如何以最大性能标准化 CUDA 中的矩阵列? [英] How to normalize matrix columns in CUDA with max performance?

问题描述

如何有效地对 CUDA 中的矩阵列进行归一化?

How to effectively normalize matrix columns in CUDA?

我的矩阵以列为主，典型大小为 2000x200.

My matrix is stored in column-major, and the typical size is 2000x200.

该操作可以用下面的matlab代码表示.

The operation can be represented in the following matlab code.

A = rand(2000,200);

A = exp(A);
A = A./repmat(sum(A,1), [size(A,1) 1]);

这可以通过 Thrust、cuBLAS 和/或 cuNPP 有效地完成吗?

Can this be done effectively by Thrust, cuBLAS and/or cuNPP?

一个包含 4 个内核的快速实现如下所示.

A rapid implementation including 4 kernels is shown as follows.

想知道这些是否可以在 1 个或 2 个内核中完成以提高性能，特别是对于 cublasDgemv() 实现的列求和步骤.

Wondering if these can be done in 1 or 2 kernels to improve the performance, especially for the column summation step implemented by cublasDgemv().

#include <cuda.h>
#include <curand.h>
#include <cublas_v2.h>
#include <thrust/device_vector.h>
#include <thrust/device_ptr.h>
#include <thrust/transform.h>
#include <thrust/iterator/constant_iterator.h>
#include <math.h>

struct Exp
{
    __host__ __device__ void operator()(double& x)
    {
        x = exp(x);
    }
};

struct Inv
{
    __host__ __device__ void operator()(double& x)
    {
        x = (double) 1.0 / x;
    }
};

int main()
{
    cudaDeviceSetCacheConfig(cudaFuncCachePreferShared);
    cublasHandle_t hd;
    curandGenerator_t rng;
    cublasCreate(&hd);
    curandCreateGenerator(&rng, CURAND_RNG_PSEUDO_DEFAULT);

    const size_t m = 2000, n = 200;
    const double c1 = 1.0;
    const double c0 = 0.0;

    thrust::device_vector<double> A(m * n);
    thrust::device_vector<double> sum(1 * n);
    thrust::device_vector<double> one(m * n, 1.0);

    double* pA = thrust::raw_pointer_cast(&A[0]);
    double* pSum = thrust::raw_pointer_cast(&sum[0]);
    double* pOne = thrust::raw_pointer_cast(&one[0]);

    for (int i = 0; i < 100; i++)
    {
        curandGenerateUniformDouble(rng, pA, A.size());


        thrust::for_each(A.begin(), A.end(), Exp());

        cublasDgemv(hd, CUBLAS_OP_T, m, n,
                &c1, pA, m, pOne, 1, &c0, pSum, 1);

        thrust::for_each(sum.begin(), sum.end(), Inv());

        cublasDdgmm(hd, CUBLAS_SIDE_RIGHT, m, n, pA, m, pSum, 1, pA, m);
    }

    curandDestroyGenerator(rng);
    cublasDestroy(hd);

    return 0;
}

推荐答案

我在 M2090 和 CUDA 5.0 上比较了 3 种方法的性能.

I compared the performance of 3 approaches on M2090 with CUDA 5.0.

1. [173.179 us] cublas 实现如问题所示
2. [733.734 us] 使用来自@talonmies 的 thrust::reduce_by_key 的纯 Thrust 实现
3. [1.508 ms] 使用 thrust::inclusive_scan_by_key
的纯 Thrust 实现

1. [173.179 us] cublas implementation as shown in the question
2. [733.734 us] pure Thrust implementation with thrust::reduce_by_key from @talonmies
3. [1.508 ms] pure Thrust implementation with thrust::inclusive_scan_by_key

可以看出，

1. cublas 在这种情况下具有最高的性能；
2. thrust::reduce_by_key 和thrust::inclusive_scan_by_key 启动多个内核，导致额外开销；
3. thrust::inclusive_scan_by_key 与 thrust::reduce_by_key 相比，向 DRAM 写入的数据要多得多，这可能是内核时间更长的原因之一；
4. cublas 和推力方法之间的主要性能差异是矩阵列求和.推力较慢可能是因为 thrust::reduce_by_key 旨在对具有不同长度的段进行归约，但 cublas_gemv() 只能适用于固定长度的段(行/列).

1. cublas has highest performance in this case;
2. both thrust::reduce_by_key & thrust::inclusive_scan_by_key launch multiple kernels, which lead to extra overhead;
3. thrust::inclusive_scan_by_key writes much more data to DRAM compared to thrust::reduce_by_key, which can be one of the reasons for longer kernel time;
4. the main performance difference between cublas and thrust approach is the matrix column summation. thrust is slower possibly because thrust::reduce_by_key is designed to do reduction on segments with variant length, but cublas_gemv() can only apply to fixed length segments (row/col).

当矩阵 A 大到足以忽略内核启动开销时，cublas 方法仍然表现最好.A_{20,000 x 2,000} 上的分析结果如下所示.

When the matrix A is large enough to ignore the kernel launching overhead, the cublas appoach still performs best. The profiling result on A_{20,000 x 2,000} is shown as follows.

将第一个 for_each 操作与@talonmies 指示的 cublasSgemv 调用融合可能会进一步提高性能，但我认为应该使用手工编写的内核而不是 thrust::reduce_by_key.

Fusing the first for_each operation with the cublasSgemv call as indicated by @talonmies may further improve the performance, but I think kernel written by hand should be used instead of thrust::reduce_by_key.

三种方法的代码如下所示.

The code for the 3 approaches is shown as follows.

#include <cuda.h>
#include <curand.h>
#include <cublas_v2.h>
#include <thrust/device_vector.h>
#include <thrust/device_ptr.h>
#include <thrust/transform.h>
#include <thrust/reduce.h>
#include <thrust/scan.h>
#include <thrust/iterator/counting_iterator.h>
#include <thrust/iterator/transform_iterator.h>
#include <thrust/iterator/discard_iterator.h>
#include <thrust/iterator/permutation_iterator.h>
#include <math.h>

struct Exp: public thrust::unary_function<double, double>
{
    __host__ __device__ double operator()(double x)
    {
        return exp(x);
    }
};

struct Inv: public thrust::unary_function<double, double>
{
    __host__ __device__ double operator()(double x)
    {
        return (double) 1.0 / x;
    }
};

template<typename T>
struct MulC: public thrust::unary_function<T, T>
{
    T C;
    __host__ __device__ MulC(T c) :
        C(c)
    {
    }
    __host__ __device__ T operator()(T x)
    {
        return x * C;
    }
};

template<typename T>
struct line2col: public thrust::unary_function<T, T>
{
    T C;
    __host__ __device__ line2col(T C) :
            C(C)
    {
    }

    __host__ __device__ T operator()(T i)
    {
        return i / C;
    }
};

int main()
{
    cudaDeviceSetCacheConfig(cudaFuncCachePreferShared);
    cublasHandle_t hd;
    curandGenerator_t rng;
    cublasCreate(&hd);
    curandCreateGenerator(&rng, CURAND_RNG_PSEUDO_DEFAULT);

    const size_t m = 2000, n = 200;
    const double c1 = 1.0;
    const double c0 = 0.0;

    thrust::device_vector<double> A(m * n);
    thrust::device_vector<double> B(m * n);
    thrust::device_vector<double> C(m * n);
    thrust::device_vector<double> sum1(1 * n);
    thrust::device_vector<double> sum2(1 * n);
    thrust::device_vector<double> one(m * n, 1);

    double* pA = thrust::raw_pointer_cast(&A[0]);
    double* pB = thrust::raw_pointer_cast(&B[0]);
    double* pSum1 = thrust::raw_pointer_cast(&sum1[0]);
    double* pSum2 = thrust::raw_pointer_cast(&sum2[0]);
    double* pOne = thrust::raw_pointer_cast(&one[0]);

    curandGenerateUniformDouble(rng, pA, A.size());

    const int count = 2;

    for (int i = 0; i < count; i++)
    {
        thrust::transform(A.begin(), A.end(), B.begin(), Exp());
        cublasDgemv(hd, CUBLAS_OP_T, m, n, &c1, pB, m, pOne, 1, &c0, pSum1, 1);
        thrust::transform(sum1.begin(), sum1.end(), sum1.begin(), Inv());
        cublasDdgmm(hd, CUBLAS_SIDE_RIGHT, m, n, pB, m, pSum2, 1, pB, m);
    }

    for (int i = 0; i < count; i++)
    {
        thrust::reduce_by_key(
                thrust::make_transform_iterator(thrust::make_counting_iterator(0), line2col<int>(m)),
                thrust::make_transform_iterator(thrust::make_counting_iterator(0), line2col<int>(m)) + A.size(),
                thrust::make_transform_iterator(A.begin(), Exp()),
                thrust::make_discard_iterator(),
                sum2.begin());
        thrust::transform(
                A.begin(), A.end(),
                thrust::make_permutation_iterator(
                        sum2.begin(),
                        thrust::make_transform_iterator(thrust::make_counting_iterator(0), line2col<int>(m))),
                C.begin(),
                thrust::divides<double>());
    }

    for (int i = 0; i < count; i++)
    {
        thrust::inclusive_scan_by_key(
                thrust::make_transform_iterator(thrust::make_counting_iterator(0), line2col<int>(m)),
                thrust::make_transform_iterator(thrust::make_counting_iterator(0), line2col<int>(m)) + A.size(),
                thrust::make_transform_iterator(A.begin(), Exp()),
                C.begin());
        thrust::copy(
                thrust::make_permutation_iterator(
                        C.begin() + m - 1,
                        thrust::make_transform_iterator(thrust::make_counting_iterator(0), MulC<int>(m))),
                thrust::make_permutation_iterator(
                        C.begin() + m - 1,
                        thrust::make_transform_iterator(thrust::make_counting_iterator(0), MulC<int>(m))) + n,
                sum2.begin());
        thrust::transform(
                A.begin(), A.end(),
                thrust::make_permutation_iterator(
                        sum2.begin(),
                        thrust::make_transform_iterator(thrust::make_counting_iterator(0), line2col<int>(m))),
                C.begin(),
                thrust::divides<double>());
    }

    curandDestroyGenerator(rng);
    cublasDestroy(hd);

    return 0;
}

这篇关于如何以最大性能标准化 CUDA 中的矩阵列?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！