如何使CUDA中的矩阵列标准化并获得最大性能？-Java 学习之路

如何有效地规范化CUDA中的矩阵列？

我的矩阵存储在column-major中，典型大小为2000x200 .

该操作可以用以下matlab代码表示 .

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？

包括4个内核的快速实现如下所示 .

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

#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;
}

3 回答

我将M2090的3种方法的性能与CUDA 5.0进行了比较 .

[173.179 us] cublas实现如问题所示
[733.734 us]来自@talonmies的 thrust::reduce_by_key 的纯粹推力实施
[1.508 ms]纯推力实现 thrust::inclusive_scan_by_key

$Performance on A_{2,000 x 200}$

可以看出，

cublas在这种情况下表现最好;
thrust::reduce_by_key ＆ thrust::inclusive_scan_by_key 启动多个内核，这会导致额外的开销;
与 thrust::reduce_by_key 相比，
thrust::inclusive_scan_by_key 向DRAM写入更多数据，这可能是内核时间更长的原因之一;
cublas和推力方法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 旨在减少具有变体长度的段，但 cublas_gemv() 只能应用于固定长度的段（行/列） .

当矩阵A足够大以忽略内核启动开销时，cublas appoach仍然表现最佳 . A_ {20,000 x 2,000}的分析结果如下所示 .

$Performance on A_{20,000 x 2,000}$

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写的内核而不是 thrust::reduce_by_key .

3种方法的代码如下所示 .

#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;
}

回复于 2024-05-13T11:27:15+08:00

您应该能够将第一个 for_each 操作与 cublasSgemv 呼叫融合到一个 reduce_by_key 呼叫中 . 如果您将仿函数定义/重新定义为：

struct Accessor : public thrust::unary_function<int,int>
{
    int lda;
    __host__ __device__ Accessor(int _lda) : lda(_lda) {};
    __host__ __device__ int operator()(const int& idx)
    {
        return idx/lda;
    }
};

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

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

然后，您可以将标准化输出计算为

Accessor columns(m);
thrust::reduce_by_key(
        thrust::make_transform_iterator(thrust::make_counting_iterator(int(0)), columns),
        thrust::make_transform_iterator(thrust::make_counting_iterator(int(m*n)), columns),
        thrust::make_transform_iterator(A.begin(), Exp()),
        thrust::make_discard_iterator(),
        sum.begin());

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

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

[免责声明：所有代码均以浏览器编写，未经测试，使用风险自负]

除了减少内核调用的数量之外，使用花哨的迭代器消除了对大单位矩阵的需要，这应该减少内存占用和内存事务的总数以进行求和和取幂操作 .

回复于 2024-05-13T11:27:15+08:00

3
您可以通过以下方式使用ArrayFire
```
array A = randu(2000, 2000);
A = exp(A);
A /= tile(sum(A, 0), A.dims(0), 1);
```
你也可以用推力做到这一点 . 但是如果你打算使用矩阵（而不是简单的向量），你必须在for循环中进行，这样效率会不高 .

DISCLAIMER 我是Accelereyes的开发人员，正在开展阵火工作 .

EDIT 我正在按照要求制作新的基准 .

EDIT 由于此基准测试，我们在代码中发现了 exp 的性能错误 . 我们正在审查和修复它 .
回复于 2024-05-13T11:27:15+08:00

如何使CUDA中的矩阵列标准化并获得最大性能？

3 回答

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