kan*_*yin 7 performance cuda matrix thrust cublas
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]);
Run Code Online (Sandbox Code Playgroud)
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;
}
Run Code Online (Sandbox Code Playgroud)
我比较了M2090和CUDA 5.0在3种方法上的性能。
thrust::reduce_by_key来自@talonmies的纯Thrust实现thrust::inclusive_scan_by_key
可以看出,
thrust::reduce_by_key与thrust::inclusive_scan_by_key启动多个内核,从而导致额外的开销;thrust::inclusive_scan_by_key与相比thrust::reduce_by_key,将更多的数据写入DRAM 可能是内核时间更长的原因之一;thrust::reduce_by_key其目的是减少长度可变的分段,但cublas_gemv()只能应用于固定长度的分段(行/列)。当矩阵A足够大而忽略内核启动开销时,cublas方法仍然表现最佳。A_ {20,000 x 2,000}上的分析结果如下所示。

如@talonmies所示,将第一个for_each操作与cublasSgemv调用融合在一起可以进一步提高性能,但是我认为应该使用手写的内核代替thrust::reduce_by_key。
三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;
}
Run Code Online (Sandbox Code Playgroud)