Question

我需要总结一下存储在数组中的100000值，但有条件。

有没有办法在CUDA中做到这一点以产生快速结果？

任何人都可以发布一个小代码吗？

Answer 1

我认为，要执行条件缩减，您可以直接将条件引入加法0（false）或1（true）到加数。换句话说，假设您要满足的条件是加数小于10.f。在这种情况下，借用Optimizing Parallel Reduction in CUDA by M. Harris的第一个代码，然后上面的意思是

__global__ void reduce0(int *g_idata, int *g_odata) {

    extern __shared__ int sdata[];

    // each thread loads one element from global to shared mem
    unsigned int tid = threadIdx.x;
    unsigned int i = blockIdx.x*blockDim.x + threadIdx.x;
    sdata[tid] = g_idata[i]*(g_data[i]<10.f);
    __syncthreads();

    // do reduction in shared mem
    for(unsigned int s=1; s < blockDim.x; s *= 2) {
        if (tid % (2*s) == 0) {
            sdata[tid] += sdata[tid + s];
        }
        __syncthreads();
    }

    // write result for this block to global mem
    if (tid == 0) g_odata[blockIdx.x] = sdata[0];
}

如果您希望使用CUDA Thrust执行条件缩减，则可以使用thrust::transform_reduce执行相同操作。或者，您可以创建一个新的向量d_b，复制d_a的所有元素，使thrust::copy_if满足谓词，然后在thrust::reduce上应用d_b。我没有检查哪种解决方案表现最佳。也许，第二种解决方案在稀疏阵列上表现更好。下面是一个实现这两种方法的例子。

#include <thrust/host_vector.h>
#include <thrust/device_vector.h>
#include <thrust/reduce.h>
#include <thrust/count.h>
#include <thrust/copy.h>

// --- Operator for the first approach
struct conditional_operator { 
    __host__ __device__ float operator()(const float a) const {
    return a*(a<10.f);
    }
};

// --- Operator for the second approach
struct is_smaller_than_10 {
    __host__ __device__ bool operator()(const float a) const {
        return (a<10.f);
    }
};

void main(void) 
{
    int N = 20;

    // --- Host side allocation and vector initialization
    thrust::host_vector<float> h_a(N,1.f);
    h_a[0] = 20.f;
    h_a[1] = 20.f;

    // --- Device side allocation and vector initialization
    thrust::device_vector<float> d_a(h_a);

    // --- First approach
    float sum = thrust::transform_reduce(d_a.begin(), d_a.end(), conditional_operator(), 0.f, thrust::plus<float>());
    printf("Result = %f\n",sum);

    // --- Second approach
    int N_prime = thrust::count_if(d_a.begin(), d_a.end(), is_smaller_than_10());
    thrust::device_vector<float> d_b(N_prime);
    thrust::copy_if(d_a.begin(), d_a.begin() + N, d_b.begin(), is_smaller_than_10());
    sum = thrust::reduce(d_b.begin(), d_b.begin() + N_prime, 0.f);
    printf("Result = %f\n",sum);

    getchar();

}

有条件减少CUDA

1 个答案: