我想在网格中的每个点评估一个函数。问题是,如果我在CPU端创建网格,将其传输到GPU的行为需要比实际计算更长的时间。我可以在GPU端生成网格吗?
下面的代码显示了在CPU端创建网格并评估GPU端的大部分表达式(我不知道如何让atan2在GPU上工作,所以我把它放在CPU端)。我应该事先道歉并说我还在学习这些东西,所以我确信下面的代码还有很大的改进空间!
谢谢!
import math
from numba import vectorize, float64
import numpy as np
from time import time
@vectorize([float64(float64,float64,float64,float64)],target='cuda')
def a_cuda(lat1, lon1, lat2, lon2):
return (math.sin(0.008726645 * (lat2 - lat1))**2) + \
math.cos(0.01745329*(lat1)) * math.cos(0.01745329*(lat2)) * (math.sin(0.008726645 * (lon2 - lon1))**2)
def LLA_distance_numba_cuda(lat1, lon1, lat2, lon2):
a = a_cuda(np.ascontiguousarray(lat1), np.ascontiguousarray(lon1),
np.ascontiguousarray(lat2), np.ascontiguousarray(lon2))
return earthdiam_nm * np.arctan2(a,1-a)
# generate a mesh of one million evaluation points
nx, ny = 1000,1000
xv, yv = np.meshgrid(np.linspace(29, 31, nx), np.linspace(99, 101, ny))
X, Y = np.float64(xv.reshape(1,nx*ny).flatten()), np.float64(yv.reshape(1,nx*ny).flatten())
X2,Y2 = np.float64(np.array([30]*nx*ny)),np.float64(np.array([101]*nx*ny))
start = time()
LLA_distance_numba_cuda(X,Y,X2,Y2)
print('{:d} total evaluations in {:.3f} seconds'.format(nx*ny,time()-start))
让我们建立一个绩效基准。为earthdiam_nm
添加定义(1.0),并在nvprof
下运行代码,我们有:
$ nvprof python t38.py
1000000 total evaluations in 0.581 seconds
(...)
==1973== Profiling result:
Type Time(%) Time Calls Avg Min Max Name
GPU activities: 55.58% 11.418ms 4 2.8544ms 2.6974ms 3.3044ms [CUDA memcpy HtoD]
28.59% 5.8727ms 1 5.8727ms 5.8727ms 5.8727ms cudapy::__main__::__vectorized_a_cuda$242(Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>)
15.83% 3.2521ms 1 3.2521ms 3.2521ms 3.2521ms [CUDA memcpy DtoH]
(...)
因此,在我的特定设置中,“内核”本身在我的(小型,慢速)QuadroK2000 GPU上运行~5.8ms,从主机到设备的4个副本的数据复制时间总共为11.4ms,对于结果转回主机。重点是从主机到设备的4个副本。
让我们先看看低调的果实。这行代码:
X2,Y2 = np.float64(np.array([30]*nx*ny)),np.float64(np.array([101]*nx*ny))
除了将值30和101传递给每个“工人”之外,其他任何事情都没有。我在这里使用“worker”来指代在大型数据集中“广播”vectorize
函数的numba过程中的特定标量计算的想法。 numba矢量化/广播过程不要求每个输入都是相同大小的数据集,而只需要提供的数据是“广播”的。因此,可以创建一个适用于数组和标量的vectorize
ufunc。这意味着每个worker将使用其数组元素和标量来执行计算。
因此,悬而未决的成果就是简单地删除这两个数组,并将值(30,101)作为标量传递给ufunc a_cuda
。当我们追求“低悬的果实”时,让我们将你的arctan2
计算(替换为math.atan2
)和earthdiam_nm
的最终缩放合并到vectorize代码中,这样我们就不必在python / numpy中的主机上进行:
$ cat t39.py
import math
from numba import vectorize, float64
import numpy as np
from time import time
earthdiam_nm = 1.0
@vectorize([float64(float64,float64,float64,float64,float64)],target='cuda')
def a_cuda(lat1, lon1, lat2, lon2, s):
a = (math.sin(0.008726645 * (lat2 - lat1))**2) + \
math.cos(0.01745329*(lat1)) * math.cos(0.01745329*(lat2)) * (math.sin(0.008726645 * (lon2 - lon1))**2)
return math.atan2(a, 1-a)*s
def LLA_distance_numba_cuda(lat1, lon1, lat2, lon2):
return a_cuda(np.ascontiguousarray(lat1), np.ascontiguousarray(lon1),
np.ascontiguousarray(lat2), np.ascontiguousarray(lon2), earthdiam_nm)
# generate a mesh of one million evaluation points
nx, ny = 1000,1000
xv, yv = np.meshgrid(np.linspace(29, 31, nx), np.linspace(99, 101, ny))
X, Y = np.float64(xv.reshape(1,nx*ny).flatten()), np.float64(yv.reshape(1,nx*ny).flatten())
# X2,Y2 = np.float64(np.array([30]*nx*ny)),np.float64(np.array([101]*nx*ny))
start = time()
Z=LLA_distance_numba_cuda(X,Y,30.0,101.0)
print('{:d} total evaluations in {:.3f} seconds'.format(nx*ny,time()-start))
#print(Z)
$ nvprof python t39.py
==2387== NVPROF is profiling process 2387, command: python t39.py
1000000 total evaluations in 0.401 seconds
==2387== Profiling application: python t39.py
==2387== Profiling result:
Type Time(%) Time Calls Avg Min Max Name
GPU activities: 48.12% 8.4679ms 1 8.4679ms 8.4679ms 8.4679ms cudapy::__main__::__vectorized_a_cuda$242(Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>, Array<double, int=1, A, mutable, aligned>)
33.97% 5.9774ms 5 1.1955ms 864ns 3.2535ms [CUDA memcpy HtoD]
17.91% 3.1511ms 4 787.77us 1.1840us 3.1459ms [CUDA memcpy DtoH]
(snip)
现在我们看到复制HtoD操作已从总共11.4ms减少到总共5.6ms。内核已从~5.8ms增长到~8.5ms,因为我们在内核中做了更多的工作,但python报告该函数的执行时间已从~0.58s下降到~0.4s。
我们可以做得更好吗?
我们可以,但为了这样做(我相信),我们需要使用不同的numba cuda方法。 vectorize
方法对于标量元素操作很方便,但它无法知道整个数据集在哪里执行操作。我们需要这些信息,我们可以在CUDA代码中获取它,但我们需要切换到@cuda.jit
装饰器才能这样做。
以下代码将以前的vectorize
a_cuda
函数转换为@cuda.jit
设备函数(基本上没有其他更改),然后我们创建一个CUDA内核,根据提供的标量参数生成网格,并计算结果:
$ cat t40.py
import math
from numba import vectorize, float64, cuda
import numpy as np
from time import time
earthdiam_nm = 1.0
@cuda.jit(device='true')
def a_cuda(lat1, lon1, lat2, lon2, s):
a = (math.sin(0.008726645 * (lat2 - lat1))**2) + \
math.cos(0.01745329*(lat1)) * math.cos(0.01745329*(lat2)) * (math.sin(0.008726645 * (lon2 - lon1))**2)
return math.atan2(a, 1-a)*s
@cuda.jit
def LLA_distance_numba_cuda(lat2, lon2, xb, xe, yb, ye, s, nx, ny, out):
x,y = cuda.grid(2)
if x < nx and y < ny:
lat1 = (((xe-xb) * x)/(nx-1)) + xb # mesh generation
lon1 = (((ye-yb) * y)/(ny-1)) + yb # mesh generation
out[y][x] = a_cuda(lat1, lon1, lat2, lon2, s)
nx, ny = 1000,1000
Z = cuda.device_array((nx,ny), dtype=np.float64)
threads = (32,32)
blocks = (32,32)
start = time()
LLA_distance_numba_cuda[blocks,threads](30.0,101.0, 29.0, 31.0, 99.0, 101.0, earthdiam_nm, nx, ny, Z)
Zh = Z.copy_to_host()
print('{:d} total evaluations in {:.3f} seconds'.format(nx*ny,time()-start))
#print(Zh)
$ nvprof python t40.py
==2855== NVPROF is profiling process 2855, command: python t40.py
1000000 total evaluations in 0.294 seconds
==2855== Profiling application: python t40.py
==2855== Profiling result:
Type Time(%) Time Calls Avg Min Max Name
GPU activities: 75.60% 10.364ms 1 10.364ms 10.364ms 10.364ms cudapy::__main__::LLA_distance_numba_cuda$241(double, double, double, double, double, double, double, __int64, __int64, Array<double, int=2, A, mutable, aligned>)
24.40% 3.3446ms 1 3.3446ms 3.3446ms 3.3446ms [CUDA memcpy DtoH]
(...)
现在我们看到: