t2_time = dtime_usec(0);
thrust::device_vector<float> keys(DSIZE);
thrust::device_vector<int> vals(DSIZE);
thrust::copy(thrust::make_transform_iterator(d_data2.begin(), f4_to_fw()), thrust::make_transform_iterator(d_data2.end(), f4_to_fw()), keys.begin());
thrust::sequence(vals.begin(), vals.end());
thrust::sort_by_key(keys.begin(), keys.end(), vals.begin());
thrust::device_vector<float4> result(DSIZE);
thrust::copy(thrust::make_permutation_iterator(d_data2.begin(), vals.begin()), thrust::make_permutation_iterator(d_data2.begin(), vals.end()), result.begin());
cudaDeviceSynchronize();
t2_time = dtime_usec(t2_time);
if (!validate(d_data1, result)){
std::cout << "Validation failure " << std::endl;
}
}
std::cout << "thrust t1 time: " << t1_time/(float)USECPSEC << "s, t2 time: " << t2_time/(float)USECPSEC << std::endl;
}
$ nvcc -o t686 t686.cu
$ ./t686
thrust t1 time: 0.731456s, t2 time: 0.149959
$
在任一幼崽或推力,我們可以排序的'.w'「關鍵字「,做一個鍵值排序,其中的值只是一個線性遞增索引(小熊和推力提供了花哨的迭代器來自動生成索引序列)。然後,我們可以使用索引序列的結果重新排列來對「float4」數組重新排序,以「.w」排序。這將允許您保留基數分類速度(無論是Cub還是Push),並且也可能相當有效,因爲'float4'數量只需移動/重新排列一次,而不是在分類操作過程中連續移動。 – 2015-04-02 13:59:23
@RobertCrovella你能提供一個這樣的代碼示例嗎?我不知道如何指定w組件是關鍵?這樣我可以將你標記爲答案。 – Kinru 2015-04-02 14:41:31