[英]Count reduction using Thrust
Given some input keys and values, I am trying to count how many consecutive values with the same key exist. 给定一些输入键和值,我试图计算存在多少具有相同键的连续值。 I will give an example to make this more clear.
我将举一个例子来说明这一点。
Input keys: { 1, 4, 4, 4, 2, 2, 1 }
输入键:
{ 1, 4, 4, 4, 2, 2, 1 }
Input values: { 9, 8, 7, 6, 5, 4, 3 }
输入值:
{ 9, 8, 7, 6, 5, 4, 3 }
Expected output keys: { 1, 4, 2, 1 }
预期的输出键:
{ 1, 4, 2, 1 }
1,4,2,1 { 1, 4, 2, 1 }
Expected output values: { 1, 3, 2, 1 }
预期输出值:
{ 1, 3, 2, 1 }
1,3,2,1 { 1, 3, 2, 1 }
I am trying to solve this problem on a GPU using CUDA. 我试图使用CUDA在GPU上解决这个问题。 The reduction capabilities of the Thrust library seemed like a good solution for this and I got to the following:
Thrust库的减少功能似乎是一个很好的解决方案,我得到了以下内容:
#include <thrust/reduce.h>
#include <thrust/functional.h>
struct count_functor : public thrust::binary_function<int, int, int>
{
__host__ __device__
int operator()(int input, int counter)
{
return counter + 1;
}
};
const int N = 7;
int A[N] = { 1, 4, 4, 4, 2, 2, 1 }; // input keys
int B[N] = { 9, 8, 7, 6, 5, 4, 3 }; // input values
int C[N]; // output keys
int D[N]; // output values
thrust::pair<int*, int*> new_end;
thrust::equal_to<int> binary_pred;
count_functor binary_op;
new_end = thrust::reduce_by_key(A, A + N, B, C, D, binary_pred, binary_op);
for (int i = 0; i < new_end.first - C; i++) {
std::cout << C[i] << " - " << D[i] << "\n";
}
This code is pretty similar to an example from the Thrust documentation . 此代码与Thrust文档中的示例非常相似。 However, instead of the
plus
operation, I am trying to count. 但是,我试图计算,而不是
plus
操作。 The output from this code is the following: 此代码的输出如下:
1 - 9
4 - 7
2 - 5
1 - 3
However, I would expected the second column to contain the values 1, 3, 2, 1
. 但是,我希望第二列包含值
1, 3, 2, 1
。 I think the counts are off because the reduction starts with the first value it finds and does not apply the operator until it has a second value, but I am not sure this is the case. 我认为计数是关闭的,因为减少从它找到的第一个值开始,并且在它有第二个值之前不应用运算符,但我不确定是这种情况。
Am I overlooking something about the reduce_by_key
function that could solve this problem or should I use a completely different function to achieve what I want? 我是否忽略了一些可以解决这个问题的
reduce_by_key
函数,或者我应该使用一个完全不同的函数来实现我想要的东西?
For your use case you don't need the values of B
, the values of D
are only dependent on the values of A
. 对于您的用例,您不需要
B
的值, D
的值仅取决于A
的值。
In order to count how many consecutive values are in A
you can supply a thrust::constant_iterator
as the input values and apply thrust::reduce_by_key
: 为了计算
A
有多少个连续值,你可以提供一个thrust::constant_iterator
作为输入值并应用thrust::reduce_by_key
:
#include <thrust/reduce.h>
#include <thrust/functional.h>
#include <iostream>
#include <thrust/iterator/constant_iterator.h>
int main()
{
const int N = 7;
int A[N] = { 1, 4, 4, 4, 2, 2, 1 };
int C[N];
int D[N];
thrust::pair<int*, int*> new_end;
thrust::equal_to<int> binary_pred;
thrust::plus<int> binary_op;
new_end = thrust::reduce_by_key(A, A + N, thrust::make_constant_iterator(1), C, D, binary_pred, binary_op);
for (int i = 0; i < new_end.first - C; i++) {
std::cout << C[i] << " - " << D[i] << "\n";
}
return 0;
}
output 产量
1 - 1
4 - 3
2 - 2
1 - 1
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