Question

当需要推力时，我需要知道每个元素映射到的位置。

例如：

arr:       5  2  -1  3  -1  6 -1  7
compacted: 5  2   3  6  7     --------------(remove all -1 element)
map Arr:   0  1  -1  2  -1  3 -1  4

映射数组，这里我的意思是指示每个元素移动到哪个位置的数组，抱歉我不能用更好的名字来描述这个，我希望我做的事情很清楚。

紧凑的推力很容易，但我想知道我是否可以在推力压缩时获得该映射阵列。

Answer 1

以下推力调用序列可以生成所需的映射数组。

您的示例的输出是：

compacted: 5 2 3 6 7 
map: 0  0 -1  0 -1  0 -1  0 
map: 0  0 -1 -1 -2 -2 -3 -3 
map: 0  1  1  2  2  3  3  4 
map: 0  1 -1  2 -1  3 -1  4

#include <iostream>
#include <string>

#include <thrust/scan.h>
#include <thrust/scatter.h>
#include <thrust/remove.h>
#include <thrust/iterator/counting_iterator.h>
#include <thrust/iterator/constant_iterator.h>

void print(const std::string& name, int* begin, int* end)
{
  std::cout << name << ": ";
  thrust::copy(begin, end, std::ostream_iterator<int>(std::cout, " "));
  std::cout <<std::endl;
}

struct is_marker
{
  __host__ __device__
  bool operator()(const int x) const
  {
    return (x == -1);
  }
};

int main()
{
  const int N = 8;
  int arr[N] = {5,2,-1,3,-1,6,-1,7};

  int compacted[N] = {0};
  int* compacted_end = thrust::remove_copy(arr, arr+N, compacted, -1);
  print("compacted", compacted, compacted_end);

  int map[N] = {0};
  thrust::scatter_if(thrust::make_constant_iterator(-1), thrust::make_constant_iterator(-1)+N, thrust::make_counting_iterator(0), arr, map, is_marker());
  print("map", map, map+N);

  thrust::inclusive_scan(map, map+N, map);
  print("map", map, map+N);

  thrust::transform(map, map+N, thrust::make_counting_iterator(0), map, thrust::plus<int>());
  print("map", map, map+N);

  thrust::scatter_if(thrust::make_constant_iterator(-1), thrust::make_constant_iterator(-1)+N, thrust::make_counting_iterator(0), arr, map, is_marker());
  print("map", map, map+N);

  return 0;
}

如何在推力紧凑的情况下获得映射数组

1 个答案: